Apollonius (Pergaeus), Lawson, John, The two books of Apollonius Pergaeus, concerning tangencies, as they have been restored by Franciscus Vieta and Marinus Ghetaldus : with a supplement to which is now added, a second supplement, being Mons. Fermat's Treatise on spherical tangencies, 1771

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Author: Apollonius (Pergaeus), Lawson, John
Title: The two books of Apollonius Pergaeus, concerning tangencies, as they have been restored by Franciscus Vieta and Marinus Ghetaldus : with a supplement to which is now added, a second supplement, being Mons. Fermat's Treatise on spherical tangencies
Year: 1771
City: London
Publisher: Bigg
Edition: 2. ed.
Number of Pages: VII, 29 S., 7 Taf., XI, 40 S., 5 Taf., II, II, 16 S. : Ill.

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Document ID: MPIWG:NQC24YTK
Permanent URL: http://echo.mpiwg-berlin.mpg.de/MPIWG:NQC24YTK

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Copyright: Max Planck Institute for the History of Science (unless stated otherwise)
License: CC-BY-SA (unless stated otherwise)
Table of contents
1. Page: 0
2. THE TWO BOOKS OF APOLLONIUS PERGÆUS, CONCERNING TANGENCIES, As they have been Reſtored by FRANCISCUSVIET A and MARINUSGHETALDUS. WITH A SUPPLEMENT. Page: 5
3. THE SECOND EDITION. TO WHICH IS NOW ADDED, A SECOND SUPPLEMENT, BEING Monſ. FERMAT’S Treatiſe on Spherical Tangencies. LONDON: Printed by G. BIGG, Succeſſor to D. LEACH. And ſold by B. White, in Fleet-Street; L. Davis, in Holborne, J. Nourse, in the Strand; and T. Payne, near the Mews-Gate. MDCCLXXI. Page: 5
4. PREFACE. Page: 7
5. EXTRACT from PAPPUS’s Preſace to his Seventh Book in Dr. HALLEY’s Tranſlation. DE TACTIONIBUS II. Page: 9
6. Synopsis of the PROBLEMS. Page: 11
7. PROBLEMS CONCERNING TANGENCIES. PROBLEM I. Page: 13
8. PROBLEM II. Page: 13
9. PROBLEM III. Page: 14
10. The GENERAL Solution. Page: 15
11. PROBLEM IV. Page: 15
12. PROBLEM V. Page: 16
13. The general Solution. Page: 16
14. PROBLEM VI. Page: 17
15. The general Solution. Page: 17
16. PROBLEM VII. Page: 18
17. LEMMA I. Page: 19
18. PROBLEM VIII. Page: 19
19. Mr. Simpſon conſtructs the Problem thus. Page: 19
20. PROBLEM IX. Page: 19
21. LEMMA II. Page: 20
22. LEMMA III. Page: 20
23. PROBLEM X. Page: 20
24. PROBLEM XI. Page: 21
25. PROBLEM XII . Page: 22
26. LEMMA IV. Page: 23
27. LEMMA V. Page: 23
28. PROBLEM XIII. Page: 23
29. PROBLEM XIV. Page: 24
30. SUPPLEMENT. PROBLEM I. Page: 26
31. PROBLEM II. Page: 26
32. PROBLEM III. Page: 26
33. PROBLEM IV. Page: 28
34. PROBLEM V. Page: 28
35. PROBLEM VI. Page: 29
36. General Solution. Page: 29
37. A SECOND SUPPLEMENT, BEING Monſ. DE FERMAT’S Treatiſe on Spherical Tangencies. PROBLEM I. Page: 31
38. PROBLEM II. Page: 32
39. PROBLEM III. Page: 33
40. PROBLEM IV. Page: 33
41. PROBLEM V. Page: 34
42. PROBLEM VI. Page: 34
43. PROBLEM VII. Page: 35
44. LEMMA I. Page: 36
45. LEMMA II. Page: 36
46. LEMMA III. Page: 36
47. LEMMA IV. Page: 37
48. LEMMA V. Page: 37
49. PROBLEM VIII. Page: 38
50. PROBLEM IX. Page: 39
51. PROBLEM X. Page: 39
52. PROBLEM XI. Page: 39
53. PROBLEM XII. Page: 40
54. PROBLEM XIII. Page: 40
55. PROBLEM XIV. Page: 40
56. PROBLEM XV. Page: 40
57. Synopſis of the PROBLEMS. Page: 41
58. THE TWO BOOKS OF APOLLONIUS PERGÆUS, CONCERNING DETERMINATE SECTION, As they have been Reſtored by WILLEBRORDUS SNELLIUS. By JOHN LAWSON, B. D. Rector of Swanſcombe, Kent. TO WHICH ARE ADDED, THE SAME TWO BOOKS, BY WILLIAM WALES, BEING AN ENTIRE NEW WORK. LONDON: Printed by G. BIGG, Succeſſor to D. LEACH. And ſold by B. White, in Fleet-Street; L. Davis, in Holborne; J. Nourse, in the Strand; and T. Payne, near the Mews-Gate. MDCC LXXII. Page: 64
59. ADVERTISEMENT. Page: 66
60. EXTRACT from PAPPUS's Preface to his Seventh Book in Dr. HALLEY's Tranſlation. DE SECTIONE DETERMINATA II. Page: 68
61. THE PREFACE. Page: 70
62. PROBLEMS CONCERNING DETERMINATE SECTION. PROBLEM I. Page: 78
63. LEMMA I. Page: 79
64. LEMMA II. Page: 80
65. LEMMA III. Page: 80
66. PROBLEM II. Page: 80
67. LEMMA IV. Page: 83
68. LEMMA V. Page: 83
69. PROBLEM III. Page: 84
70. PROBLEM IV. Page: 86
71. DETERMINATE SECTION. BOOK I. PROBLEM I. (Fig. 1.) Page: 92
72. PROBLEM II. (Fig. 2 and 3.) Page: 93
73. PROBLEM III. (Fig. 4. and 5.) Page: 94
74. PROBLEM IV. (Fig. 6. 7. and 8.) Page: 95
75. PROBLEM V. (Fig. 9. 10. 11. 12. 13. 14. 15. 16.) Page: 96
76. PROBLEM VI. (Fig. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.) Page: 98
77. THE END OF BOOK I. Page: 102
78. DETERMINATE SECTION. BOOK II. LEMMA I. Page: 103
79. LEMMA II. Page: 103
80. LEMMA III. Page: 104
81. LEMMA IV. Page: 105
82. LEMMA V. Page: 106
83. PROBLEM VII. (Fig. 32, 33, 34, &c.) Page: 108
84. PROBLEM I. (Fig. 32 to 45.) Page: 109
85. PROBLEM II. (Fig. 46 to 57.) Page: 113
86. PROBLEM III. Page: 116
87. THE END. Page: 117
88. A SYNOPSIS OF ALL THE DATA FOR THE Conſtruction of Triangles, FROM WHICH GEOMETRICAL SOLUTIONS Have hitherto been in Print. Page: 133
89. By JOHN LAWSON, B. D. Rector of Swanscombe, in KENT. ROCHESTER: Page: 133
90. MDCCLXXIII. [Price One Shilling.] Page: 133
91. ADVERTISEMENT. Page: 135
92. AN EXPLANATION OF THE SYMBOLS made uſe of in this SYNOPSIS. Page: 137
93. INDEX OF THE Authors refered to in the SYNOPSIS. Page: 139
94. Lately was publiſhed by the ſame Author; [Price Six Shillings in Boards.] Page: 140
95. SYNOPSIS. Page: 141
96. Continuation of the Synopsis, Containing ſuch Data as cannot readily be expreſſed by the Symbols before uſed without more words at length. Page: 149
97. SYNOPSIS Page: 152
98. FINIS. Page: 156
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THE
TWO BOOKS
OF
APOLLONIUS PERGÆUS,
CONCERNING
TANGENCIES,
As they have been Reſtored by
FRANCISCUSVIET A and MARINUSGHETALDUS.
WITH A
SUPPLEMENT.
By JOHN LAWSON, B. D. Rector of Swanſcombe in Kent.
THE SECOND EDITION.
TO WHICH IS NOW ADDED,
A SECOND SUPPLEMENT,
BEING
Monſ. FERMAT’S Treatiſe on Spherical Tangencies.
LONDON:
Printed by G. BIGG, Succeſſor to D. LEACH.
And ſold by B. White, in Fleet-Street; L. Davis, in Holborne, J. Nourse, in the
Strand; and T. Payne, near the Mews-Gate.
MDCCLXXI.
6
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7(iii)
PREFACE.
OF the twelve Analytical Treatiſes recited by Pappus in
his Preface to the 7th Book of his Mathematical Col-
lections, we have very little of the Originals remaining, viz.
only Euclid’s Data, and part of Apollonius’s Conics. The
loſs of the reſt is very much to be lamented by all Lovers of
the Mathematics. “ Valde quidem dolendum eſt quod re-
liqui tractatus Veterum Analytici, a Pappo memorati,
aut perierint, aut nondum lucem conſpexerint. Nam
minime dubito quin eorum nonnulli, Arabicè ſaltem
verſi, alicubi terrarum lateant, pulvere magis quam tene-
bris ſuis involuti. ” Dr. Halley’s Preface to his Apol-
LONIUS DE Sectione Rationis et Spatii.
Some ingenious men have attempted, from the account of
them given by Pappus, to reſtore ſome of theſe loſt Trea-
tiſes. Snellius has endeavoured to give us the Books De
Sectione Rationis, De Sectione Spatii, and De Sec-
TIONE Determinata. Fermat and Schooten have laboured
in the Treatiſe De Locis Planis; and Marinus Ghetaldus
in that De Inclinationibus. But thoſe who have ſuc-
ceeded beſt, and done the moſt this way, are two
8(iv) parable Mathematicians of our own Country, Dr. Halley
and Dr. Simſon, to whom the World is very much obliged
for their Geometrical Labours. The firſt of theſe, from an
Arabic MS in the Bodleian Library, has reſtored the Books
De Sectione Rationis; and from his own Sagacity ſup-
plied thoſe De Sectione Spatii: and the other has with
equal pains and ingenuity completed thoſe De Locis Planis.
As to the Treatiſe De Tactionibus, which I now give
the Engliſh Reader, it has been reſtored by Vieta under the
Title of Apollonius Gallus, and his Deficiencies ſupplied by
Marinus Ghetaldus. I have endeavoured to do Juſtice to my
Authors by all poſſible Care both in the Text and in the
Figures; and have added a few Propoſitions of my own, by
way of Supplement, in which I have propoſed Ghetaldus’s
Problems over again without a Determination, and have
found the Locus of the center of the circle required, which
I have not ſeen done before in any Author.
9(v)
EXTRACT from PAPPUS’s Preſace to his Seventh Book
in Dr. HALLEY’s Tranſlation.
DE TACTIONIBUS II.
HIS ordine ſubnexi ſunt libri duo DE Tactionibus, in
quibus plures ineſſe propoſitiones videntur; ſed & ex
his unam etiam faciemus, ad hunc modum ſe habentem. “ E
punctis rectis & circulis, quibuſcunque tribus poſitione
datis, circulum ducere per ſingula data puncta, qui, ſi fieri
poſſit, contingat etiam datas lineas. ” Ex hac autem ob mul-
titudinem in Hypotheſibus datorum, tam ſimilium quam diſſi-
milium GENERUM, fiunt neceſſario decem propoſitiones di-
verſæ; quia ex tribus diſſimilibus generibus fiunt diverſæ
triades inordinatæ numero decem. Data etenim eſſe poſſunt
vel tria puncta; vel tres rectæ; vel duo puncta & recta; vel
duæ rectæ & punctum; vel duo puncta & circulus; vel duo
circuli & punctum; vel duo circuli & recta; vel punctum,
recta & circulus; vel duæ rectæ & circulus; vel tres circuli.
Horum duo quidem prima problemata oſtenduntur in libro
quarto primorum Elementorum. Nam per tria data puncta,
quæ non ſint in linea recta, circulum ducere, idem eſt ac
circa datum triangulum circumſcribere. Problema autem in
tribus datis rectis non parallelis, ſed inter ſe occurrentibus,
idem eſt ac dato triangulo circulum inſcribere. Caſus vero
duarum rectarum parallelarum cum tertiâ occurrente,
10(vi) pars eſſet ſecundæ ſubdiviſionis, cæteris permittitur. Deinde
proxima ſex problemata continentur in primo libro. Reliqua
duo, nempe de duabus rectis datis & circulo, & de tribus
datis circulis, ſola habentur in ſecundo libro; ob multas di-
verſaſque poſitiones circulorum & rectarum inter ſe, quibus
fit ut etiam plurium determinationum opus ſit. Prædictis his
Tactionibus congener eſt ordo problematum, quæ ab edito-
ribus omiſſa fuerant. Nonnulli autem priori horum librorum
illa prefixerunt: Compendioſus enim & introductorius erat
tractatus ille, & ad plenam de Tactionibus doctrinam abſol-
vendam maxime idoneus. Hæc omnia rurſus una propoſitio
complectitur, quæ quidem quoad Hypotheſim magis quam
præcedentia contracta eſt, ſuperaddita autem eſt conditio ad
conſtructionem: eſtque hujuſmodi. “ E punctis, rectis, vel
circulis, datis duobus quibuſcunque, deſcribere circulum
magnitudine datum, qui tranſeat per punctum vel puncta
data, ac, ſi fieri poſſit, contingat etiam lineas datas. ” Con-
tinet autem hæc propoſitio ſex problemata: ex tribus enim
quibuſcunque diverſis generibus fiunt Duades inordinatæ di-
verſæ numero ſex. Vel enim datis duobus punctis, vel duabus
rectis, vel duobus circulis, vel puncto & rectâ, vel puncto &
circuìo, vel rectâ & circulo, opportet circulum magnitudine
datum deſcribere, QUI DATA CONTINGAT; hæc autem re-
ſolvenda ſunt & componenda ut & determinanda juxta Caſus.
Liber primus Tactionum problemata habet ſeptem; ſe-
cundus vero quatuor. Lemmata autem ad utrumque librum
ſunt XXI; Theoremata LX.
11(vii)
Synopsis of the PROBLEMS.
11
Prob. 1. ● ● # Prob. 2. | | # Prob. 3. ○ ○
4. ● | # 6. | ●
5. ● ○
Euclid ● ● ● # Euclid | | | # Prob. 14. ○ ○ ○
Prob. 7. ● ● | # Prob. 9. | | ○
12. ● ● ○ # 11. | ○ ○
8. ● | |
13. ● ○ ○
10. ● ○ ○
12
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13
PROBLEMS
CONCERNING
TANGENCIES.
PROBLEM I.
THROUGH two given points A and B to deſcribe a circle whoſe radius
ſhall be equal to a given line Z.
Limitation. 2 Z muſt not be leſs than the diſtance of the points A and B.
Construction. With the centers A and B, and diſtance Z, deſcribe two
arcs cutting or touching one another in the point E, ( which they will neceſſarily
do by the Limitation ) and E will be the center of the circle required.
PROBLEM II.
Having two right lines AB CD given in poſition, it is required to draw 2
circle, whoſe Radius ſhall be equal to the given line Z, which ſhall alſo touch
both the given lines.
Case 1ſt. Suppoſe AB and CD to be parallel.
Limitation. 2Z muſt be equal to the diſtance of the parallels, and the
conſtruction is evident.
Case 2d. Suppoſe AB and CD to be inclined to each other, let them be
produced till they meet in E, and let the angle BED be biſected by EH, and
through E draw EF perpendicular to ED, and equal to the given line Z; through
F draw FG parallel to EH, meeting ED in G, and through G draw GH paral-
lel to EF. I ſay that the circle deſcribed with H center, and HG radius,
touches the two given lines: it touches CD, becauſe EFGH is a
14(2) and hence the angle FEG is equal to EGH, but FEG is a right one by Con-
ſtruction. Let now HI be drawn from H perpendicular to AB: then the two
triangles EHI and EHG having two angles in one HEI and EIH reſpectively
equal to two angles in the other HEG and EGH, and alſo the ſide EH com-
mon, by Euc. I. 26. HI will be equal to HG, and therefore the circle will
touch alſo the other line AB: and HG or HI equals the given line Z, becauſe
EF was made equal to Z, and HG and EF are oppoſite ſides of a paral-
lelogram.
PROBLEM III.
Having two circles given whoſe centers are A and B, it is required to draw
another, whoſe Radius ſhall be equal to a given line Z, which ſhall alſo touch
the two given ones.
This Problem has various Caſes, according to the various poſition of the
given circles, and the various manner of deſcribing the circle required: but there
are ſix principal ones, and to the conditions of theſe all the reſt are ſubject.
Case 1ſt. Let the circle to be deſcribed be required to be touched outwardly
by the given circles.
Limitation. Then it is neceſſary that 2Z, or the given Diameter, ſhould
not be leſs than the ſegment of the line joining the centers of the given circles
which is intercepted between their convex circumferences, viz. not leſs than CD
in the Figure belonging to Caſe 1ſt.
Case 2d. Let the circle to be deſcribed be required to be touched inwardly by
the given circles.
Limitation. Then it’s Diameter muſt not be given leſs than the right line,
which drawn through the centers of the given circles, is contained between their
concave circumferences; viz. not leſs than CD.
Case 3d. Let the circle to be deſcribed be required to be touched outwardly
by one of the given circles, and inwardly by the other.
Limitation. Then it’s Diameter muſt not be given leſs than the ſegment
of the right line, joining the centers of the given circles, which is intercepted
between the convex circumference of one and the concave circumference of the
other; viz. not leſs than CD.
Case 4th. Let one of the given circles include the other, and let it be re-
quired that the circle to be deſcribed be touched outwardly by them both.
Limitation. Then it’s Diameter muſt not be given greater than the greater
ſegment of the right line, joining the centers of the given circles, which is in-
tercepted between the concave circumference of one and the convex circumference
of the other; nor leſs than the leſſer ſegment; viz. not greater than CD, nor
leſs than MN.
15(3)
Case 5th. Let one of the given circles include the other, and let be required
that the circle to be deſcribed be touched outwardly by one of the given circles,
and inwardly by the other.
Limitation. Then it’s Diameter muſt not be given greater than the greater
ſegment of the right line, joining the centers of the given circles, which is in-
tercepted between the two concave circumferences of the ſaid circles, nor leſs
than the leſſer ſegment; viz. not greater than CD, nor leſs than MN.
Case 6th. Let the two given circles cut each other, and let it be required
that the circle to be deſcribed, and to be touched by them both, ſhall alſo be
included in each of them.
Limitation. Then it’s Diameter muſt not be given greater than the ſeg-
ment of the right line, joining the centers of the given circles, intercepted by
their concave circumferences, which lies in the ſpace common to both the given
circles; viz. not greater than CD.
There may be alſo three other Caſes of this Problem, when the given circles
cut each other; but becauſe they are ſimilar to the 1ſt, 2d, and 4th Caſes already
propoſed, and ſubject to juſt the ſame Limitations; except that which is ſimilar
to the 1ſt, which is ſubject to no Limitation at all, they are here omitted; as are
likewiſe thoſe Caſes where the given circles touch each other; becauſe they are
the ſame as the preceding, and ſolved in the ſame manner.
The GENERAL Solution.
Join the given centers A and B, and where the Caſe requires, let AB be pro-
duced to meet the given circumferences in C and D: and let CI and DH be
taken equal to the given line Z: and let two circles be deſcribed; one with cen-
ter A and diſtance AI, and the other with center B and diſtance BH: and theſe
two circles will neceſſarily cut or touch each other by the Limitations given. Let
the point of concourſe be E: from E draw the right line EAF cutting the circle
whoſe center is A in F; as alſo EBG cutting the circle whoſe center is B in G:
then with center E and diſtance EF deſcribe a circle FK, this will be the circle
required: becauſe AF and AC are equal as alſo AI and AE; therefore FE
and CI are alſo equal: but CI was made equal to Z, therefore FE is equal to
Z. Again, becauſe BD and BG are equal, as alſo BH and BE, therefore DH
and EG are alſo equal: but DH was made equal to Z, therefore EG is equal to
Z. Hence it appears that the circle FK, paſſing through F will alſo paſs thro’
G, and likewiſe that it will alſo touch the given circles in F and G, becauſe EAF
and EBG are right lines paſſing through the centers.
PROBLEM IV.
Having a given point A, and a given right line BC, it is required to draw a
circle, whoſe Radius ſhall be equal to a given line Z, which ſhall paſs through
the given point, and alſo touch the given line.
16(4)
Limitation. 2Z muſt not be given leſs than the perpendicular let fall from
the given point A upon the given line BC.
From the point A let AD be drawn perpendicular to BC, and in this perpen-
dicular take DE equal to the given line Z: and through E draw EF parallel to
BC, and from A upon this line EF ſet off AF equal to Z, which may be done,
for by the Limitation Z is not leſs than AE: then with center F and diſtance
FA deſcribe a circle, and I ſay it will touch the line BC: for through F drawing
FG parallel to AD, FGDE will be a Parallelogram, and FG will be equal to
DE, that is to Z, and at right angles to BC.
PROBLEM V.
Having a given point A, and alſo a given circle whoſe center is B, it is re-
quired to draw a circle whoſe Radius ſhall be equal to a given line Z, which ſhall
paſs through the given point, and alſo touch the given circle.
This Problem has three Caſes, each of which is ſubject to a Limitation.
Case Iſt. Let the circle to be doſcribed be required to be touched outwardly
by the given circle.
Limitation. Then the Diameter muſt not be given leſs than the ſegment
of the right line, joining the given point and the center of the given circle,
which is intercepted between the given point and the convex circumference; viz.
not leſs than AC.
Case 2d. Let the circle to be deſcribed be required to be touched inwardly
by the given circle.
Limitation. Then the Diameter muſt not be given leſs than the right line
which, drawn from the given point through the center of the given circle, is con-
tained between the given point and the concave circumſerence; viz. not leſs
than AC.
Case 3d. Let the given point lie in the given circle.
Limitation. Then a diameter of the given circle being drawn through the
given point, it is divided into two ſegments by the ſaid point, and the Diameter
of the circle required muſt not be given greater than the greater of them, nor
leſs than the leſſer; viz. not greater than AC, nor leſs than AG.
The general Solution.
Let A and B be joined, and in the line AB take CF equal to Z, and then
with center A and diſtance Z, let an arc be drawn, and with center B, and
diſtance BF let another be drawn, which by the Limitations will neceſſarily
either touch or cut the former; let the point of their concourſe be D; then with
D center and DA diſtance let a circle be drawn, and I ſay it will touch the
given circle whoſe center is B: for DB being drawn meeting the circumference
of the circle whoſe center is B in E, BC is equal to BE, and hence CF equals
ED, and they are both equal to the given line Z.
17(5)
PROBLEM VI.
Having a right line given BC, and alſo a circle whoſe center is A, it is re-
quired to draw another circle, whoſe Radius ſhall be equal to a given right line
Z, and which ſhall touch both the given line and alſo the given circle.
This Problem has alſo three caſes, each of which is ſubject to a Limitation.
Case Iſt, Let the circle to be deſcribed be required to be touched outwardly
by the given circle.
Limitation. Then the Diameter of the circle required muſt not be given
leſs than the ſegment of a line, drawn from the center of the given circle, per-
pendicular to the given line, which is intercepted between the ſaid line and the
convex circumference; viz. not leſs than BD.
Case 2d. Let the circle to be deſcribed be required to be touched inwardly by
the given circle.
Limitation. Then the given line muſt not be in the given circle, neither
muſt the Diameter of the circle required be given leſs than that portion of the
perpendicular, drawn from the center of the given circle to the given line, which
is intercepted between the ſaid line and the concave circumference; viz. not leſs
than BD.
Case 3d. Let the circle to be deſcribed be required to be both touched and
included in the given circle.
Limitation. Then the given right line muſt be in the given circle, and
when a Diameter of this given circle is drawn cutting the given line at right an-
gles, the Diameter of the circle required muſt not be given greater than the
greater ſegment ; viz. not greater than BD.
The general Solution.
From A draw AB perpendicular to BC, cutting the given circumference in D;
and in this perpendicular let BG and DF be taken each equal to the given line
Z; and through G draw GE parallel to BC; and with center A and diſtance
AF let an arc be ſtruck, which by the Limitations will neceſſarily either touch
or cut GE; let the point of concourſe be E, let AE be joined, and, if neceſſary,
be produced to meet the given circumference in H; then with E center and
EH diſtance deſcribe a circle, and I ſay it will be the required circle; it is evi-
dent it will touch the given circle: and becauſe AD and AH are equal, as alſo
AF and AE, therefore DF (which was made equal to Z) will be equal to HE:
let now EC be drawn perpendicular to BC, then GBCE will be a Parallelogram,
and EC will be equal to GB, which was alſo made equal to Z: hence the
circle will alſo touch the given line BC, becauſe the angle ECB is a right
one.
18(6)
PROBLEM VII.
Having two points A and B given in poſition, and likewiſe a right line EF
given in poſition, it is required to find the center of a circle, which ſhall paſs
through the given points and touch the given line.
Case Iſt. When the points A and B are joined, ſuppoſe AB to be parallel to
EF: then biſecting AB in D, and through D drawing DC perpendicular to it,
DC will alſo be perpendicular to EF: draw a circle therefore which will paſs
through the three points A, B, andC, (by Euc. IV. 5.) and it will be the circle
required: (by a Corollary from Euc. III. 1. and another from Euc.
III. 16.)
Case 2d. Suppoſe AB not parallel to EF, but being produced meets it in
E: then from EF take the line EC ſuch, that its ſquare may be equal to the
rectangle BEA, and through the points A, B, C, deſcribe a circle, and it will be
the circle required by Euc. III. 37.
This is Vieta’s Solution. But Mr. Thomas Simpſon having conſtructed this,
and ſome of the following, both in the Collection of Problems at the end of his
Algebra, and alſo among thoſe at the end of his Elements of Geometry, I ſhall
add one of his Conſtructions.
Let A and B be the points given, and CD the given line: drawing AB and
biſecting it in F, through E let EF be drawn perpendicular to AB and meeting
CD in F: and from any point H in EF draw HG perpendicular to CD, and
having drawn BF, to the ſame apply HI = HG, and parallel thereto draw BK
meeting EF in K: then with center K and radius BK let a circle be deſcribed,
and the thing is done: join KA, and draw KL perpendicular to CD, then be-
cauſe of the parallel lines, HG: HI: : KL: KB; whence as HG and HI are
equal, KL and KB are likewiſe equal. But it is evident from the Conſtruction
that KA = KB, therefore KB = KL = KA.
Because two equal lines HI and Hi may be applied from H to BF each
equal to HG, the Problem will therefore admit of two Solutions, as the Figure
ſhews: except in the caſe when one of the given points, A for inſtance, is given
in the line CD, for then the Problem becomes more ſimple, and admits but of
one conſtruction, as the center of the circle required muſt be in the line EF con-
tinued, as alſo in the perpendicular raiſed from A to CD, and therefore in their
common interſection: and this is the limit of poſſibility; for ſhould the line CD
paſs between the given points, the Problem is impoſſible.
N. B. Tho’ Vieta does not take notice that this Problem is capable of two
anſwers, yet this is as evident from his conſtruction, as from Mr. Simpſon’s, for
EC (the mean proportional between E B and EA) may be ſet off upon the given
line EF either way from the given point E.
19(7)
LEMMA I.
Apoint A being given between the two right lines BC and DE, it is requir-
ed through the point A to draw a line cutting the two given ones at equal
angles.
If the given lines be parallel, then a perpendicular to them through the point
A is the line required. But if not, then let them be produced to meet in the
point F: and let FG be drawn biſecting the Angle BFD, and through A draw
a perpendicular to FG, and it will be the line required by Euc. I. 26.
PROBLEM VIII.
Having a point A given, and alſo two right lines BC and DE, to draw a circle
which ſhall paſs through the given point, and touch both the given right lines.
By the preceding Lemma draw a line IAH’ through the point A, which ſhall
make equal angles with the two given lines BC and DE: biſect IH in K; and
taking KL = KA, by means of the preceeding Problem draw a circle which ſhall
paſs through the points A and L, and likewiſe touch one of the given lines, BC
for inſtance, in the point M. I ſay this circle will alſo touch the other given
line DE: for from the center N letting fall the perpendicular NO, and joining
NI, NH, NM; in the triangles NKH, NKI, NK being common, and HK =
KI, and the Angles at K right ones, by Euc. I. 4. NH = NI likewiſe the
angle NHK = angle NIK, from hence it follows: that the angle NHM = angle
NIO; and the angles at M and O, being both right, and NH being proved
equal to NI, NM will be equal to NO by Euc. I. 26.
Mr. Simpſon conſtructs the Problem thus.
Let BD and BC be the given lines meeting in B, and A the given point,
join AB, and draw BN biſecting the given angle DBC: and from any point E
in BN upon BC let fall the perpendicular EF, and to BA apply EG = EF, pa-
rallel to which draw AH meeting BN in H: then from center H with Interval
AH let a circle be deſcribed, and the thing is done. Upon BC and BD let fall
the perpendiculars HI, HK, which are manifeſtly equal, becauſe by Conſtruction
the angle HBI = HBK; moreover as EF: EG: : HI: HA: but EF and EG
are equal, therefore alſo HI and HA.
PROBLEM IX.
Having a circle whoſe center is A given in magnitude and poſition, and alſo
two right lines BD and ZC given in poſition, to draw a circle which ſhall touch
all three.
20(8)
From A draw two perpendiculars to the right lines DB, ZC; viz. ADF and
AZX; and in theſe perpendiculars take DF, ZX, on “either ſide of D and Z,
equal to the Radius of the given circle: and through F and X draw lines pa-
rallel to DB, ZC; viz. FG, XH; and then by the preceding Problem draw
a circle which ſhall paſs through the given center A and touch the two lines
FG, XH; and E the center of this circle will alſo be the center of the circle
required, as appears by ſubtracting equals from equals in Figure 1: and by adding
equals to equals in Figure 2.
LEMMA II.
If the two circles CEB and CED cut one another C, then I ſay a line drawn
from the point of ſection CBD cutting both circles, will cut off diſſimilar ſeg-
ments from thoſe circles.
1ſt. Suppose CB to be the Diameter of one of them: then draw to the
other point of ſection E the line CE, and joining EB, ED, the angle CEB will
be a right one, and the angle CED either greater or leſs than a right one, and
conſequently CD cannot be a Diameter of the other.
2dly. Suppose CBD not to paſs through the center of either: then through
C draw a Diameter CAG, and continue it to meet the other circle in F, and
join BG, DF: then the angle CBG is a right one, and the angle CDF is either
greater or leſs than a right one: and therefore the lines BG and DF are not
parallel: let H be the center of the other circle, and let a Diameter CHI be
drawn: draw DI and continue it meet to meet CG in K: then DIK will be pa-
rallel to BG: hence CB: CD: : CG: CK. But CI and CK are unequal,
(being both applied from the ſame point in a right angle) and therefore it cannot
be 2s CB: CD: : CG: CI: and hence it appears that the Segments CB and
CD are diſſimilar.
LEMMA III.
If through the legs of any triangle EDF (ſee Figure to Problem 10.) a line
BI be drawn parallel to the baſe DF, ſo that there be conſtituted two ſimilar
triangles about the ſame vertex; and a circle be circumſcribed about each of
theſe triangles; theſe circles will touch one another in the common vertex E.
It is plain that they will either touch or cut each other in the point E: if
they cut each other, then by the preceding Lemma the Segments BE and DE
would be diſſimilar; but they are ſimilar, and they muſt therefore touch each
other.
PROBLEM X.
Having a point A, and alſo a right line BC, given in poſition; together
with a circle whoſe center is G given both in mde and poſition; to
21(9) a circle which ſhall paſs through the given point, and likewiſe touch both the
given line and the given circle.
The given right line may either 1ſt cut, touch, or be entirely without the
given circle, and the given point be without the ſame or in the circum-
ference; or 2dly, it may cut the given circle, and the given point be within
the ſame, or in the circumference.
Case Iſt. Suppoſe the given line BC to cut, touch, or fall entirely with-
out the given circle; and the given point A to be without the ſame, or in the
circumference: through G the center of the given circle draw DGFC per-
pendicular to BC, and joining DA, take DH a 4th proportional to DA, DC,
DF, ſo that DA X DH = DC X DF: then through the points A and H draw
a circle touching the line CB by Problem VII, and I ſay it will alſo touch the
given circle.
Draw DB cutting the given circle in E, and join FE. Now becauſe the
triangles DEF, DCB are ſimilar, DF: DE: : DB: DC, and therefore DC
X DF = DB X DE. But DC X DF = DA X DH by Conſtruction. Hence
DB X DE = DA X DH, and therefore the points B, E, H, A, will be alſo in
a circle: but the point E is alſo in the given circle; therefore theſe circles either
touch or cut one another in that point. Let now BI be drawn from the point
of contact B perpendicular to the touching line BC to meet the circumference
again in I, and it will be a Diameter: and let EI be joined: then becauſe the
angles FED and BEI are vertical and each of them right ones, FEI will be a
continued ſtraight line: and it appears that the two circles will touch each
other by the preceding Lemmas.
Case 2d. Suppoſe the given line BC to cut the given circle, and the given
point to be within the ſame, or in the circumference; the Conſtruction and De-
monſtration are exactly the ſame as before, except that the angles FED and
BEI are not vertical but coincident, and ſo EI is coincident with EF.
N. B. In either of theſe caſes if the point A coincide with E or be given in
the circumference, draw DEB, and erect BI perpendicular to CB to meet FE in
I, then upon BI as diameter deſcribe a circle, and the thing will manifeſtly be done.
PROBLEM XI.
Having two circles given in magnitude and poſition, whoſe centers are A
and B, as likewiſe a right line CZ; to draw a circle which ſhall touch all three.
From the center of the leſſer circle B let BZ be drawn perpendicular to CZ,
and in BZ (or in BZ continued as the caſe requires) let be taken ZX = AL the
Radius of the other circle; and through X let XH be drawn parallel to CZ,
and with center B and Radius BG, equal to the difference ( or ſum as the Caſe
requires) of the Radii of the two given circles, let a circle be deſcribed; and
22(10) let another circle be deſcribed, touching this laſt, and alſo the line XH and paſſ-
ing through the point A by Problem X, and I ſay that E the center of this circle
will alſo be the center of the circle required; as will appear by taking equals from
equals, or adding equals to equals, as the aſſigned Caſe and Data ſhall require.
The Caſes are four, though Vieta makes but three.
Case Iſt. If it be required that the circle ſhould touch both the others ex-
ternally then BG muſt be taken equal to the difference of the Semidiameters of
the two given circles, and ZX muſt be taken in BZ produced.
Case 2d. If it be required that the circle ſhall touch and include both the
given ones; then BG muſt be taken equal to the difference, as in Caſe 1ſt, but
ZX muſt be taken in BZ itſelf.
Case 3d. If it be required that the circle ſhould touch and include the greater
of the given circles, and touch externally the other whoſe center is B; then BG
muſt be taken equal to the ſum of the Radii of the given circles, and ZX muſt
be taken in BZ itſelf.
Case 4th. If it be required that the circle ſhould touch the greater of the
given circles externally, and touch and include the leſſer; then BG muſt be taken
equal to the ſum of the Radii, and ZX muſt be taken in BZ produced.
PROBLEM XII .
Having two points given B and D, and like wiſe a circle whoſe center is A; to de-
ſcribe another circle which ſhall paſs through the given points, and touch the
given circle.
Let DB be joined, as alſo AB, and let AB be produced to cut the given
circle in the points I and K, then let BH be taken a 4th proportional to DB,
BK, BI; ſo that BD X BH = BI X BK: from H let a Tangent HF be drawn
to the given circle; and BF be joined and cut the circle again in G: and let DG
be drawn cutting the given circle again in E, and laſtly through the points D,
B, G, let a circle be drawn, I ſay it will touch the given circle in G.
For joining EF; becauſe the rectangle DBH = the Rectangle KBI, i. e. the
Rectangle GBF, thereforethefourpoints D, H, F, G, are in a circle; and hence the
angle HFB = the angle GDB: (for in the two firſt ſigures one is theexternal angle
of a quadrilateral figure, and the other is the internal and oppoſite, and in the two
laſt figures theſe angles are in the ſame ſegment.) But the angle HFB = the angle
GEF by Eu. III. 32. hence GEF = GDB: therefore the triangles GEF and
GDB are ſimilar and under the ſame vertex, and therefore by Lemma 3. the cir-
cles deſcribed about them will touch each other in the common vertex G.
1
11There are other Conſtructions of this Problem in Hugo de Omerique, Simpſon, and the Mathematician.
See alſo Monthly Review for Oct. 1764, where the Reviewer is pleaſed to ſpeak favourably of the 1ſt Edi-
tion of this work, but wiſhes that ſome modern ſolutions of theſe Problems had been inſerted, which, he
ſays, are more conciſe and elegant than any which are to be met with in the works of the Antients. The
Editor acknowledges that the conſtruction there given is more ſimple than Vieta’s; but Vieta is not an
Antient, and he knows of no others that exceed his.
23(11)
LEMMA IV.
Having two circles ABCI and EFGH given, it is required to find a point
M, in the line joining their centers, or in that line continued, ſuch, that any
line drawn through the ſaid point M, cutting both the circles, ſhall always cut
off ſimilar ſegments.
Let the line KL joining the centers be ſo cut or produced, that KM may be
to LM in the given ratio of the Radii AK to EL [in the caſe of producing
KL we muſt make it as AK-EL: EL: : KL: LM, for then by compoſition it
will be AK: EL: : KM: LM] and then I ſay that the point M will be the
point required. For from it drawing any line MGFCB cutting the circle
ABCI in B and C, and the circle EFGH in F and G; and let B and F be the
correſpondent points moſt diſtant from M, and C and G the correſpondent
points that are nearer; and let be joined BK, CK, FL, GL, and thereby two
triangles will be formed BKC, FLG. Now becauſe by Conſtriction KM: LM
: : KB: LF, KB and LF will be parallel by Euc. VI. 7. and for the ſame rea-
ſon KC and LG will be parallel; and therefore the triangles BKC and FLG
will be equiangled, and hence the ſegment BC will be ſimilar to the ſegment FG.
LEMMA V.
The point M being found as in the preceding Lemma, I ſay that it is a pro-
perty of the ſaid point, that MG x MB = MH x MA: as alſo that MF x
MC = ME x MI.
For joining CI and GH, it is evident that theſe lines will alſo be parallel.
Hence MI: MC: : MH: MG, but MI: MC: : MB: MA, therefore
MH: MG: : MB: MA, and MG x MB = MH x MA. Again MH: MG
: : MF: ME, but MH: MG: : MI: MC, therefore MF: ME: : MI: MC
and MF x MC = ME x MI.
PROBLEM XIII.
Having two circles given in magnitude and poſition, whoſe centers are K
and L, and alſo a point D; to draw a circle which ſhall touch the two given
ones, and alſo paſs through the point D.
Join the given centers by drawing KL, and in KL or KL produced ſind the
point M (by Lemma 4.) ſuch, that all the lines drawn from it cutting the given
circles ſhall cut off ſimilar ſegments; and let KL cut the circumferences, one
of them in the points A and I, and the other in the points E and H; and join-
ing MD, make it as MD: MA: : MH: MN. Then through the points D and
N draw a circle which ſhall alſo touch the given circle whoſe center is K, by
Prob. XII. and I ſay that this circle will alſo touch the other given circle whoſe
center is L. For let B be the point of contact and BM be drawn cutting the
circle K in C, and the circle L in F and G; then by Lemma 5. MB x MG
24(12) MA x MH: but MA x MH = MD x MN by Conſtruction; therefore MB x
MG = MD x MN, and the points B, G, N, D, are in a circle. But the point
G is alſo in the circle L; therefore theſe circles either touch or cut each other in
the point G. Now the circles BND and BCI touch one another in B by con-
ſtruction; therefore the ſegment BC is ſimilar to the ſegment BG; and alſo by
conſtruction the ſegment BC is ſimilar to the ſegment FG; and therefore the
ſegment FG is ſimilar to the ſegment BG; and hence the circles FGE and BGD
touch one another in the point G.
The Caſes are three.
Case Iſt. Iſ the circle be required to touch and include both the given ones;
then M muſt be taken in KL produced; and MN muſt be taken a fourth pro-
portional to MD, MA, MH, A being the moſt diſtant point of interſection in
the circle K, and H the neareſt point of interſection in the circle L.
Case 2d. If the circle be required to touch both the given ones externally;
then alſo M muſt be taken in KL produced; and MN taken a fourth proportional
to MD, MA, MH, A being the neareſt point of interſection in the circle K,
and H the moſt diſtant in the circle L.
Case 3d. If the circle be required to touch and include the circle K, and to
touch L externally; then M muſt be taken in KL itſelf; and MN a fourth pro-
portional to MD, MA, MH, A being the moſt diſtant point in K, and H the
neareſt in L.
PROBLEM XIV.
Having three circles given whoſe centers are A, B, and D; to draw a fourth
which ſhall touch all three.
Let that whoſe center is A be called the 1ſt, that whoſe center is B the 2d,
and that whoſe center is D the 3d. Then with center B, and radius equal to
the differcnce, or ſum, as the caſe requires, of the ſemidiameters of the 1ſt and
2d circles, let an auxiliary circle be deſcribed; and likewiſe with D center, and
radius equal to the difference, or ſum, as the caſe requires, of the ſemidiameters
of the 1ſt and 3d circles, let another auxiliary circle be deſcribed; and laſtly by
the preceding Problem draw a circle which ſhall touch the two auxiliary ones,
and likewiſe paſs through the point A which is the center of the firſt circle.
Let the center of this laſt deſcribed circle be E, and the ſame point E will like-
wiſe be the center of the circle required; as will appear by adding equals to
equals, or taking equals from equals, as the caſe requires.
The Caſes are theſe.
Case 1ſt. Iſ it be required that the circle ſhould touch and include all the other
three; then let A be the center of the greateſt given circle, B of the next, and D
of the leaſt: and let BG = the difference of the ſemidiameters of the 1ſt and
25(13) and DF = the Difference of the ſemidiameters of the 1ſt and 3d, and through A
deſcribe a circle which ſhall touch the two auxiliary ones externally.
Case 2d. If it be required that the circle ſhould touch all the other three
externally; then the circles being the ſame as before in reſpect to their magnitude,
let BG and DF be alſo the ſame as in the 1ſt Caſe, but through A deſcribe a circle
which ſhall touch and include the two auxiliary ones.
Case 3d. If it be required that the circle ſhould touch and include one of the
given circles A, and touch the other two externally; then let BG and DF = the
the ſum of the ſemidiameters of the 1ſt and 2d, and ſum of the ſemidiameters of
of the 1ſt and 3d reſpectively; and through A deſcribe a circle which ſhall touch
the two auxiliary ones externally.
Case 4th. If it be required that the circle ſhould touch externally one of the
given circles A, and ſhould touch and include the other two; then let BG and
DF = the ſums as in Caſe 3d, but through A deſcribe a circle which ſhall touch
and include the two auxiliary ones.
26(14)
SUPPLEMENT.
PROBLEM I.
HAVING two points given A and B, to determine the Locus of the cen-
ters of all ſuch circles as may be drawn through A and B.
Join AB and biſect it in the point C, and through C, draw a perpendicular
to it CE, and continue it both ways in infinitum, and it is evident that this line
CE will be the Locus required.
PROBLEM II.
Having two right lines given AB and CD, to determine the Locus of the
centers of all ſuch circles as may be drawn touching both the ſaid lines.
Case 1ft. Suppoſe AB and CD to be parallel; then drawing GI perpendicu-
lar to them both; biſect it in H, and through H draw EHF parallel to the two
given lines, and it will be the Locus required.
Case 2d. Suppoſe the given lines being produced meet each other in E,
then biſecting the angle BED by the line EHF, this line EHF will be the Locus
required.
PROBLEM III.
Having two circles given whoſe centers are A and B; to determine the Locus
of the centers of all ſuch circles as ſhall touch the two given ones.
Cases 1ſt and 2d. Suppoſe it be required that the circles be touched outwardly
by both the given ones; or that they be touched inwardly by both the given
ones.
Then joining the centers A and B, let AB cut the circumferences in C and
D and produced in P and O: let CD which is intercepted between the convex
circumſerences be biſected in E: ſet off from E towards B the center of
27(15) greater of the given circles the line EH = the difference of the Radii of the given
circles, and with A and B as Foci and EH Tranverſe Axis, let two oppoſite
Hyperbolas be deſcribed KEK and LHL: then I ſay that KEK will be the
locus of the centers of all the circles which can be drawn ſo as to be touched
outwardly by both the given circles; and LHL will be the locus of the centers
of all the circles which can be drawn ſo as to be touched inwardly by both the
given circles. For taking any point K in the Hyperbola KEK, and drawing KA
and KB, let theſe lines cut the given circumferences in F and G reſpectively:
and make KQ = KA: then from the nature of the curve QB = EH, but by con-
ſtruction EH = BG - AF, therefore QB = BG - AF and hence QG = AF, and
then taking equals from equals KG = KF, which is a demonſtration of the 1ſt
Caſe.
Then with regard to the 2d Caſe, taking any point L in the Hyperbola
LHL, and drawing LB and LA and producing them to meet the concave cir-
cumferences in M and N, let alſo LR be taken equal to LB; then from the pro-
perty of the curve AR = EH, but EH (by conſtruction) = BM - NA; there-
fore AR = BM - NA, and NR = BM, and then adding equals to equals LN =
LM, which is a demonſtration of the 2d Caſe.
Case 3d. Suppoſe it be required that the circles to be deſcribed be touched
outwardly by one of the given circles, and inwardly by the other.
Then drawing AB, let it cut the convex circumferences in C and D, and the
concave ones in P and O, biſect PD in E, and from E towards B ſet off EH =
the ſum of the given radii. Then with A and B foci and EH tranſverſe axis,
let two oppoſite hyperbolas be deſcribed KEK and LHL: and KEK will be the
locus of the centers of the circles which are touched inwardly by the circle A
and outwardly by the circle B; and LHL will be the locus of the centers of
thoſe circles which are touched inwardly by B and outwardly by A. The demon-
ſtration mutatis mutandis is the ſame as before.
Case 4th. Suppoſe the given circle A to include B, and it be required that
the circles to be deſcribed be touched outwardly by them both.
Let AB cut the circumferences in C and D, P and O: and biſecting CD in
I, and ſetting off from I towards P, IL = the ſum of the ſemidiameters of the
given circles, and with A and B foci, and IL tranſverſe axis, deſcribing an
ellipſe LKI, it will be the locus of the centers of the circles required. For
taking any point K in the ellipſe, and drawing AK and BK, let AK be con-
tinued to meet one of the given circumferences in G, and let BK meet the other
F. Then from the property of the curve AK + BK = IL = AG + BF (by
28(16) ſtruction:) hence by ſubſtraction BK = KG + BF, and by ſubtraction again
FK = KG.
Case 5th. Suppoſe the given circle A to include B, and it be required that
the circles to be deſcribed be touched outwardly by A and inwardly by B.
Then let AB cut the circumferences in C and D, P and O: and biſecting
CO in I, and ſetting off from I towards P, IL = the difference of the ſemidia-
meters of the given circles, and with A and B foci and IL tranſverſe axis de-
ſcribing an ellipſe LKI, it will be the locus of the centers of the circles deſcribed,
and the demonſtration, mutatis mutandis, is the ſame as in the laſt caſe.
Cases 6th and 7th. Suppoſe the two given circles cut each other, and it be
required that the circles to be deſcribed either be touched and included in them
both, or be touched by them both and at the ſame time include them both.
These two caſes are ſimilar to caſes 1ſt and 2d, and as there, ſo alſo here,
the tranſverſe axis of the two oppoſite hyperbolas, which are the loci required,
muſt be taken = the difference of the ſemidiameters of the given circles. The
demonſtration is ſo alike, it need not be repeated.
PROBLEM IV.
Having a given point A, and a given right line BC, to determine the locus
of the centers of thoſe circles which ſhall paſs through A and touch BC.
From A draw AG perpendicular to BC, then with focus A and directrix BC
let a parabola be deſcribed, and it will be the locus required; for by the propert
of the curve FA always equals FG drawn perpendicular to the directrix.
PROBLEM V.
Having a given point A, and a given circle whoſe center is B, to determine
the locus of the centers of all thoſe circles, which paſs through A, and at the
ſame time are touched by the given circle.
Cases 1ſt and 2d. Suppoſe the point A to lie out of the given circle, and
it be required that the circles to be deſcribed be either touched outwardly by the
given circle, or inwardly by it.
Let AB be drawn, and let it cut the given circumference where it is convex
towards A in the point C, and where it is concave in the point O: then biſecting
AC in E, and ſetting off from E towards B, EH = BC the given radius, and
with A and B foci and EH tranſverſe Axis deſcribing two oppoſite Hyperbolas
KEK and LHL, it is evident that KEK will be the locus of the centers of thoſe
circles which paſs thrugh A and are touched outwardly by the given circle, and
LHL will be the locus of the centers of thoſe circles which paſs through A and
are touched inwardly by the given circle.
29(17)
Case 3d. Suppoſe the given point A to lie in the given circle, whoſe center
is B.
Then joining AB and continuing it to meet the given circumference in C and
O, biſect AC in E, and from E towards O ſetting off EH = BC the given
Radius, and with A and B Foci and EH tranſverſe Axis deſcribing an Ellipſe
EKH, it will evidently be the Locus required.
PROBLEM VI.
Having a right line BC given, and alſo a circle whoſe center is A, to deter-
mine the Locus of the centers of the circles which ſhall be touched both by the
given right line and alſo by the given circle.
There are three Caſes, but they are all comprchended under one general
ſolution.
Case 1ſt. Let the given right line be without the given circle, and let it be
required that the circles to be deſcribed be touched outwardly by the given circle.
Case 2d. Let the given right line be without the given circle, and let it be
required that the circles to be deſcribed, be touched inwardly by the given
circle.
Case 3d. Let the given right line be within the given circle, and then the
circles to be deſcribed muſt be touched outwardly by the given circle.
General Solution.
From the given center A let fall a perpendicular AG to the given line BC,
which meets the given circumference in D [or in Caſes 2d and 3d is produced
to meet it in D] and biſecting DG in F, and ſetting off FM = FA (which is the
ſame thing as making GM = AD the given Radius) and through M drawing
MLK parallel to the given line BC, with A Focus and LK Directrix deſcribe a
Parabola, and it will be the Locus of the centers of the circles required; for
from the property of the Curve FA = FM, and adding equals to equals, or
ſubtracting equals from equals, as the Caſe requires, FD = FG.
30
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31(19)
A SECOND
SUPPLEMENT,
BEING
Monſ. DE FERMAT’S Treatiſe on
Spherical Tangencies.
PROBLEM I.
HAVING four points N, O, M, F, given, to deſcribe a ſphere which
ſhall paſs through them all.
Taking any three of them N, O, M, ad libitum, they will form a triangle,
about which a circle ANOM may be circumſcribed, which will be given in
magnitude and poſition. That this circle is in the ſurface of the ſphere
ſought appears from hence; becauſe if a ſphere be cut by any plane, the
ſection will be a circle; but only one circle can be drawn to paſs through the
three given points N, O, M; therefore this circle muſt be in the ſurface of
the ſphere. Let the center of this circle be C, from whence let CB be
erected perpendicular to it’s plane; it is evident that the center of the ſphere
ſought will be in this line CB. From the fourth given point F let FB be
drawn perpendicular to CB, which FB will be alſo given in magnitude and
poſition. Through C draw ACD parallel to FB, and this line will be
32(20) diameter of the circle given in poſition, and therefore the points A and D
will alſo be given.
Letus now ſuppoſe the thing done, and that the center of the ſphere
ſought is E, which, as obſerved before, muſt be in the line CB. Drawing
EF, EA, ED, theſe lines muſt be equal, ſince the points F, A, D, have
been ſhewn to be in the ſurface of the ſphere. But theſe lines EF, EA, ED,
are in the ſame plane, ſince FB and AD are parallel, and BC perpendicular
to each of them. If therefore a circle be deſcribed to paſs through the three
points F, A, D, whoſe center is E, it will be in the line CB, and will be
the center of the ſphere required.
PROBLEM II.
Having three points N, O, M, given, and a plane AD, to deſcribe a-
ſphere which ſhall paſs through the three given points; and alſo touch the
given plane.
Let a circle ENOM be deſcribed to paſs through the three given points,
it will be in the ſurface of the ſphere ſought, from what was ſaid under the
former Problem. From it’s center I let a perpendicular to it’s plane IA be
erected; the center of the ſphere ſought will be in this line IA; let the line
IA meet the given plane in the point A, which point will be therefore given.
From the center of the given circle ENOM, let a perpendicular to the given
plane ID be drawn, the point D will then be given, and therefore the line
AD both in poſition and magnitude, as likewiſe the lines ID, IA, and the
plane of the triangle ADI. But the plane of the circle NOM is alſo given
in poſition, and therefore alſo their common ſection EIF, and hence the
points E and F.
Suppoſe now the thing done, and that the center of the ſphere ſought is B.
Draw BE, BF, and BC parallel to ID. Since the triangle ADI, and the
line EIF are in the ſame plane, therefore BE, BF, BC, will be in the ſame
plane. But the line ID is perpendicular to the given plane, therefore the
line BC parallel to it, will alſo be perpendicular to the given plane. Since
then a ſphere is to be deſcribed to touch the plane AD, a perpendicular BC
from it’s center B will give the point of contact C; and BC, BE, BF will be
equal, and it has been proved that they are in a plane given in poſition, in
which plane is alſo the right line AD. The queſtion is then reduced to this,
Having two points E and F given, as alſo a right line AD in the ſame
33(21) to find the center of a circle which will paſs through the two points, and like-
wiſe touch the right line, which is the VIIth of the preceeding Problems.
PROBLEM III.
Having three points N, O, M, given, as likewiſe a ſphere IG, to de-
ſcribe a ſphere which will paſs through the three given points, and likewiſe
touch the given ſphere.
The circle NOM in the ſurface of the ſphere ſought is given, and a per-
pendicular to its plane from it’s center FA being drawn, the center of the
ſphere required will be in this line. From I the center of the given ſphere
let IB be drawn perpendicular to FA, and through F, ED parallel to IB,
which, from what has been before proved, will be in the plane of the circle
NOM, and the points E and D will be given.
Suppoſe now the thing done, and that the center of the ſphere required is
C. Then the lines CI, CE, CD, will be in the ſame plane, which is given, as
the points I, E, and D are given. But the point of contact of two ſpheres is
in the line joining their centers; therefore the ſphere ſought will touch the
ſphere given in the point G, and the line IC will exceed the lines EC, ED, by
IG the radius of the given ſphere: with center I therefore and this diſtance
IG let a circle be deſcribed in the plane of the lines CI, CE, CD, and it
will paſs through the point G and be given in magnitude and poſition; but
the points D and E are alſo in the ſame plane; and therefore the queſtion is
reduced to this, Having two points E and D given, as likewiſe a circle
IGH, to find the center of a circle which will paſs through the two points
and likewiſe touch the circle, which is the XIIth of the preceeding Problems.
PROBLEM IV.
Having four planes AH, AB, BC, HG, given; it is required to de-
ſcribe a ſphere which ſhall touch them all four.
If two planes touch a ſphere, the center of that ſphere will be in a plane
beſecting the inclination of the other two. And if the planes be parallel, it
will be in a parallel plane beſecting their interval. This being allowed,
which is too evident to need further proof; the center of the ſphere ſought
will be in a plane biſecting the inclination of two planes CB and BA; it will
likewiſe be in another plane biſecting the inclination of the two planes BA and
AH; and therefore in a right line, which is the common ſection of theſe
34(22) biſecting planes; let this right line be EF. Moreover, the center of the
ſphere ſought will alſo be in a plane biſecting the inclination of the two planes
AH and GH, and the interſection of this laſt biſecting plane with the right
line EF will give a point D, which will be the center of the ſphere required.
PROBLEM V.
Having three planes AB, BC, CD, given, and alſo a point H; to find
a ſphere which ſhall paſs through the given point, and likewiſe touch the
three given planes.
Suppose it done. The three planes, by what was ſaid under the laſt pro-
poſition, will give a right line in poſition, in which will be the center of the
ſphere required. Let this right line be GE, perpendicular to which from H
the given point let HI be drawn, which therefore will be given in magnitude
and poſition. Let HI be produced, and FI taken equal to HI; the point
F will then be given. Now ſince the center of the ſphere required is in the
line GE, and FH is perpendicular thereto and biſected thereby, and one
extreme H is by hypotheſis in the ſurface of the ſaid ſphere, the other extreme
F will be ſo too. Nay even a circle deſcribed with I center and IH radius
in a plane perpendicular to GE will be in the ſaid ſpherical ſurface. Here
then we have a circle given in magnitude and poſition, and taking any one
of the given planes AB, by an evident corollary from Problem II. of this
Supplement, a ſphere may be deſcribed which will touch the given plane,
and likewiſe have the given circle in it’s ſurface; and ſuch a ſphere will
anſwer every thing here required.
PROBLEM VI.
Having three planes ED, DB, BC, given, and alſo a ſphere RM, to
conſtruct a ſphere which ſhall touch the given one, and likewiſe the three
given planes.
Suppose it done, and that the ſphere ERCA is the required one, viz.
touches the ſphere in R, and the planes in E, A, C. Let the center of this
ſphere be O; then drawing RO, EO, AO, CO, they will all be equal, and
RO will paſs through M the center of the given ſphere; and EO, AO, CO,
will be perpendicular to the planes ED, DB, BC. Let OU, OG, OI, be
made each equal to OM; and through the points U, G and I, let the planes
UP, GH, IN, be ſuppoſed drawn parallel to the given ones ED, DB, BC,
reſpectively. Since OR is equal to OE, and OM equal to OU, RM will
35(23) equal to UE: but RM is given in magnitude, being the radius of the given
ſphere, therefore UE is alſo given in magnitude. And ſince OE is perpen-
dicular to the plane DE, it will be alſo to the plane PU which is parallel
thereto. UE then being given in magnitude, and being the interval be-
tween two parallel planes DE, PU, whereof DE is given in poſition by hypo-
theſis, the other PU will alſo be given in poſition. In the ſame manner it
may be proved that the planes GH, IN, are given in poſition, and that the
lines OG, OI, are perpendicular thereto reſpectively, and each alſo equal to
OM. A ſphere therefore deſcribed with center O and OM radius will touch
the three planes PU, GH, IN, given in poſition: but the point M is given,
being the center of the given ſphere. The queſtion is then reduced to this,
Having three planes given PU, GH, IN, and a point M, to find the radius
of a ſphere which ſhall touch the given planes, and paſs through the given
point; which is the ſame as the preceeding Problem. [And this radius be-
ing increaſed or diminiſhed by MR, according as R is taken in the further or
nearer ſurface of the given ſphere, will give the radius of a ſphere which will
touch the three given planes DE, DB, BC, and likewiſe the given
ſphere. ]
By a like method, when among the Data there are no points, but only
planes and ſpheres, we ſhall always be able to ſubſtitute a given point in the
place of a given ſphere.
PROBLEM VII.
Having two points H, M, as alſo two planes AB, BC, given, to find a
ſphere which ſhall paſs through the given points, and touch the given
planes.
Draw HM and biſect it in I, the point I will be given, through the
point I let a plane be erected perpendicular to the right line HM, this plane
will be given in poſition, and the center of the ſphere required will be in this
plane. But becauſe it is alſo to touch the planes AB, BC, its center will be
alſo in another plane given in poſition (by what has been proved, Prob. IV.)
and therefore in a right line which is their interſection, given in poſition,
which let be GE; to which line GE from one of the given points M demit-
ting a perpendicular MF, it will be given in magnitude and poſition, and
being continued to D ſo that FD equals MF, the point D will be given;
and, from what has been proved before, will be in the ſpherical ſurface.
36(24) Thercfore there are given three points H, M, D, as likewiſe a plane AB,
or AC, through which points the ſphere is to paſs, and alſo touch the given
plane. Hence it appears that this Problem is reduced to the IId of this
Supplement.
Before we proceed, the following eaſy Lemmas muſt be premiſed.
LEMMA I.
Let there be a circle BCD, and a point E taken without it, and iſ from
E a line EDOB be drawn to paſs through the center, and another line ECA
to cut it any ways; we know from the Elements that the rectangle AEC is
equal to the rectangle BED. Let us now ſuppoſe a ſphere whoſe center is O,
and one of whoſe great circles is ACDB; if from the ſame point E a line
ECA be any-how drawn to meet the ſpherical ſurface in the points C and A,
I ſay the rectangle AEC will ſtill be equal to the rectangle BED. For if we
ſuppoſe the circle and right line ECA to revolve upon EDB as an immove-
able axis, the lines EC and EA will not be changed, becauſe the points C
and A deſcribe circles whoſe planes are perpendicular to that axis; and
therefore the rectangle AEC will in any plane be ſtill equal to the rectangle
BED.
LEMMA II.
By the ſame method of reaſoning, the Vth Lemma immediately preceed-
ing Problem XIII, in the Treatiſe of Circular Tangencies, may be extended
alſo to ſpheres, viz. that in any plane (ſee the Figures belonging to that
Lemma) MG X MB = MH X MA. And alſo that MF X MC = ME X MI.
LEMMA III.
Let there be two ſpheres YN, XM, through whoſe centers let the right
line RYNXMU paſs, and let it be as the radius YN to the radius XM, ſo
YU to XU; and from the point U let a line UTS be drawn in any plane,
and let the rectangle S U T be equal to the rectangle RUM; I fay that if
any ſphere OTS be deſcribed to paſs through the points T and S, and to
touch one of the given ſpheres XM as in O, it will alſo touch the other
given ſphere YN. For joining UO, and producing it to meet the ſurface of
the ſphere OTS in Q; the rectangle QUO = the rectangle SUT, by
Lemma I. but the rectangle SUT = the rectangle RUM. by
37(25) which RUM by Lemma II. is equal to a rectangle under UO and a line
drawn through the points U and O to the further ſurface of the ſphere YN.
Therefore the point Q is in the ſurface of the ſphere YN; it is therefore
common to the ſpheres YN and OTS; and I ſay that theſe ſpheres touch in
the ſaid point Q. For from the point U let a line UZ be drawn in any
plane of the ſphere OTS, and being produced let it cut the three
ſpheres in the points Z, D, H, K, P, B. The rectangle ZUB in the
ſphere OTS is by Lemma I. and II. equal to the rectangle DUP terminated
by the ſpheres XM and YN. But DU is greater than ZU, becauſe the
ſpheres XM and OTS touch in the point O, and therefore any other line from
U but UO muſt meet the ſurface of OTS before it meets the ſurface XM.
Since then ZUB = DUP, and DU is greater than ZU, UP muſt be leſs
than UB, and the point B will fall without the ſphere YN; and by the
fame reaſon, all other points in the ſurface of the ſphere OTS, except the
point Q.
The Demonſtration is ſimilar and equally eaſy in all caſes, whether the
ſpheres touch exlernally or internally.
LEMMA IV.
Let there be a plane AC, and a ſphere FGD through whoſe center O let
FODB be drawn perpendicular to the plane, and from F any right line
FGA cutting the ſphere in G and the plane in A; I ſay that the rectangle
AFG = the rectangle BFD. For let the given ſphere and plane be cut by
the plane of the triangle ABF, the ſection of the one will be the circle GDF,
and of the other the right line ABC. Since the line FB is perpendicular
to the plane AC, it will be alſo to the right line AC. Having then a circle
FDB, and a right line AC in the ſame plane; and a line FDB paſſing thro'
the center perpendicular to AC, join D and G, and in the quadrilateral
figure ABDG the angles at B and G being both right ones, it will be in a
circle, and the rectangle AFG = the rectangle BFD; and the ſame may be
proved in any other ſection of the ſphere.
LEMMA V.
Let there be a plane ABD and a ſphere EGF, through whoſe center O
let FOEC be drawn perpendicular to the plane, and in any other plane let
FHI be drawn ſo that the rectangle IFH = the rectangle CFE: if
38(26) the points I and H a ſphere be deſcribed which touches the plane AC, I ſay
it will alſo touch the ſphere EGF. From F draw FB to the point of contact
of the ſphere and plane, and make the rectangle BFN = the rectangle CFE,
and the point N will be in the ſurface of the ſphere EGF, by Lemma IV.
But the rectangle CFE, by conſtruction, = the rectangle IFH; therefore
IFH = BFN, and the point N will be alſo in the ſurſace of the ſphere IHB.
It remains then to be proved that theſe ſpheres touch in N, which is very eaſy
to be done. For from the point F through any point R in the ſpherical ſur-
face EGF let the line FR be drawn, which may cut the ſpherical ſurface
IBH in L and P, and the plane AC in K. The rectangle KFR = the rect-
angle CFE, by Lemma IV. = the rectangle IFH, by conſtruction, = the
rectangle PFL. Since then KFR = PFL, and KF is greater than PF, be-
cauſe the ſphere IHB touches the plane AC in B, therefore FR is leſs than
FL, and the point R is without the ſphere IHB, and the ſame may be ſhewn
of every other point in the ſpherical ſurface EGF, except the point N.
Theſe Lemmas, though they be very eaſy, are very elegant and valuable,
eſpecially the IIId and Vth. In the IIId. though there be an inſinite num-
ber of ſpheres which, paſſing through the points T and S, may touch the
ſphere XM, yet they will all alſo touch the ſphere YN, by what is there
proved. In the Vth, though there be an infinite number of ſpheres which,
paſſing through the points I and H, may touch the plane AC, yet they will
all alſo touch the ſphere EGF, by what is there proved.
We ſhall now be able to go through the remaining Problems with eaſe.
PROBLEM VIII.
Let there be given a plane ABC, and two points H and M, and alſo a
ſphere DFE; to find a ſphere which ſhall paſs through the given points, and
touch the given plane, and likewiſe the given ſphere.
Through the center O of the given ſphere let EODB be demitted perpen-
dicular to the given plane ABC, and let HE be drawn, and make the rect-
angle HEG equal to the rectangle BED, and G will then be given. Find
then a ſphere, by Problem II. which ſhall paſs through the three points M,
H, G, and touch the plane ABC, and it will be the ſphere here required.
For it paſſes through the points M and H, and touches the plane ABC,
by conſtruction; it likewiſe touches the ſphere DFE, by Lemma V. For
ſince the rectangle HEG = the rectangle BED, every ſphere which
39(27) through the points H and G, and touches the plane ABC, touches likewiſe
the ſphere DFE.
PROBLEM IX.
Let there be given two ſpheres AB, DE, as alſo two points H and M;
to find a ſphere which ſhall paſs through the two given points, and likewiſe
touch the two given ſpheres.
Let the right line AF be drawn paſſing through the centers of the
ſpheres, and as the radius AB is to the radius DE, ſo make BF to EF, and
the point F will be given. Make the rectangle HFG = the rectangle NFA,
and the point G will be given. Now having given three points M, H, G,
as alſo a ſphere DE; find a ſphere by Problem III, which ſhall paſs through
the given points, and touch the given ſphere; and, by Lemma III, it will be
the ſphere here required.
PROBLEM X.
Let there be given two planes AB, BD, a point H, and a ſphere
EGF; to find a ſphere which ſhall paſs through the given point, and touch
the given ſphere, as alſo the two given planes.
Through the center O of the given ſphere let a perpendicular to either of
the given planes CEOF be demitted, and make the rectangle HFI = the
rectangle CFE. Then having given the two points H and I, as alſo the
two planes AB, BD; find a ſphere, by Problem VII, which ſhall paſs
through the two given points, and likewiſe touch the two given planes; and,
by Lemma V, it will be the ſphere required.
PROBLEM XI.
Let there be given a point, a plane, and two ſpheres; to find a ſphere
which ſhall paſs through the point, touch the plane, and alſo the two
ſpheres.
This Problem, by a like method of reaſoning, is immediately reduced to
the VIIIth, where two points, a plane, and a ſphere are given, and that by
means of the Vth Lemma. But if you chuſe to uſe the IIId Lemma, it will
be reduced to the ſame Problem by a different method, and a different
conſtruction.
40(28)
PROBLEM XII.
Let there be given a point and three ſpheres, to ſind a ſphere which ſhall
paſs through the point, and touch all the three ſpheres.
We aſſign no figure to this Problem alſo, becauſe, by help of Lemma III,
it may immediately be reduced to Problem IX, where two points and two
ſpheres are given.
PROBLEM XIII.
Let there be two planes, and alſo two ſpheres given; to find a ſphere
which ſhall touch the planes, as alſo the ſpheres.
Suppoſe the thing done. If therefore we imagine another ſpherical ſurface
parallel to that which is required, and which we now ſuppoſe found, and
whoſe radius is leſs than it's by the radius of the leſſer of the two given
ſpheres; this new ſpherical ſurface will touch two planes parallel to the two
given ones, and whoſe diſtance therefrom will be equal to the radius of the
leſſer of the given ſpheres; it will alſo touch a ſphere concentric to the
greater given one whoſe radius is leſs than it's by the radius of the leſſer given
one; and it will likewife paſs through the center of the leſſer given one.
The Queſtion is then reduced to Problem X, where a point, two planes and
a ſphere are given.
PROBLEM XIV.
Having three ſpheres and a plane given; to find a ſphere which ſhall
touch them all.
By a like method to what is uſed in the preceeding, and in the VIth Pro-
blem, this is reduced to Problem XI, where a point, a plane, and two
ſpheres are given.
PROBLEM XV.
Having four ſpheres given; to ſind a ſphere which ſhall touch them all.
Suppose the thing done. As, in the treatiſe of Circular Tangencies, the
laſt Problem, where it is required, having three circles given, to find a fourth
which ſhall touch them all, is reduced to another, where a point and two
circles are given; ſo alſo this, by a like method, and ſimilar to what has been
uſed in the preceding Problems, is reduced to Problem XII, where three
ſpheres and a point are given.
41(29)
The various Caſes, Determinations and other Minuliæ we have taken no
notice of: for if we had, this Treatiſe would have very much exceeded that
to which it was intended as a Supplement.
Synopſis of the PROBLEMS.
N. B. A point is repreſented by. , a plane by 1, and a ſphere by 0.
11
1. .... # 4. 1111 # 15. 0000
2. ...1 # 5. 111. # 12. 000.
3. ...0 # 6. 1110 # 14. 0001
7. ..11 # 13. 1100 # 9. 00..
8. ..10 # 10. 11.0 # 11. 00.1
42
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43 1[Figure 1]
44 2[Figure 2]
45
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46 3[Figure 3]
47 4[Figure 4]
48
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49 5[Figure 5]
50 6[Figure 6]
51
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52 7[Figure 7]
53 8[Figure 8]
54
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55 9[Figure 9]
56 10[Figure 10]
57
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58 11[Figure 11]
59 12[Figure 12]
60
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61 13[Figure 13]
62 14[Figure 14]
63
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64
THE
TWO BOOKS
OF
APOLLONIUS PERGÆUS,
CONCERNING
DETERMINATE SECTION,
As they have been Reſtored by
WILLEBRORDUS SNELLIUS.
By JOHN LAWSON, B. D. Rector of Swanſcombe, Kent.
TO WHICH ARE ADDED,
THE SAME TWO BOOKS, BY WILLIAM WALES,
BEING
AN ENTIRE NEW WORK.
LONDON:
Printed by G. BIGG, Succeſſor to D. LEACH.
And ſold by B. White, in Fleet-Street; L. Davis, in Holborne; J. Nourse, in the
Strand; and T. Payne, near the Mews-Gate.
MDCC LXXII.
65
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66[i]
ADVERTISEMENT.
SINCE the publication of the preceding Tract on
Tangencies, the Tranſlator thereof has obſerved,
that thoſe pieces of Willebrordus Snellius, which he
mentioned in his Preface thereto, are exceeding ſcarce
in England. His Reſuſcitata Geometria de ſectione rationis
& ſpatii, 1607, he has never once had an opportunity
of ſeeing; but ſuppoſing this ſhould in a ſhort time be
loſt, more than ample amends is made by what
Dr. Halley has done on the ſame ſubject. Leſt the
other Tract, De Sectione Determinatâ, ſhould undergo
the ſame fate with the original Apollonius, he was
determined to reſcue it therefrom, or reſpite it at leaſt
for ſome time, by putting it into an Engliſh dreſs.
While he was doing this, he happened to communicate
the piece to ſome friends; one of whom has ventured,
after Snellius, on this ſubject, and he preſumes with ſome
ſucceſs, as every Reader will allow, when he peruſes the
Propoſitions here printed after thoſe of Snellius. Yet,
notwithſtanding this, the Editor perſiſted in his reſolu-
tion of printing his tranſlation of Snellius, as the
67[ii] has much merit, and was in danger of being loſt, and as he
was the firſt that conſtructed Quadratic Equations after
this particular manner, as Dr. Simson obſerves in his
Note on Euc. VI. 28 and 29.
The Editor leaves his Friend to ſpeak for himſelf
in relation to what he has done, and truſts that the
candid Reader will not think more meanly of his
performance from the modeſt manner in which he
ſpeaks of it himſelf.
68[iii]
EXTRACT from PAPPUS's Preface to his Seventh Book
in Dr. HALLEY's Tranſlation.
DE SECTIONE DETERMINATA II.
HIS ſubjiciuntur libri duo de Sectione Determinatâ,
quas etiam ad modum præcedentium unam pro-
poſitionem dicere liceat, ſed disjunctam: quæ hujuſ-
modi eſt. “ Datam rectam infinitam in uno puncto ſe-
care, ita, ut è rectis interceptis inter illud & puncta
in illâ data, vel quadratum ex unâ, vel rectangulum
ex duabus interceptis, datam habeat rationem, vel ad
contentum ſub aliâ unâ interceptâ & datá quâdum;
vel etiam ad contentum ſub duabus aliis interceptis:
idque ad quam partem velis punctorum datorum. ”
Hujus autem, quaſi bis disjunctæ, & intricatos Dio-
riſmos habentis, per plura neceſſario facta eſt demon-
ſtratio. Hanc autem dedit Apollonius communi methodo
tentamen faciens, ac ſolis rectis lineis uſus, ad exemplum
ſecundi libri Elementorum primorum Euclidis: ac rur- ſus idem demonſtravit ingenioſe quidem, & magis ad
inſtitutionem accomodate, per ſemicirculos. Habet
1
11From hence it appears that Euclid's were called the firſt Elements,
and that the other Analytical Tracts, recited by Pappus, were called the
ſecond Elements.
69[iv] autem primus liber Problemata ſex, Epitagmata, ſive
Diſpoſitiones punctorum, ſedecim; Dioriſmos quinque:
quorum quatuor quidem Maximi ſunt, Minimus vero
unus. Sunt autem maximi, ad ſecundum Epitagma ſe-
cundi problematis; item ad tertium quarti problematis;
ad tertium quinti & ad tertium ſexti. Minimus vero
eſt ad tertium Epitagma tertii problematis. -Secundus
liber de Sectione Determinatâ tria habet Problemata,
Diſpoſitiones novem, Determinationes tres; e quibus
Minima ſunt ad tertium primi, ut & ad tertium ſecun-
di; Maximum autem eſt ad tertium tertii problematis.
-Lemmata habit liber primus XXVII, ſecundus vero
XXIV. Inſunt autem in utroque libro de Sectione
determinatâ Theorenata octoginta tria.
70[v]
THE
PREFACE.
HAD not a motive more prevalent than Cuſtom induced me to
ſay ſomething by way of Preface to the Performance which
I herewith offer to the Public, the difficulty I find in doing it with
propriety, would have determined me to remain entirely ſilent.
The ſubject has employed the Pen of one of the ableſt Geometers
of the laſt Century; it may therefore ſeem very preſumptuous, in
me at leaſt, who am but young in theſe matters, to attempt it after
him. To obviate, if poſſible, this Cenſure is my only intention
here; and I hope I ſhall not be deemed impertinent, if I attempt to
ſhew wherein I apprehend I have come nearer to the great original
than he hath done.
71[vi]
Pappus, in his preface to the ſeventh Book of Matbematical
Collections, tells us that this Tract of Apollonius was divided into
two Books; that the firſt Book contained ſix Problems, and the
ſecond three: now Snellius' whole work contains but four; and
it ſeemed to me difficult to ſhew how thoſe could contain the ſub-
ſtance of nine, and yet the ſix firſt have ſixteen Epitagmas, or ge-
neral Caſes, and the three laſt nine. I firſt, therefore began with
inquiring whether, or no, other Problems could not be found,
wherein the ſection of an indefinite ſtraight line is propoſed to be
effected, “So, that of the ſegments contained between the point of
ſection ſought, and given points in the ſaid line, either the ſquare
on one of them, or the rectangle contained by two of them, may
have a given ratio to the rectangle contained by one of them and a
given external line, or to the rectangle contained by two of them; “
as is deſcribed by PAPPUS.
In this inquiry it ſoon occurred to me, that the three problems
which make my firſt, ſecond and fourth, come within the account
given by PAPPUS; and therefore are properly Problems in Determi-
ate Section, to be added to the four given by Snellius: and it does
not appear to me that more can be found which ſhould. Hence I
concluded, that there were in theſe, ſome, more general than thoſe
of Apollonius, which ought therefore to be divided.
My next buſineſs was, if poſſible, to find out the order in which
Apollonius had arranged them: and here, with reſpect to the
firſt Book, I had no other information to guide me, but what is
72[vii] be met with in the above mentioned Preface of Pappus; where he
tells us that in the ſix Problems of Book I. there were “Sixteen
Epitagmas, or general Caſes, five Determinations; and of theſe,
four were Maxima, and one a minimum: That the maxima are at the
ſecond Epitagma of the ſecond Problem, at the third of the fourth,
the third of the fifth, and the third of the ſixth; but that the minimum
was at the third Epitagma of the third problem. ” It moreover ſeem- ed reaſonable to me, that theſe Problems wherein the feweſt points
are given, would be antecedent to thoſe wherein there were more;
and of theſe wherein the number of given points are the ſame, that
thoſe would be prior to the others, wherein there was a given ex-
ternal line concerned: and laſtly, that when the number of given
points were two, the ſecond Caſe, or Epitagma, would naturally
be when the required point O is ſought between the two given
ones.
Now the three new Problems, together with the three firſt of
Snellius, making exactly ſixteen Epitagmas, viz. one in the firſt,
and three in each of the others; it ſeemed highly probable, that
theſe compoſed the firſt book. Alſo that the Problem, wherein
only one point was given, would be the firſt; and it ſeemed eaſy
to aſſign the ſecond, becauſe it is the only one wherein the limita-
1
11The words which are in Italics were entirely omitted in Snellius's Extract
from Pappus, both in the Greek and Latin, and are read with ſome variation in
Commandine's tranſlation; but are here printed according to Dr. Halley: and
though I know not whether in this particular place he had the Authority of either
of the Savilian MSS, yet I hope I run no great riſk in ſubſcribing to the opinion
of ſo excellent a Geometer.
73[viii] tion is at the ſecond Epitagma; and farther, the Limiting Ratio is
therein a maximum, as it ought. Again, the Problem, wherein
it is propoſed to make the ſquare on AO in a given ratio to the rect-
angle contained by EO and P, has its limiting ratio a minimum
when the required point is ſought beyond (E) that of the given ones
which bounds the ſegment concerned in the conſequent term of
the ratio; which, therefore, I apprehend muſt have been the third
Epitagma, and if ſo, this of courſe muſt have been the third Pro-
blem: and as there remains only one wherein the number of given
pointsare two, I make that the fourth. With reſpect to the fifth and
ſixth Problems, in which three points are given, it ſhould ſeem
that that would be the firſt in order, wherein there is a given ex-
ternal line concerned.
But it ſhould, by no means, be diſſembled that objections may be
brought againſt the identity, and arrangment of ſome of theſe Pro-
blems. For firſt, Pappus no where expreſsly ſays that Apollo-
NIUS compared together two ſquares, wherefore, if this cannot be
implied, the identity of the fourth Problem is deeply ſtruck at:
and moreover, this fourth Problem perhaps cannot with propriety,
be ſaid to have its limiting ratio either maximum or minimum, un-
leſs the ratio of equality, can be admitted as ſuch. Laſtly, in the
fifth Problem, the ſaid limiting ratio is a minimum, and not a maxi-
mum as it is ſaid to have been by Pappus: either, therefore,
74[ix] miſtake muſt be admitted in this Author, or the fifth Problem is
wrong placed. I am not prepared farther to obviate theſe objec-
tions, and only mention them to ſhew that although I ſaw them
in their full force, I could by no means agree, that they are pow-
erful enough to overturn thoſe already advanced for what I have
11[Handwritten note 1]22[Handwritten note 2]done.
I come now to Book II, which if I am not much miſtaken, was
entirely employed about what Snellius makes his fourth Problem.
In this I am confirmed not only by the account which Pappus gives
in his Preface, but much more by the Lemmas of Apollonius
11[Handwritten note 1]22[Handwritten note 2] which he hath left us. For we there find that Lemma 21, where-
in is aſſigned the leaſt ratio which the rectangle contained by AO
and UO can bear to that contained by EO and IO, when O is ſought
between the two mean points of the four given ones, is ſaid to be
concerned in determining the μοναχὴ, or ſingle Caſe , of Problem 1. This Problem therefore of Apollonius contained only thoſe
Caſes of the general one, where O is ſought between the two mean
points. In like manner, we gather from Lemma 22, that his ſe-
cond Problem was concerned in determining the point O when ſought
between a mean point, and an extreme one. And laſtly, from
Lemma 24, that the third Problem of Book II. determined
the point O when required without all the given ones.
1
11So called, I conceive, becauſe in every other Caſe of the third Epitagma,
except this extreme, or limiting one, there are two points which will ſatisſy
the Problem.
75[x] The Limitations of the two former are ſaid by Pappus to have
been minimums, and that of the third a maximum, in conformity,
to which, I have here made them ſo; although I cannot ſee with
what propriety: each of them admitting, in ſome Caſes, of a maxi-
mum and in others of minimum, as I have intimated in a ſcbolium at
the end of each Problem. But notwithſtanding I have conformed to
the manner of Apollonius in dividing this Problem into three,
which it muſt be confeſſed contributes much to order in enumera-
ting ſuch a multitude of Caſes, yet have I previouſly ſhewn how
the whole may be generally conſtructed at once; and that by a me-
thod, which I flatter myſelf will not be found inferior to any that
hath heretofore been given of this very intricate and general
Problem.
Such are the things that I have attempted, and ſuch the reaſons
for what I have done in the following little Tract. The merit due
to each of them I chearfully ſubmit (where every one ought) to
the deciſion of the impartial Reader. In the Conſtructions, my
chief Aim was novelty and uniformity: I could have given more
ſimple conſtructions to one or two of them; in particular the ſixth
of Book I: but it was not my intention to give any thing that I
knew had been done before. I know of many imperfections, but
no falſe reaſonings, and hope none will be found; but if there
ſhould, I hope the candid Geometer will be more inclined to ex-
cuſe than exult, when I aſſure him the greateſt part of the work
has been executed at different times, amidſt the hurry and perplexi-
ties which it may eaſily be conceived attend the fitting out for a
three years Voyage to the ſouth ſeas.
76[xi]
I cannot conclude without acknowledging, in the warmeſt man-
ner, the obligations I am under to my truly worthy and ingenious
friend, the Tranſlator of Snellius; for the great pains and trou-
ble he hath taken to furniſh me with tranſlations from various Au-
thors, which my utter want of the Greek, and little acquaintance
with the Latin Language made abſolutely neceſſary to me: And
after all, had it not been for his kindneſs, this attempt might ſtill
have remained in as great obſcurity as its Author.
77
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78
PROBLEMS
CONCERNING
DETERMINATE SECTION.
PROBLEM I.
TO cut a given indefinite right line in one point, ſo that of the ſegments
intercepted between that point and two other points given in the inde-
finite right line, the ſquare of one of them may be to the rectangle under the
other and a given external right line, in a given ratio.
In the given indefinite right line let be aſſigned the points A and E, it is then
required to cut it in the point O, ſo that AO2 may be to OE into a given
line AU in the ratio of R to S; which ratio let be expreſſed by AI to AU,
ſetting off AI from A either way, either towards E or the contrary; and
then from A and I erect two perpendiculars AY equal to AE, and IR
equal to AI, and theſe on the ſame ſide of the given indefinite line, if AI
was ſet off towards E; but on oppoſite ſides, if AI was ſet off the other way.
The former conſtruction I will beg leave to call Homotactical, and the latter
Antitactical. Let now the extremities of theſe perpendiculars Y and R be
joined, and upon YR as a diameter let a circle be deſcribed, I ſay that the
interſection of this circle with the given indefinite line ſolves the Problem.
If it interſects the line in two places, the Problem admits of two Solutions;
79[2] but if it only touches, then only of one; if it neither touches nor cuts, it is
then impoſſible.
Demonstration. Let a point of interſection then be O, and join O Y
and OR. The angles AYO and IOR are equal, the angle AOY being the
complement of each of them to a right one, and hence the triangles AOY and
IOR are ſimilar.
Hence AY = AE: AO: : OI: IR = AI
And by div . or comp . EO: AO: : AO: AI
And AO2 = EO x AI
Therefore AO2 (= EO x AI): EO x AU: : AI: AU: : R: S
Q. E. D.
This Problem admits of two Caſes. The 1ſt determinate or limited, the 2d
unlimited.
Case I. Is when OE the co-efficient of the given external line AU is part of
AO the ſide of the required ſquare [fig. 1. 2. ] and here the Llmitation is,
that AI muſt not be given leſs than four times AE, as appears from fig. 2.
for AE: AO: : OI: AI; and here OI being the half of AI, AE will be
the half of AO, or the fourth part of AI. In this Caſe the Homotactical Con-
ſtruction is uſed.
Case II. Is when AO the ſide of the required ſquare is part of OE the co-
efficient of the given external line AU, [fig. 3. ] and this is unlimited, for here
the Anlitactical Conſtruction is uſed. Or if O be required between A and E,
this is effected by the ſame Conſtruction.
LEMMA I.
If from the extremes of any diameter perpendiculars be let fall upon any
Chord, I ſay that the ſegments of theſe perpendiculars intercepted by like Arcs
are equal, and moreover alſo the ſegments of the Chords themſelves.
That YO is equal to IU may be thus ſhewn. Having joined YI, the
angle IYE is a right one, being in a ſemicircle, and the angle at O is right by
hypotbeſis; hence YI is parallel to the Chord, and YOUI is a parallelogram,
and the oppoſite ſides YO and IU will be equal. In the ſame manner OE is
proved equal to UL. And as to the ſegments of the Chord, it is thus ſhewn.
By Euc. III. 35. and 36, the rect. EOY = rect. SOR, and rect. LUI = rect.
SUR. But, by what has been juſt proved, rect. EOY = rect. LUI; hence
rect. SOR = rect. SUR, and the ſegments SO and OR are reſpectively equal
to the ſegments UR and SU.
80[3]
LEMMA II.
If of four proportionals the ſum of two, being either extremes or means, be
greater than the ſum of the other two; then I ſay theſe will be greateſt and
leaſt of all.
This is the converſe of Euc. V. 25. and may be thus demonſtrated. Draw
a circle whoſe diameter may be equal to the greater ſum; and in it inſcribe the
leſſer ſum IO, which will therefore not paſs through the center, and let the
parts be IU and UO; then through U draw a diameter AUE, and the other
two terms will be AU and EU, of which AU is greateſt of all and EU leaſt of
all, and IU and UO of intermediate magnitude, by Euc. III. 7.
LEMMA III.
If of four proportionals the difference of two, being either extremes or
means, be greater than the difference of the other two, then I ſay theſe will be
the greateſt and leaſt of all.
This is demonſtrated in the ſame manner as the preceding by Euc. III. 8.
PROBLEM II.
To cut a given indefinite right line in one point, ſo that, of the three ſeg-
ments intercepted between the ſaid point and three points given in the ſame in-
definite right line, the rectangle under one of them and a given external right
line may be to the rectangle under the other two in a given ratio.
In the given indefinite line let the aſſigned points be A, E, I. It is then
required to cut it again in the point O, ſo that AO into a given external line
R may be to EO x IO as R to S. If A be an extreme point and E the
middle one, then ſet off IU = AE the contrary way from A; but if A be the
middle point, then ſet it off towards A. Then from U ſet off UN = S the
conſequent of the given ratio, either towards A, or the contrary way; for as
the Caſes vary, it’s poſition will vary. From A and N erect perpendiculars
AY and NM to the given indefinite right line equal to AE and AI re-
ſpectively, and theſe bomotactical if A be an extreme point, but antitactical if
A be the middle point of the three given ones. Join the extremes of theſe
perpendiculars Y and M, and upon YM as a Diameter deſcribe a circle. I ſay
that the interſection of this circle with the given indefinite line ſolves the
Problem. If it interſects the line in two points, then the Problem admits
81[4] two ſolutions; if it only touches, then but of one; if it neither cuts nor
touches, it is then impoſſible.
Demonstration. Let the point of interſection then be O or o. We ſhall
have, by Lemma I. AO x ON = MN x NK = MN x AY = AI x AE, by
conſtruction. Let now from N be ſet off NL = AI in the ſame direction as A
is from I; then by what has been demonſtrated NL: ON: : AO: AE.
And by Diviſion or Compoſition OL: ON: : OE: AE
And by Permutation OL: OE: : ON: AE
But by what has been proved ON: AE: : AI: AO
Therefore by Equality OL: OE: : AI: AO
And by Diviſion or Compoſition LE: OE: : OI: AO
And LE x AO = OE x OI
But LE = NU. for NL was put = AI, and IU = AE
Hence NU x AO = S x AO = OE x OI
And R: S: : R x AO: S x AO or OE x OI
Q. E. D.
This Problem may be conſidered as having two Epitacmas, the firſt,
when the ſegment aſſigned for the coefficient of the given external line R is
terminated by an extreme point of the three given ones and the point ſought;
and this again admits of three Caſes. The other is when the aforeſaid ſeg-
ment is terminated by the middle point of the three given ones and the
point ſought.
Epitagma I. Case I. Let the aſſigned points be A, E, I. A an extreme
and E the middle one. And let the point O ſought (ſuch that AO x R: OE
x OI: : R: S) be required to lie between A and E, or elſe beyond I, which
will ariſe from the ſame conſtruction.
Here the Homotactical conſtruction is uſed, and IU as likewiſe UN is ſet off
in the ſame direction as AI. And ſince AO: AE: : AI: ON, and AO + ON
is greater than AE + AI or AU, by Lemma II. AO and ON will be the
leaſt and greateſt of all; and AO will therefore be leſs than AE, as likewiſe
Ao (being equal ON by Lemma I.) greater than AI. This Caſe is
unlimited.
Case II. Let the aſſigned points be in the ſame poſition as before, and let
the point O ſought be required between E and I.
Here the conſtruction is Homotactical, and UN is ſet off the contraty way, viz.
in the direction IA. And ſince AO: AE: : AI: ON, and AO + ON is leſs
than AE + AI or AU, by Lemma II. AE and AI will be the leaſt
82[5] greateſt of all, and AE will therefore be leſs than AO, and AI greater. And
the ſame will hold with regard to Ao.
Here is a Limitation, which is this; that UN or S the conſequent of the
given ratio, ſet off from R, muſt not be given greater than the difference of
the ſum of AE and AI and of a line whoſe ſquare is equal to four times their
rectangle [i. e. to expreſs it in the modern manner, UN muſt not exceed AI +
AE - √4 AI x AE*. ] This appears by Fig. 2. to this Caſe, the circle there
touching the given indefinite line, and pointing out the Limit.
Case III. Let the aſſigned points be ſtill in the ſame poſition, and let the
point ſought be now required on the contrary ſide of A.
Here the conſtruction is ſtill Homotactical, and UN is ſet off the ſame way as
in the laſt Caſe; and the Limitation is, that UN muſt not be given leſs than
the ſum of AI, AE, and a line whoſe ſquare is equal to four times their rect-
angle [or expreſſing it Algebraically, UN muſt not be leſs than AI + AE +
√4 AI x AE*. ]
Epitagma II. Case IV. Let now A be the middle point of the given
ones, and let O the point ſought be required either between A and one of the
extremes, or beyond either of the extremes.
Here having ſet off IU = AE toward A, you may ſet off UN either way,
and uſing the Antitactical conſtruction, the ſolution will be unlimited. The
only difference is, that if UN be in the direction UI, two ſolutions will ariſe,
whereof in one the point O will fall between A and E, and in the other be-
yond I; but if UN be in the direction IU, two ſolutions will ariſe, whereof
in one the point will fall between A and I, and in the other beyond E. In
proof of which Lemma III. is to be uſed, as Lemma II. was in Caſe I. II.
Corollary I. If then the given ratio be that of AT to TI, or of AE to
EP ſet off from A the other way, ſo that EP be leſs than AE, I ſay then
that O will fall between E and P, as likewiſe ο between T and I, provided o
falls beyond I.
For by conſtruction IU = AE, and UN = PE. therefore IN = AP. But by
Lemma I. oN = AO. therefore (o falling beyond I by hypotbeſis) O will fall
beyond P; but by hypotbeſis it falls ſhort of E; therefore O falls between
P and E.
Next to ſhew that ο will fall between T and I, we have AT: TI: : AE: EP
And by Diviſion AT: AI: : AE: AP
Hence AT x AP = IAE or o AO
83[6]
Therefore AT x AO is greater than o AO
Or AT greater than Ao.
Corollary II. If the three given points be I, A, E; and O falls between
A and I, ſo as to make AO x PE: IOE: : AL: LI, I ſay then O will fall
beyond L.
For let us ſuppoſe that O and L coincide; then by hypotbeſis AL: LI: :
AL x PE: IL x LE
And by the next following Lemma IV. AL x IL: IL x PE: : AL: LE
i. e. AL: PE: : AL: LE
Hence PE is equal to LE, a part to the whole, which is manifeſtly abſurd.
LEMMA IV.
If it be as a line to a line ſo a rectangle to a rectangle; then I ſay it will be
as the flrſt line into the breadth of the ſecond rectangle to the ſecond line into
the breadth of the firſt rectangle, ſo the length of the firſt rectangle to the
length of the ſecond.
Suppoſition. AE: IO: : UYN: SRL.
Concluſion. AE x RL: IO x YN: : UY: SR.
Dem. AE: IO: : AE x YN: IO x YN: : UYN: SRL
And by Permutation AE x YN: UYN: : AE: UY: : IO x YN: SRL
But SR: AE: : SRL: : AE x RL
Therefore ex æquo perturbatè SR: UY: : IO x YN: AE x RL
Q. E. D.
LEMMA V.
If a right line be cut in two points, I fay the rectangle under the alternate
ſegments is equal to that under the whole and the middle ſegment, together
with the rectangle under the extremes.
Dem. AI x IE + IO x IE = AO x IE.
Hence {AI x IE + IO x IE + AE x IO \\ i. e. AI x IE + AI x IO \\ i. e. AI x EO} = AO x IE + AE x IO.
Q. E. D.
N. B. Theſe two Lemmas ſave much Circumlocution and Tautology in
the two following Propoſitions, and indeed are highly uſeful in all caſes where
compound ratios are concerned.
84[7]
PROBLEM III.
To cut a given indefinite right line in one point, ſo that of the three ſeg-
ments intercepted between the ſame, and three points given, the rectangle
under two of them may be to the ſquare of the remaining one in a given ratio.
In the indefinite line let the three points be A, E, I. it is then required to be
cut again in O, ſo that OA x OE may be to OI2 (let the ſituation of I be
what it may) in a given ratio, which ratio let be expreſſed by EL to LI.
[And here I cannot but obſerve with Hugo D'Omerique, page 113. that
this Problem, viz. ‘To exhibit two lines in a given ratio whoſe ſum, or whoſe
difference is given,’ ought to have had a place in the Elements as a Propoſition;
or at leaſt to have been annext as a Scholium to the 9th or 10th of the VIth
Book. ] And be the ſituation of L alſo what it may, either between A and E,
or between A and I, or between E and I, or beyond either extreme. To the three
points E, L, I, and the right line AI, let be found, by Problem II, a fourth
point O ſuch, that AI x OE: OI x OL: : EI: IL. And let ſuch a Caſe be
choſen of Problem II, that, according as AO is greater or leſs than AI, ſo of
the three rectangles, deſcribed in Lemma V, made by the four points E, O,
I, L, that of IO x EL may accordingly be greater or leſs than that of EI x OL.
DEMONSTRATION . On ſuppoſition then that ſuch a Caſe of Problem II. is
made uſe of, we have
AI x OE: OI x OL: : EI: IL
And by Lemma IV, OL x EI: OE x IL: : AI: OI
And by Diviſion or Compoſition EL x OI: OE x IL: : AO: OI
This appears from Lemma V.
Then again by Lemma IV, AO x OE: OI2: : EL: IL.
Q. E. D.
This Problem has two Epitagmas. The firſt wherein OI, whoſe ſquare is
ſought, is bounded by I an extreme point of the three given ones. And this
again admits of three Caſes. The ſecond is when the point I is the middle
point. And this again has three caſes. And there remain two Anomalous
Caſes, wherein Problem II. is of no uſe, which muſt therefore be conſtructed
by themſelves.
Epitagma I. Case I. Let the ratio given, EL to LI, be inequalitatis
majoris, i. e. of a greater to a leſs; and the point O ſought be required to lie
between I and the next point to it E, or elſe to lie beyond I the other way;
for the ſame conſtruction ſerves for both. Here Case I. of Problem II. is
85[8] be uſed, and the point O will fall between E and I, and the point o beyond
L, much more beyond I.
Case II. Let the given ratio, EL to LI, be inæqualitatis minoris, i. e. of a
leſs to a greater, and the point O ſought be required to lie between I and the
next point to it E; or elſe to fall beyond A the other extreme. For the ſame
conſtruction ſerves for both. Here Case IV. of Problem II. is to be uſed, and
the point O will fall between E and I, and o beyond A, if we uſe one of the
conſtructions there recited: but if we uſe the other, the points will ſhift places,
as was obſerved under that Caſe, viz. O will fall beyond I the other way, and
o between L and E.
Case III. Let now the point O be ſought between A and E. Here ſet off
the given ratio in ſuch a manner that EI may be the ſum of the terms, and
make uſe of the IIId Case of Problem II. and the Limitation here will
be evident from the Limitation there given, viz. making EI: IL: : AI: X,
the Limitation here is that X muſt not be leſs than IE + EL + √4 IEL*.
Epitagma II. Case IV. Here OI the line whoſe ſquare is concerned is
to be bounded by I the middle point of the three given ones, and O or o, its
other bound is to be ſought between I and either extreine A or E. the ſame
conſtruction ſerving for both. The given ratio muſt here be ſet off in ſuch a
manner that EI may be the ſum of the terms of it; and make uſe
of Iſt Case of the IId Problem; with this caution, that of the two ſegments
AI, IE, you choſe the leſſer IE whereon to exhibit the given ratio; for then
it will appear by the work itſelf that O falling between E and L, o will alſo
fall between A and I: otherwiſe, if AI was leſs than IE, there would want
ſome proof of this. Therefore of the two extreme given points call that E
which bounds the leſſer ſegment, and then the general Demonſtration will fit
this Caſe as well as the reſt.
Case V. Let the given ratio of EL to LI be inæqualitatis minoris; and let
the point ſought be required to lie beyond either extreme. The ſame con-
ſtruction ſerves for both. Here we muſt uſe the IVth Case of the IId Pro-
BLEM, and O being made to fall between E and L, o will fall always beyond
A, provided we call that point E which bounds the bigger ſegment. I have
in the Figure made AI = IE on purpoſe to ſhew that in that caſe the point N
will coincide with A. But if IE be greater than AI, the point N will
always fall beyond A, and conſequently the point o more ſo.
86[9]
Case VI. Let the given ratio of EL to LI be inæqualitatis majoris, and
let the point ſought be required to lie beyond either extreme. Here we muſt
uſe the IIId Case of the IId Problem; and the Determination is that
UN (found in the ſame ratio to AI as IL is to IE) muſt not be leſs than
IE + EL + √4 IEL*.
Case VII. Let the ſituation of O be required the ſame as in the two laſt
Caſes, but let the given ratio be that of equality, which was there ſuppoſed
of inequality. Here the IId Problem will be of no uſe, and this Caſe
requires a particular conſtruction.
Let then the three Points be A, I, E, and I the middle one; and let it
be required to find a fourth O beyond E, ſuch that AO x OE may equal
OI2.
Construction. Upon AE diameter deſcribe a circle,, and let another YS
cut the former at right angles. Join SI, and continue it to meet the circum-
ference in R. From R draw a tangent to meet the given line in O, and I
ſay O is the point required.
Demonstration. Joining YR, the triangles SUI and SYR will be ſimi-
lar, and the angle UIS or RIO = SYR. But the angle IRO made by the
tangent and ſecant = SYR in the alternate ſegment. Therefore RIO = IRO,
and OR = OI. But by the property of the circle AOE = OR2. And
therefore AOE = OI2.
Q. E. D.
The Determination is that AI muſt be greater than IE.
Case VIII. Whereas in the Iſt and IId Cases the given ratio was that of
inequality, let us now ſuppoſe it that of equality; and let the three points be
A, E, I, and E the middle one; and let a fourth O be ſought between E and
I, ſuch that AOE may equal OI2.
The Construction and Demonstration of this Caſe is in every reſpect
the ſame as that of the preceeding, as will appear by comparing the figures.
PROBLEM IV.
To cut a given indefinite right line ſo in one point that, of the four ſeg-
ments intercepted between the ſame and four points given in the indefinite
line, the rectangle under any two aſſigned ones may be to the rectangle under
the two remaining ones in a given ratio.
87[10]
In the indefinite line let the four points be A, E, I, U. It is then required
to be cut again in O ſo that OA x OU may be to OE x OI (be the Poſition
of the four given points what they may) in the ratio of AL to LE, let the
point L fall alſo as it may.
Construction. To the three points E, L, U, and the right line UI, let
be found by the IId Problem a fourth point O, ſo that UI x OE may be to
OU x OL as AE to AL. And let ſuch a Case be choſen of the IId Problem
that, according as UO is required greater or leſs than UI, or their ſum ſhall
conſtitute OI, ſo of the three rectangles deſcribed in the Vth Lemma made
by the four points E, O, A, L, that of OE x AL may accordingly be greater
or leſs than OL x AE, or their ſum conſtitute that of OA x EL.
N. B. U being now uſed to repreſent one of the given points, in all the
following Diagrams I have ſubſtituted V in the place where U was uſed
before.
Demonstration. On ſuppoſition therefore that ſuch a Case of the
IId Problem is made uſe of,
We have UI x OE: OU x OL: : AE: AL
And by Inverſion OU x OL: UI x OE: : AL: AE
And by Lemma IV. AL x OE: AE x OL: : OU: UI
Hence by Compoſition or Diviſion, & c. AL x OE: OA x LE: : OU
: OI as appears by Lemma V.
Then again by Lemma IV. OU x OA: OI x OE: : AL: LE
Q. E. D.
This Problem has three Epitagmas. The Iſt whereof is when of the
two aſſigned points A and U, the one of them is an extreme, and the other an
alternate mean; and this admits of three Cases. The IId is when A and U
are both of them extremes; and this has four Cases. The IIId is when of
A and U one of them is an extreme, and the other is the point next to it; and
this has three Caſes. And there remain three more Anomalous Caſes, wherein
the IId Problem is of no uſe, but which may be reduced to one, as ſhall be
ſhewn in it's proper place.
Epitagma I. Case I. Let A the firſt aſſigned point be an extreme, and
U the ſecond aſſigned point be an alternate mean; and let the point O be
ſought between the firſt aſſigned A and the next point to it E; or between
the ſecond aſſigned U and the laſt I. For the ſame Conſtruction ſerves
for both.
88[11]
Here AE is to be made the ſum of the terms of the given ratio, and we are
to uſe the IVth Case of the IId Problem, whereby O falling between
L and E, o will fall beyond U; and that it will fall ſhort of I appears from
the Iſt Corollary from the IVth Case of the IId Problem.
Case II. The given ratio being inæqualitatis minoris, let the point ſought
be required between the ſecond aſſigned U and the ſecond in order E, or be-
yond the firſt A, which ariſes from the ſame Conſtruction. Here AE is to be
made the difference of the terms of the given ratio, and we are to uſe the
IVth Case of the IId Problem, where O being made to fall between U and E,
o will fall beyond L, much more beyond A.
Case III. The given ratio being inæqualitatis majoris, let the point ſought
be required between the ſecond aſſigned U and the ſecond in order E, or be-
yond the laſt I, which ariſes from the ſame Conſtruction. Here AE is to be
made the difference of the terms of the given ratio, and L is to be ſet off the
contrary way to what it was in the laſt Case; and we are to uſe the Iſt Case
of the IId Problem, whereby O being made to fall between E and L, or
between E and U, according as L or U is neareſt to the point E, o will fall
beyond I, as any one will ſee who conſiders the Conſtruction of that Case with
due attention.
Epitagma II. Case IV. Let the aſſigned points now be the extremes
A and U, and let O the point ſought be required now between the firſt
aſſigned A and the next to it E, or, which is effected by the ſame Conſtruction,
between the ſecond aſſigned U and the next to it I. Here AE is to be made
the ſum of the terms of the given ratio, and the IVth Case of the IId Pro-
BLEM is to be uſed, ſo that of the three points L, E, U, O being made to fall
beyond L one of the extremes, and o within U the other extreme, it will appear
from the Iſt Corollary from the IVth Case of the ſaid Problem that O
will fall between A and E, and o between U and I.
Case V. The given ratio being inæqualitatis majoris, let the point ſought be
required between the ſecond and third in order, viz. between E and I. Here
AE muſt be the difference of the terms of the given ratio, and L ſet off to-
wards I, and the IId Case of the IId Problem uſed, and then O, as like-
wiſe o, will fall between E and I, if the Problem be poſſible.
As to the Determination, ſee Lemma VII. following.
Case VI. The given ratio being inæqualitatis majoris, let the point ſought
be required beyond the laſt aſſigned, that is the laſt in order, U. Here AE
muſt be the difference of the terms of the given ratio, [and L muſt
89[12] fall beyond U, but for a more particular Determination ſee Lemma VII.
following] and the IId Case of the IId Problem is to be uſed, and then O,
as likewiſe o, will fall beyond U.
Case VII. The given ratio being inæqualitatis minoris, let the point ſought
be required to lie beyond either of the aſſigned ones, i. e. beyond either ex-
treme, the ſame Conſtruction ſerving for both. Here AE is to be the diſ-
ference of the terms of the given ratio, and L to be ſet off backwards beyond
A; and the IVth Case of the IId Problem uſed, that ſo O being made to
fall beyond U, it will appear, by the IId Corollary from the IVth Case of
the ſaid Problem, that o will alſo fall beyond A.
Epitagma III. Case VIII. Let the aſſigned points A and U be now
one an extreme, and the other the point next it: and let the point ſought be re-
quired to fall between the two aſſigned ones. Here AE muſt be the ſum of
the terms of the given ratio, and the IId Case of the IId Problem uſed.
And ſo O, as likewiſe o, being made to fall between L and U, they will
much more fall between A and U.
The Limitation is, that VN (found in the ſame ratio to UI as AL to AE)
muſt not exceed LE + EU - √4 LEU*.
Case IX. The given ratio being inæqualitatis minoris, let the point ſought
be required between the ſecond aſſigned U and the third in order E, or elſe
beyond the firſt aſſigned A, the ſame Conſtruction ſerving for both. Here AE
is to be the difference of the terms of the given ratio, and L to be ſet off be-
yond A, and the Iſt Case of the IId Problem uſed: and ſo O being made
to fall between E and U, o will fall beyond L, and much more be-
yond A.
Case X. The given ratio being inæqualitatis majoris, let the point ſought
be required between the ſecond aſſigned U and the third in order E, or elſe
beyond the laſt in order I, the ſame conſtruction ſerving for both. Here AE
is to be the difference of the terms of the given ratio, and L to be ſet off be-
yond E, and the IVth Case of the IId Problem uſed, ſo that O being
made to fall between U and E, o will fall beyond I, as any one will ſee who
conſiders the conſtruction of that Case with due attention.
Case XI. As to the three Anomalous Caſes, in which the IId Problem is
of no uſe, and which I ſaid before might be reduced to one, they are theſe:
whereas in the IId and IIId Cases, as alſo in the VIth and VIIth, and like-
wiſe in the IXth and Xth the ratio given was that of inequality, let us
90[13] ſuppoſe it that of equality, and they may all be ſolved by one conſtruction, viz.
the rectangle AOU made equal to the rectangle EOI.
Let it be made as UI: AE: : UO: EO
Then by permutation UO: UI: : EO: AE
And by comp. or diviſ. UO: OI: : EO: AO
Hence AOU = EOI.
Lemma VI. Let there be two ſimilar triangles IAE, UAO, having their
baſes IE and UO parallel; I ſay Iſt when they are right-angled, that the ex-
ceſs of the rectangle EAO, under the greater ſides of each, above the rect-
angle IAU, under the leſſer ſides of each, will be equal to the rectangle
IE x OU, under their baſes. IIdly, When they are obtuſe-angled, that the
ſaid exceſs will be equal to the rectangle under the baſe of one and the ſum
of the diſtances of the angles at the baſe of the other from the perpendicular,
viz. EI x OS + US. IIIdly, When they are acute-angled, that then the ſaid
exceſs will be equal to the rectangle under the baſe of one and the difference
of the ſegments of the baſe of the other made by the perpendicular, viz.
OU x EL.
Demonstration. Since EA: AO: : IA: AU: : EI: OU, the rect-
angles EAO, IAU, and EI x OU will be ſimilar, and when Iſt the triangles
are right-angled EAO = IAU + EI x OU by Euc. VI. 19. and I. 47. But if
they be oblique-angled, draw the perpendicular YAS. Then IIdly, in caſe
they be obtuſe-angled, EAO = YAS + EY x OS by part Iſt; and IAU =
YAS + IY x US by the ſame. And therefore EAO - IAU = EY x OS -
IY x US = EY - IY or EI x OS + US. But if IIIdly they be acute-angled,
and EY be greater than IY, then from Y ſet off YL = YI, and draw LAR
which will be equal and ſimilarly divided to IAU. Then by part IId EAO
- LAR, i. e. EAO - IAU = EL x OS + RS = EL x OU.
Q. E. D.
Lemma VII. If a right line VY, joining the tops of two perpendiculars
drawn from two points of the diameter of a circle E and I to the circum-
ference on oppoſite ſides of the diameter, cut the ſaid diameter in O, and
A and U be the extremes of the ſaid diameter, I ſay that the ratio of the
rectangle AOU to the rectangle EOI is a Minimum.
But if VY joins the tops of two perpendiculars from E and I drawn on
the ſame ſide of the diameter, and conſequently meets the diameter produced
in O, that then the ratio of AOU to EOI is a Maximum.
91[14]
Demonstration. Through S, any other point taken at pleaſure, draw
LSM parallel to VOY, and join VS and produce it to meet the perpendi-
cular in N and the circumference in R. Produce alfo the perpendicular YI
to meet the circumference again in F, and join RF; then from ſrmilar tri-
angles it appears that the rectangle LSM: ESI: : VOY i. e. AOU: EOI.
But the rectangle ASU i. e. VSR is greater than LSM. (for LV i. e. MY x
NI + MI together with VS x SN is by the preceeding Lemma equal to LSM.
But ASU or VSR is equal to VS x SN, together with VS x NR; for which
laſt rectangle we may ſubſtitute MY x NF: for the triangles VLS and NRF
are ſimilar, being each of them ſimilar to VNY; therefore VL or MY: VS
: : NR: NF and MY x NF = VS x NR. Now NF being always greater
than NI + MI, it appears from thence that ASU is greater than LSM.)
Therefore the ratio of ASU to ESI is greater than that of AOU to EOI.
And the ſame holds good with regard to any other point S taken between E
and I, ſo that the ratio of AOU to EOI is a Minimum, and ſingular, or
what the Antients called μοναχ@.
Let now VY join the tops of two perpendiculars drawn on the ſame ſide of
the diameter, and meet the diameter produced in O; I ſay that the ratio of
AOU to EOI is a Maximum.
For uſing the ſame conſtruction as before, it will appear that the rectangle
LSM: ESI: : VOY or AOU: EOI. And it may be proved in the ſame
manner that VSR or ASU is leſs than LSM. (LV i. e. MY x NI + MI or
(making IK = MI) MY x NK, together with LSM is by the preceding Lem-
MA equal to VS x SN. But VS x NR, together with VSR, is alſo equal to
VS x SN. Now VS x NR is equal to MY x NF or LV x NF from the ſimi-
larity of the triangles LVS, NRF. Therefore now alſo MY x NF together
with VSR is proved equal to VS x SN. But as NK is leſs than NF, VSR
will be a leſs rectangle than LSM) Hence the ratio of LSM to ESI or it's
equal AOU to EOI is greater than the ratio of VSR or ASU to ESI. And
the ſame holds with regard to any other point taken in the diameter pro-
duced. Therefore the ratio of AOU to EOI is a Maximum.
Q. E. D.
92[15]
DETERMINATE SECTION.
BOOK I.
PROBLEM I. (Fig. 1.)
In any indefinite ſtraight line, let the Point A be aſſigned; it is required
to cut it in ſome other point O, ſo that the ſquare on the ſegment AO
may be to the ſquare on a given line, P, in the ratio of two given ſtraight
lines R and S.
Analysis. Since, by Hypotheſis, the ſquare on AO muſt be to the
ſquare on P as R is to S, the ſquare on AO will be to the Square on P as
the ſquare on R is to the rectangle contained by R and S (Eu. V. 15.)
Let there be taken AD, a mean proportional between AB (R) and AC
(S); then the Square on AO is to the ſquare on P as the ſquare on R is
to the ſquare on AD, or (Eu. VI. 22) AO is to P as R to AD; conſe-
quently, AO is given by Eu. VI. 12.
Synthesis. Make AB equal to R, AC equal to S, and deſcribe on
BC a ſemi-circle; erect at A the indefinite perpendicular AF, meeting the
circle in D, and take AF equal to P; draw DB, and parallel thereto FO,
meeting the indefinite line in O, the point required.
For, by reaſon of the ſimilar triangles ADB, AFO, AO is to AF (P) as
AB (R) is to AD; therefore (Eu. VI. 22.) the ſquare on AO is to the
ſquare on P as the ſquare on R is to the ſquare on AD; but the ſquare on
AD is equal to the rectangle contained by AB (R) and AC (S) by Eu. VI.
13. 17; and ſo the ſquare on AO is to the ſquare on P as the ſquare on R
is to the rectangle contained by R and S; that is (Eu. V. 15.) as R is to S.
Q. E. D.
Scholium. Here are no limitations, nor any precautions whatever to be
obſerved, except that AB (R) muſt be ſet off from A that way which O
is required to fall.
93[16]
PROBLEM II. (Fig. 2 and 3.)
In any indefinite ſtraight line, let there be aſſigned the points A and E;
it is required to cut it in another point O, ſo that the rectangle contained
by the ſegments AO, EO may be to the ſquare on a given line P, in the
ratio of two given ſtraight lines R and S.
Analysis. Conceive the thing done, and O the point ſought: then
would the rectangle AO, EO be to the ſquare on the given line P as R is
to S, by hypotheſis. Make AQ to P as R is to S: then the rectangle AO,
EO will be to the ſquare on P as AQ to P; or (Eu. V. 15. and 16) the
rectangle AO, EO will be to the rectangle AQ, EO as the ſquare on P is
to the rectangle EO, P; and therefore AO is to AQ as P is to EO; con-
ſequently (Eu. VI. 16.) the rectangle AO, EO is equal to the rectangle
AQ, P; and hence, as the ſum, or difference of AO and EO is alſo given,
theſe lines themſelves are given by the 85th or 86th of the Data.
Synthesis. On AE deſcribe a circle, and erect at A the indefinite
perpendicular AK; and, having taken AQ a fourth proportional to S, R
and P, take AD a mean proportional between AQ and P; from D draw
DH, parallel to AE if O be required to fall between A and E; but through
F, the center of the Circle on AE, if it be required beyond A or E, cutting
the circle in H; laſtly, draw HO perpendicular to DH, meeting the inde-
finite line in O, the point required.
For it is manifeſt from the conſtruction that AD and HO are equal;
hence, and Eu. VI. 17, the rectangle AQ, P is equal to the ſquare on HO;
conſequently equal to the rectangle AO, EO (Eu. III. 35. 36) and ſo
(Eu. VI. 16.) AO is to P as AQ is to EO; but by conſtruction AQ is to
P as R to S, therefore by compound ratio, the rectangle AO, AQ is to the
ſquare on P as the rectangle AQ, R is to the rectangle EO, S: hence
(Eu. V. 15. 16.) the rectangle AO, EO is to the ſquare on P as the rect-
angle EO, R is to the rectangle EO, S, that is, as R is to S. Q. E. D.
Scholium. This Problem has three Epitagmas. Firſt, when O is ſought
beyond A; ſecondly, when it is ſought between A and E, and laſtly, when
it is ſought beyond E. The firſt and laſt of theſe are conſtructed by
Fig. 2, and have no limitations; but in the ſecond, (Fig. 3.) the given
ratio of R to S muſt not be greater than that which the ſquare on half
AE bears to the ſquare on P: ſince if it be, a third proportional to
94[17] and P will be greater than one to R and half AE, and of courſe, AQ
(a fourth proportional to S, R and P) greater than a third proportional
to P and half AE; in which caſe the rectangle AQ, P will be greater
than the ſquare on half AE, and ſo AD (a mean proportional between
AQ and P) greater than half AE; but when this happens, it is plain that
DH can neither cut nor touch the circle on AE, and therefore, the
problem becomes impoſſible.
PROBLEM III. (Fig. 4. and 5.)
In any indefinite ſtraight line let there be aſſigned the points A and E;
it is required to cut it in another point O, ſo that the ſquare on the ſegment
AO may be to the rectangle contained by the ſegment EO and a given line
P, in the ratio of two given ſtraight lines R and S.
Analysis. Suppoſe the thing done, and that O is the point ſought:
then will the ſquare on AO be to the rectangle EO, P as R to S. Make
AQ to P as R is to S; then will the ſquare on AO be to the rectangle EO,
P as AQ is to P; or (Eu. V. 15.) the ſquare on AO is to the rectangle EO,
P as the rectangle AQ, AO is to the rectangle P, AO; wherefore AO is to
EO as AQ to AO; conſequently by compoſition, or diviſion, AO is to AE
as AQ is to OQ, and ſo (Eu. VI. 16.) the rectangle AO, OQ is equal to
the rectangle AE, AQ; and hence, as the ſum or difference of AO and OQ
is alſo given, theſe lines themſelves are given by the 85th or 86th of the
Data.
Synthesis. Take AQ a fourth proportional to S, P and R, and
deſcribe thereon a circle; erect at A, the indefinite perpendicular AK, and
take therein AD, a mean proportional between AE and AQ; from D,
draw DH, parallel to AE, if O be required beyond E; but through F the
center of the circle on AQ, if it be ſought beyond A, or between A and
E, cutting the ſaid circle in H: Laſtly, from H draw HO perpendicular
to DH, which will cut the indefinite line in O, the point required.
For it is plain from the Conſtruction, that AD and HO are equal; and
(Eu. VI. 17) the rectangle AE, AQ is equal to the ſquare on AD, and
therefore equal to the ſquare on HO; but the ſquare on HO is equal to the
rectangle AO, OQ, (Eu. III. 35. 36) conſequently the rectangle AO,
95[18] is equal to the rectangle AE, AQ; and hence (Eu. VI. 16.) OQ is to AQ
as AE is to AO; therefore, by compoſition or diviſion, AO is to AQ as
EO is to AO; but by conſtruction, AQ is to R as P is to S, and ſo, by
compound ratio, the rectangle AO, AQ is to the rectangle R, AQ as the
rectangle EO, P is to the rectangle AO, S; or (Eu. V. 15. and 16.) the ſquare
on AO is to the rectangle EO, P as the rectangle AO, R is to the rectangle
AO, S; that is, as R to S.
Scholium. This Problem hath three Epitagmas alſo, which I ſtill enu-
merate as before. The firſt and ſecond are conſtructed by Fig. 4, where DH
is drawn through F, the center of the circle on AQ: and theſe have no limi-
tations. The third is conſtructed as in Fig. 5, where DH is drawn parallel
to AQ; and here the given ratio of R to S muſt not be leſs than the ratio
which four times AE bears to P: for if it be, AE will be greater than
one-fourth Part of AQ (a fourth proportional to S, R and P) in which
caſe the rectangle contained by AE and AQ will be greater than the ſquare
on half AQ, and conſequently AD (a mean proportional between AE and
AQ) greater than half AQ; but it is plain when this is the caſe, that DH
will neither cut nor touch the circle on AQ, and therefore the problem is
impoſſible.
PROBLEM IV. (Fig. 6. 7. and 8.)
In any indefinite ſtraight line, let there be aſſigned the points A and E;
it is required to cut it in another point O, ſo that the two ſquares on the
ſegments AO, EO, may obtain the Ratio of two given ſtraight lines,
R and S.
Analysis. Imagine the thing to be effected, and that O is really the
point required: then will the ſquare on AO be to the ſquare on EO as
R to S; or (Eu. V. 15.) the ſquare on AO will be to the ſquare on EO
as the ſquare on R is to the rectangle contained by R and S. Let DE be
made a mean proportional between EB (R) and EC (S). Then
(Eu. VI. 17.) the ſquare on AO will be to the ſquare on EO as the
ſquare on R to the ſquare on DE; and ſo (Eu. VI. 22.) AO to EO as R to
DE; and hence both AO and EO will be given by the converſe of Prop. 38.
of Eu. Data.
96[19]
Synthesis. Make EB equal to R, EC equal to S, and deſcribe on
BC a circle; erect at E the perpendicular ED, meeting the periphery of the
circle in D; alſo at A erect the perpendicular AF equal to R; draw AD,
which produce, if neceſſary, to cut the indefinite line, as in O, which will
be the point required.
For becauſe of the ſimilar triangles AOF, EOD, AO is to EO as AF
(R) is to DE; therefore the ſquare on AO is to the ſquare on EO as the
ſquare on R is to the ſquare on DE (Eu. VI. 22); but the ſquare on DE
is equal to the rectangle contained by R and S; therefore the ſquare on AO is
to the ſquare on EO as the ſquare on R is to the rectangle R, S; that is as
R is to S, by Eu. V. 15.
Q. E. D.
Scholium. This Problem alſo hath three Epitagmas, which I enumerate
as in the laſt. The firſt is conſtructed by Fig. 6, wherein the perpendicu-
lars DE and AF are ſet off on the ſame ſide of the indefinite line; the ſecond
by Fig. 7, where they are ſet off on contrary ſides, and the third by Fig. 8,
in which they are again ſet off on the ſame ſide. The ſecond has no limits;
but in the firſt R muſt be leſs, and in the third greater than S, for reaſons
too obvious to be inſiſted on; and hence, both theſe caſes are impoſſible
when the given ratio is that of equality.
PROBLEM V. (Fig. 9. 10. 11. 12. 13. 14. 15. 16.)
In any indefinite ſtraight line let there be aſſigned the points A, E and I;
it is required to cut it in another point, O, ſo that the rectangle contained
by the ſegment AO and a given ſtraight line P may be to the rectangle
contained by the ſegments EO, IO in the ratio of two given ſtraight lines
R and S.
Analysis. Conceive the thing done, and O the point ſought: then
would the rectangle AO, P be to the rectangle EO, IO as R to S. Make
IQ to P as S is to R; then will the rectangle AO, P be to the rectangle
EO, IO as P is to IQ; or (Eu. V. 15.) the rectangle AO, P be to the
rectangle EO, IO as the rectangle IO, P is to the rectangle IO, IQ; and
hence (Eu. V. 15. 16) AO is to EO as IO is to IQ; whence, by compoſition
or diviſion, AE is to EO as OQ is to IQ: therefore (Eu. VI. 16.)
97[20] rectangle EO, OQ is equal to the rectangle AE, IQ; conſequently, as the
ſum or difference of EO and OQ is alſo given, thoſe lines themſelves are
given by the 85th or 86th of the Data.
Synthesis. Take IQ a fourth proportional to R, S and P, and
deſcribe on EQ a circle; erect at E the indefinite perpendicular EK, and
take therein ED a mean proportional between AE and IQ; from D draw
DH, parallel to EQ, if O muſt lie any where between the points E and Q;
but through F, the center of the circle on EQ if it muſt fall without them,
cutting the ſaid circle in H: Laſtly, draw HO perpendicular to DH, which
will meet the indeſinite line in O, the point required.
For it is manifeſt from the conſtruction that ED and HO are equal; and
(Eu. VI. 17.) the rectangle AE, IQ is equal to the ſquare on ED, and
therefore equal to the ſquare on HO; but the ſquare on HO is equal to the
rectangle EO, OQ (Eu. III. 35. 36.) : therefore the rectangle AE, IQ is
equal to the rectangle EO, OQ; and hence (Eu. VI. 16.) AE is to OE as OQ
to IQ, whence, by compoſition or diviſion, AO is to EO as OI to IQ; but
IQ is to P as S to R, or inverſely, P is to IQ as R to S; and ſo, by compound
ratio, the rectangle AO, P is to the rectangle EO, IQ as the rectangle IO,
R is to the rectangle IQ, S; that is (Eu. V. 15 and 16.) the rectangle AO,
P is to the rectangle IO, R as EO is to S; or the rectangle AO, P is to the
rectangle IO, R as the rectangle IO, EO is to the rectangle IO, S
(Eu. V. 16.) the rectangle AO, P is to the rectangle EO, IO as the rectangle
IO, R is to the rectangle IO, S; that is (Eu. V. 15.) as R is to S. Q. E. D.
Scholium. This Problem may be conſidered as having three Epitagmas,
or general Caſes, viz. when A, the point which bounds the ſegment aſſigned
for the co efficient of the given line P being an extreme, O is ſought be-
tween it and the next thereto, or beyond all the points with reſpect to A;
ſecondly, where A is the middle point; and thirdly, when A being again
an extreme, O is ſought beyond it, or between the other two points E and
I: and each of theſe is ſubdiviſible into four more particular ones.
Epitagma I. Here the four Caſes are when E being the middle point,
O is required between A and E, or beyond I; and theſe are both con-
ſtructed at once by Fig. 9: when I is the middle point and O ſought between
A and I or beyond E; and theſe are both conſtructed at once by Fig. 10.
and in both of theſe IQ is ſet off from I contrary to that direction which
98[21] bears therefrom, and DH drawn through F, the center of the circle on EQ:
none of theſe Caſes are ſubject to any Limitations.
Epitagma II. Wherein A is the middle point, and the Caſes, when O
is ſought beyond E, between E and A, between A and I or beyond I. The
firſt and third of which are conſtructed at once by Fig. 11, wherein IQ is
ſet off from I towards A and DH drawn through F, the center of the circle
on EQ. The ſecond and fourth are conſtructed at once, alſo, by Fig. 12.
where IQ is ſet off from I the contrary way to that which A lies, and DH
drawn parallel to EQ. There are no Limitations to any of theſe Caſes.
Epitagma III. Here, E being the middle point, the Caſes are, when O
muſt lie beyond A, or between E and I; and the ſame Caſes occur when
I is made the middle point. The firſt is conſtructed by Fig. 13, the ſecond
by Fig. 14, the third by Fig. 15, and the fourth by Fig. 16: in every one
of which IQ is ſet off from I towards A, and DH drawn parallel to EQ.
The Limits are that the given ratio of R to S, muſt not be leſs than the ratio
which the rectangle AE, P bears to the ſquare on half the Sum, or half the
difference of AE, and a fourth propor tional to R, S and P; that is, to the
ſquare on half EQ: ſince if it ſhould, the rectangle contained by AE and
the ſaid fourth proportional will be greater than the ſquare on half EQ;
and of courſe ED (a mean proportional between them) greater than half
EQ, in which Caſe DH can neither cut nor touch the circle on EQ, and
ſo the problem be impoſſible. It is farther obſervable in the two laſt caſes,
that to have the former of them poſſible, AE muſt be leſs, and to have
the latter poſſible, EI muſt be greater than the above-mentioned half
ſum; for if this latter part of the Limitation be not obſerved, theſe caſes
are changed into one another.
PROBLEM VI.
(Fig. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.)
In any indefinite ſtraight line let there be aſſigned the points A, E and I;
it is required to cut it in another point O, ſo that the rectangle contained
by the ſegments AO, EO may be to the ſquare on IO in the ratio of two
given ſtraight lines, R and S.
99[22]
Analysis. Let us conceive the thing effected, and that O is really the
point ſought. Then, by ſuppoſition, the rectangle AO, EO is to the
ſquare on IO as R to S. Make EC to IC as R is to S; and the rectangle
AO, EO is to the ſquare on IO as EC to IC. Let now OB be taken a
fourth proportional to EO, EC and IO; then (Eu. V. 15.) the rectangle
AO, EO is to the ſquare on IO as the rectangle EC, OB is to the rectangle
IC, OB; and ſo by permutation, the rectangle AO, EO is to the rectangle
EC, OB as the ſquare on IO is to the rectangle IC, OB; and becauſe EO is
to EC as IO to BO, AO will be to OB as IO to IC, and ſo by compoſition,
or diviſion CO is to EC as IB to OB, and AB is to OB as CO to IC;
whence, ex æquo perturb. et permut. AB is to IB as EC to IC; that is in the
given ratio, and hence is given BC, the ſum or difference of CO and BO,
as alſo the rectangle contained by them, equal to the rectangle AB, IC,
wherefore theſe lines themſelves are given by the 85th or 86th of the Data.
Synthesis. Make AB to IB and EC to IC in the given ratio, and
deſcribe on BC a circle; erect, at B, the indefinite perpendicular BK, and
take therein BD a mean proportional between AB and IC, or between IB
and EC: from D draw DH parallel to CB, if O muſt fall between B and C;
but through F, the center of the circle on BC, if it muſt fall without them,
cutting the ſ@id circle in H; then draw HO perpendicular to DH, which
will cut the indefinite line in O, the point required.
For it is plain from the conſtruction that BD and HO are equal, and
(Eu. IV. 17.) the rectangle AB, IC, or the rectangle IB, EC is equal to the
ſquare on BD, and therefore equal to the ſquare on HO, which (Eu. III.
35. 36.) is equal to the rectangle BO, CO: conſequently (Eu. VI. 16.) AB
is to BO as CO to IC; alſo EC is to CO as BO is to IB; wherefore, by
compoſition or diviſion, AO is to BO as IO to IC, and EO to EC as IO
to BO: conſequently by compound ratio, the rectangle contained by AO and
EO is to the rectangle contained by BO and EC, as the ſquare on IO is to
the rectangle contained by BO and IC; by permutation, the rectangle
contained by AO and EO is to the ſquare on IO as the rectangle contained
by BO and EC is to the rectangle contained by BO and IC, that is (Euc.
v. 15.) as EC is to IC, or as R to S.
Q. E. D.
100[23]
Scholium. This Problem alſo hath three Epitagmas; ſirſt, when O is
ſought between I, the point which bounds the ſegment whoſe ſquare is
concerned, and either of the other given ones; ſecondly, the ſaid point
being an extreme one, when O is ſought beyond it, or beyond both the
other given points with reſpect to it; and thirdly, when O is required be-
yond the next in order to the abovementioned point I: theſe are each of
them ſubdiviſible into other more particular caſes.
Epitagma I. Here O is ſought between I, the point which bounds the
ſegment whoſe ſquare is concerned, and the next in order to it: and there
are four caſes, viz. when I is an extreme point; and the given ratio of R
to S the ratio of a greater to a leſs; when I, remaining as before, the given
ratio is of a leſs to a greater; again when I is the middle point, and O
ſought between it and either of the other given ones.
Case I. Here the points B and C are both made to fall beyond I
(Fig. 17.) and DH is drawn through the center of the circle on BC, and O
will fall between the point I, and the next in order thereto; becauſe by
conſtruction, EC is to CO as BO is to IB, and therefore when EC is greater
than CO, BO will be greater than IB, and when leſs, leſs; but it is plain
that if O ſhould fall either beyond E or I, this could not be the Caſe. It
is farther manifeſt that ſhould the points A and E change places, the con-
ſtruction would be no otherwiſe altered than that AB would then be greater
than IC.
Case II. If the given points retain their poſition, but the ratio be made
of a leſs to a greater, the conſtruction will then be by Fig. 18, where B
muſt be made to fall beyond A, and C beyond E with reſpect to I; but
DH is ſtill drawn through the center of the circle on BC: and that O will
fall as required may be made appear by reaſonings ſimilar to thoſe uſed in
Caſe I. Moreover no change will enſue in the conſtruction when the Points
A and E change places, except that B and C will change ſituations alſo.
Cases III and IV, Are conſtructed at once by Fig. 19, when B muſt
fall between A and I, C between I and E, and DH be drawn as before:
and it is here evident that the conſtruction will be the ſame let the given
ratio be what it will. None of thoſe Caſes admit of any Limitations.
Epitagma II. There are here only two Caſes, viz. when O is required
beyond I, the point which bounds the ſegment whoſe ſquare is concerned;
101[24] and ſecondly when it is ſought beyond both the other given points: in the
firſt, the ratio of R to S muſt be of a greater to a leſs, and in the latter of
a leſs to a greater. The former is conſtructed in Fig. 17. at the ſame time
with Caſe I of Epitagma I. and is repreſented by the ſmall letters h and o:
the latter in Fig. 18, and pointed out by the ſame letters. That O will
fall in both as is required needs not inſiſting on.
Epitagma III. In which there are ſix Caſes, viz. I being the middle
point, when O is ſought beyond A, or beyond E; and that whether the
given ratio be of a leſs to a greater, or of a greater to a leſs; and again, I
being an extreme point, when O is ſought between A and E, and that let
the order of the points A and E be what it will.
Cases I and II. Are when I is a mean point and the given ratio of a
leſs to a greater; and theſe are both conſtructed at once by Fig. 20,
wherein B is made to fall beyond A, and C beyond E with reſpect to the
middle point I, and DH is drawn through the center of the circle on BC.
Cases III and IV. Here, the points remaining as before, the given
ratio is of a greater to a leſs; and the conſtruction will be effected by
making B fall beyond E, and C beyond A, and drawing DH parallel to
BC, as in Fig. 21 and 22.
Case V. Wherein I is one extreme point and A the other, and O is
ſought between A and E: in conſtructing this Caſe, B muſt be made to
fall between A and I, C between E and I, and DH drawn parallel to BC,
as is done in Fig. 23. The directions for Conſtructing Caſe VI. are exactly
the ſame, as will appear by barely inſpecting Fig. 24.
Limitation. It is plain that in the four laſt Caſes, the ratio which the
rectangle contained by AO and EO bears to the ſquare on IO, or which is
the ſame thing, the given ratio of R to S cannot exceed a certain limit;
and it is farther obvious that the ſaid limit will be when the ſtraight line
DH becomes a tangent to the circle on BC, as in Fig. 25. 26, for after
that the problem is manifeſtly impoſſible. Now when DH is a tangent to
the circle on BC, HO will be equal to half BC; but the ſquare on HO
is equal to the rectangle contained by IB and EC, wherefore the ſquare on
half BC will then be equal to the rectangle contained by IB and EC.
Moreover, by the conſtruction, R is to S as AB is to IB, and as EC is to
IC; therefore by compoſition or diviſion, the ſum or difference of R and
S is to R as EI to EC, and the ſaid ſum or difference is to S as AI is
102[25] IB, as EI is to IC: and hence, by compound ratio, the ſquare on the
abovementioned ſum or difference is to the rectangle contained by R and S
as the rectangle contained by AI and EI is to the rectangle contained by
IB and EC, alſo by permutation, AI is to EI as IB is to IC; wherefore,
by compoſition or diviſion, AE is to AI as BC is to IB, by permutation,
AE is to BC as AI is to IB, therefore by equality, the ſum or difference
of R and S is to S as AE is to BC; or (Eu. V. 15.) as half AE is to
half BC; conſequently (Eu. VI. 22.) the ſquare on the above mentioned
ſum or diſſerence is to the ſquare on S as the ſquare on half AE is to the
ſquare on half BC, or to the rectangle contained by IB and EC. Hence
exæquo perturbaté, the rectangle contained by R and S is to the ſquare on
S as the ſquare on half AE is to the rectangle contained by AI and EI,
or (Eu. V. 15.) R is to S as the ſquare on half AE is to the rectangle
contained by AI and EI; and which is therefore the greateſt ratio which
R can have to S in thoſe Caſes.
It ought farther to be remarked, that to have Caſe III poſſible, where
O is ſought beyond A, and the ratio of a greater to a leſs, it is neceſſary
that AI be leſs than IE, and to have Caſe IV. poſſible, that it be greater.
For it is plain from the Conſtruction, that IB muſt in the former caſe be
leſs, and in the latter greater than I C; but as R is to S ſo is AB to
IB, and ſo is EC to IC, wherefore by diviſion, the exceſs of R above S
is to S as AI is to IB, and as EI is to IC; and ſo by permutation AI is
to EI as IB is to IC: conſequently when IB is greater than IC, AI will
be greater than EI; and when leſs, leſs.
With reſpect to thoſe caſes wherein the given ratio is that of equality,
it may be ſufficient to remark, that none of the Caſes of Epitagma II.
are poſſible under that ratio: that one of Caſes III. and IV. Epitagma III.
is always impoſſible when the given ratio of R to S is the ratio of equality;
and both are ſo if AI be at the ſame time equal to IE. Laſtly Caſes V.
and VI. are never poſſible under the ratio of equality, unleſs the ſquare on
half AE be equal to, or exceed the rectangle contained by AI and EI;
all which naturally follows from what has been delivered above.
1
THE END OF BOOK I.
11See Prop. A. Book V. of Dr. Simſon’s Euclid.
103[26]
DETERMINATE SECTION.
BOOK II.
LEMMA I.
If from two points E and I in the diameter AU of a circle AYUV (Fig.
27.) two perpendiculars EV, IY be drawn contrary ways to terminate in
the Circumference; and if their extremes V and Y be joined by a ſtraight
line VY, cutting the faid diameter in O; then will the ratio which the
rectangle contained by AO and UO bears to the rectangle contained by
EO and IO be the leaſt poſſible.
*** This being demonſtrated in the preceeding Tract of Snellius, I
ſhall not attempt it here.
LEMMA II.
If to a circle deſcribed on AU, tangents EV, IY (Fig. 28. 29.) be drawn
from E and I, two points in the diameter AU produced, and through the
points of contact V, and Y, a ſtraight line YVO be drawn to cut the line
AI in O; then will the ratio which the rectangle contained by AO and
UO bears to that contained by EO and IO be the leaſt poſſible: and
moreover, the ſquare on EO will be the ſquare on IO as the rectangle con-
tained by AE and UE is to the rectangle contained by AI and UI.
Demonstration. If the ſaid ratio be not then a minimum, let it be
when the ſegments are bounded by ſome other point S, through which and
the point V, let the ſtraight line SV be drawn, meeting the circle again in
R; draw SM parallel to OY, meeting the tangents EV and IY in L and
M, and through R and Y draw the ſtraight line RY meeting SM produced
in N: the triangles ESL and EOV, ISM and IOY are ſimilar; wherefore
LS is to SE as VO is to EO, and SM is to SI as YO is to OI;
104[27] quently, by compound ratio, the rectangle contained by LS and SM is to
that contained by SE and SI as the rectangle contained by VO and OY,
or its equal, the rectangle contained by AO and OU is to that contained
by EO and IO. Now the triangles VSL and NSR having the angles at
R and L equal, and the angle at S common, are ſimilar; and therefore SR
is to SN as SL is to SV; conſequently, the rectangle contained by SR and
SV, or its equal, the rectangle contained by SA and SU is equal to that
contained by SN and SL: but SN is neceſſarily greater than SM, in con-
ſequence whereof the rectangle contained by SN and SL, or its equal, the
rectangle contained by AS and SU is greater than that contained by SM
and SL; wherefore the ratio which the rectangle AS, SU bears to the rect-
angle ES, SI is greater than that which the rectangle SM, SL bears to it,
and of courſe, greater than the ratio which the rectangle AO, UO bears to
the rectangle EO, IO; and that, on which ſide ſoever of the point O, S is
taken.
Again, on YO produced, let fall the perpendiculars EB and IC: the
triangles EBV and ICY, EBO and ICO are ſimilar, becauſe the angles EVO
and IYO are equal, and ſo EO is to IO as EB is to IC, alſo EV is to IY
as EB is to IC; therefore by equality of ratios EO is to IO as EV is to IY,
and (Eu. VI. 22.) the ſquare on EO is to the ſquare on IO as the ſquare
on EV is to the ſquare on IY; that is (Eu. III. 36.) as the rectangle con-
tained by AE and UE is to that contained by AI and UI.
Q. E. D.
LEMMA III.
If from two points E and I, in the diameter AU, of a circle, AVYU
(Fig. 30.) two perpendiculars EV, IY be drawn on the ſame ſide thereof to
33[Handwritten note 3] terminate in the periphery, and if their extremes V and Y be joined by a
ſtraight line VY, cutting the ſaid diameter, produced, in O; then will the
ratio which the rectangle contained by AO and UO bears to the rectangle
contained by EO and IO be the greateſt poſſible.
*** This, like the ſirſt, is demonſtrated by Snellius, and needs not be
repeated.
105[28]
LEMMA IV.
If EK and IY (Fig. 27.) be any perpendiculars to the diameter AU of
a circle AYUV, terminating in the circumference, and if KY be drawn,
on which, from U, the perpendicular UF is demitted; then will KF be a
mean proportional between AI and EU, alſo YF a mean proportional
between AE and IU.
Demonstration. Draw UY, UK, KA and AY, the angles I and F
44[Handwritten note 4]55[Handwritten note 5] being right by conſtruction, and the angles IDU, and FKV equal, being
44[Handwritten note 4]55[Handwritten note 5] both equal to the angle UAY, the triangles IYU and FKU are ſimilar,
and conſequently IY is to UY as KF is to UK; or (Eu. VI. 22.) the
ſquare on IY is to the ſquare on UY as the ſquare on KF is to the ſquare
on UK: now the ſquare on IY is (Eu. VI. 8. 17.) equal to the rectangle
contained by AI and IU, the ſquare on UY to the rectangle contained by
AU and IU, and the ſquare on UK to the rectangle contained by AU and
EU; wherefore the rectangle contained by AI and IU is to that contained
by AU and IU as the ſquare on KF is to the rectangle contained by AU
and UE, whence (Eu. V. 15.) AI is to AU as the ſquare on KF is to the
rectangle contained by AU and UE, or the rectangle contained by AI and
UE is to that contained by AU and UE as the ſquare on KF is to the
rectangle contained by AU and UE; ſeeing then that the conſequents are
here the ſame, the antecedents muſt be equal, and therefore AI is to KF
as KF is to UE.
Again, the angle AKE is equal to AUK, which is equal to the angle
AYK, of which the angle UYF is the complement, becauſe AYU is a
right angle; and therefore as the angles F and E are both right, the tri-
angles AKE and YUF are ſimilar, and AK is to AE as YU is to YF,
wherefore the ſquare on AK is to the ſquare on AE as the ſquare on YU
is to the ſquare on YF: but the ſquare on AK is equal to the rectangle
contained by AU and AE, and the ſquare on YU is equal to the rectangle
contained by AU and IU, conſequently the rectangle contained by AU
and AE is to the ſquare on AE as the rectangle contained by AU and IU
is to the ſquare on YF; whence (Eu. V. 15.) AU is to AE as the rect-
angle contained by AU and IU is to the ſquare on YF, or the rectangle
contained by AU and IU is to that contained by AE and IU as the
106[29] angle contained by AU and IU is to the ſquare on YF; hence the rect-
angle AE, IU is equal to the ſquare on YF, and AE is to YF as YF is
to IU.
Q. E. D.
LEMMA V.
If in any ſtraight line four points A, U, E and I (Fig. 31.) be aſſigned,
and if the point O be ſo taken by Lemma II, that the ratio of the rect-
angle contained by AO and UO to that contained by EO and IO may be
the leaſt poſſible; alſo if through O the indefinite perpendicular FG be
drawn; and laſtly, if from E and I, EG and IF be applied to FG, the
former equal to a mean proportional between AE and UE, and the latter
to one between AI and UI: then ſhall FG be equal to the ſum of two
mean proportionals between AE and UI, AI and UE.
Demonstration. Draw AF and AG, and, through U, FV and GY,
produce GE to meet FV in H, and let fall on FV the perpendicular XI,
cutting FG in N; moreover draw UM through N, and NP through E,
and theſe two laſt will be reſpectively perpendiculars to IF and UG, be-
cauſe the three perpendiculars of every plane triangle meet in a point.
Since by conſtruction and Eu. VI. 17, the ſquare on EG is equal to the
rectangle contained by AE and UE, and the ſquare on IF to that con-
tained by AI and UI, and becauſe (Lem. II.) the ſquare on EO is to the
ſquare on IO as the rectangle AE, UE is to the rectangle AI, UI; the
ſquare on EO is to the ſquare on IO as the ſquare on EG is to the ſquare
on IF, and (Eu. VI. 22.) EO is to IO as EG is to IF; from whence it
appears that the triangles EOG and IOF are ſimilar, and HG parallel to
IF, and the angle UHE equal to the angle UFI. Again, becauſe AI is
to IF as IF is to UI, the triangles AIF, and UFI are ſimilar, and ſo, for
like reaſons, are the triangles AEG and GEU, wherefore the angles UFI
and FAE are equal, and alſo the angles UGE and UAG; hence there-
fore (Eu. I 32.) the angle YUF is equal to the angle UAF (UHE) toge-
ther with the angle UAG (UGE) and conſequently, the angles VAY and
YUV are together equal to two right angles; wherefore the points AYUV
are in a circle: hence, and becauſe AO is perpendicular to FG, GY 1
11See Pap. Math. Collect. B. vii. prop. 60.
107[30] be perpendicular to AF, and FV to AG; whence it follows that GA is
parallel to XI, and the angle NIU equal to the angle UAG; but UAG is
equal to UGE, which is equal to CNE; wherefore, in the triangles UNI,
UNE, the angles at I and N being equal, and that at U common, they
are ſimilar, and UN is to UI as UE is to UN, conſequently the ſquare on
UN is equal to the rectangle contained by UI and UE. Moreover, ſince
IX paſſes through N, and is perpendicular to FU, by Eu. I. 47, the dif-
ference of the ſquares on IF and IU is equal to the difference of the
ſquares on NF and NU: now the ſquare on IF being equal to the rect-
angle contained by AI and UI, that is (Eu. II. I.) to the rectangle con-
tained by AU and UI together with the ſquare on UI, the difference of
the ſquares on IF and UI, and conſequently the difference of the ſquares on
NF and NU is equal to the rectangle contained by AU and UI; but the
ſquare on NU has been proved equal to the rectangle contained by UI
and UE, therefore the ſquare on NF is equal to the rectangle contained
by EU and IU together with that contained by AU and IU, that is (Eu.
II. 1.) to the rectangle contained by AE and UI; wherefore AE is to NF
as NF is to UI.
Laſtly, for the like reaſons which were urged above, the difference of
the ſquares on NU and NG is equal to the difference of thoſe on GE and
UE: now the ſquare on GE is equal to the rectangle contained by AE
and UE, that is, to the rectangle contained by AU and UE together
with the ſquare on UE; therefore the difference of the ſquares on GE
and UE, or the difference of thoſe on NU and NG, is equal to the rect-
angle contained by AU and UE; but the ſquare on NU is equal to the
rectangle contained by UI and UE, therefore the ſquare on NG is
equal to the rectangle contained by AU and UE together with that
contained by UI and UE; that is, to the rectangle contained by AI and
UE, and ſo AI is to NG as NG is to UE. Now FG is equal to the ſum
of NF and NG; therefore FG is equal to the ſum of two mean propor-
tionals between AE and UI, AI and UE.
Q. E. D.
108[31]
PROBLEM VII. (Fig. 32, 33, 34, &c.)
In any indeſinite ſtraight line let there be aſſigned the points A, E, I
and U; it is required to cut it in another point, O, ſo that the rectangle
contained by the ſegments AO, UO may be to that contained by the ſeg-
ments EO, IO in the ratio of two given ſtraight lines, R and S.
Analysis. Imagine the thing done, and O the point ſought: then will
the rectangle AO, UO be to the rectangle EO, IO as R is to S. Make
UC to EC as R is to S; and the rectangle AO, UO will be the rectangle
EO, IO as UC is to EC. Let now OB be taken a fourth proportional to
UO, UC and IO: then (Eu. V. 15.) the rectangle AO, UO will be to
the rectangle EO, IO as the rectangle UC, OB is to the rectangle EC, OB;
or (Eu. V. 16.) the rectangle AO, UO is to the rectangle UC, OB as the
rectangle EO, IO is to the rectangle EC, OB; wherefore ſince UO is to UC
as IO to OB, by conſtruction, AO will be to BO as EO to EC; and ſo by
compoſition or diviſion, CO is to CU as IB to BO, and AB is to BO as
CO to EC: wherefore ex æquo perturb. & permut. AB is to IB as UC to
EC, that is, in the given ratio; and hence is given BC, the ſum or dif-
ference of CO and BO, as alſo the rectangle contained by them, equal to
the rectangle CU, IB, whence thoſe lines themſelves are given by the 85th
or 86th of the Data.
Synthesis. Make AB to IB, and UC to EC in the given ratio, and de-
ſcribe on BC a circle; erect, at B the indeſinite perpendicular BK, and take
therein BD a mean proportional between AB and EC, or between IB and
and UC: from D, draw DH, parallel to BC, if O be required any where
between B and C; but through F, the center of the circle on BC, if it be
ſought any where without them, cutting the circle on BC in H. Laſtly,
draw HO perpendicular to DH, which will cut the indeſinite line in O,
the point required.
For it is plain from the conſtruction that HO and BD are equal, and
(Eu. VI. 17.) the rectangle AB, EC, or the rectangle IB, UC is equal to
the ſquare on BD, and therefore equal to the ſquare on HO, which (Eu.
III. 35. 36.) is equal to the rectangle BO, OC. Hence (Eu. VI. 16.) AB
is to BO as CO is to CE, and CO is to CU as IB is to BO; whence,
by compoſition or diviſion, AO is to BO as EO is to CE, and UO is
109[32] CU as IO is to BO; and ſo, by compound ratio, the rectangle AO, UO
is to the rectangle BO, CU as the rectangle EO, IO is to the rectangle
BO, CE; by permutation, the rectangle AO, UO is to the rectangle EO,
IO, as the rectangle BO, CU is to the rectangle BO, CE; or (Eu. V. 15.)
as CU is to CE; that is, by conſtruction as R to S.
Q. E. D.
Scholium. In enumerating the ſeveral Caſes of this Problem I ſhall en-
deavour to follow the method which I conceive Apollonius did: and there-
fore, notwithſtanding the preceding Analyſis and Conſtruction are general
for the whole, divide it into three Problems, each Problem into three Epi-
tagmas, or general Caſes, and theſe again into their ſeveral particular ones.
PROBLEM I. (Fig. 32 to 45.)
Here O is ſought between the two mean points of the four given ones:
and the three Epitagmas are, firſt, when A and U, the points which bound
the ſegments containing the antecedent rectangle, are one an extreme, and
the other an alternate mean; ſecondly, when thoſe points are one an ex-
treme and the other an adjacent mean; thirdly, when they are both means,
or both extremes.
Epitagma I, Conſiſts of eight Caſes, viz. when the order of the given
points is A, I, U, E; U, E, A, I; A, E, U, I; or U, I, A, E, and the
given ratio of a leſs to a greater, and four others wherein the order of the
points is the ſame as in thoſe, but the ratio of R to S, the ratio of a greater
to a leſs.
Case I. Let the order of the given points be A, I, U, E, and the given
ratio of a leſs to a greater; and the Conſtruction will be as in Fig. 32, where
B is made to fall beyond A, with reſpect to I, and C beyond U with re-
ſpect to E, and DH is drawn through F, the center of the circle on BC.
That O, when this conſtruction is uſed, will fall between I and U is
plain, becauſe CO is to CU as IB is to BO; and therefore if CU be
greater than CO, BO will be greater than IB, and if leſs, leſs; but this,
it is manifeſt, cannot be the Caſe if O falls either beyond I or U, and
therefore it falls between them.
110[33]
Case II. If the order oſ the given points be retained, but the ratio be of
a greater to a leſs, B muſt ſall beyond I with reſpect to A, and C beyond
E, as in Fig. 33, and DH muſt be drawn as before: and it may be proved
by reaſonings ſimilar to thoſe uſed in the ſirſt Caſe that O will fall between
I and U, as was required.
Cases III and IV. Theſe caſes are conſtructed exactly in the ſame man-
ner as Caſes I and II reſpectively; and the reaſonings to prove that O will
fall as it ought arethe ſame with thoſe made uſe of in Caſe I, as will appear
by inſpecting Fig. 34. and 35.
Case V. Let now the order of the given points be A, E, U, I, and the
given ratio of a leſs to a greater. Then muſt B be made to fall beyond A
and C beyond U, as in Fig. 36, and DH is here to be drawn parallel to
to BC: and that O will fall as required may be made to appear thus. Draw
EF and UG perpendicular to BC. Now (Eu. VI. 13. 17.) the ſquare on
EF is equal to the rectangle contained by EC and EB, the ſquare on UG
to the rectangle contained by UC and UB; and, by conſtruction, the
ſquare on HO to the rectangle contained by EC and AB, or to the rect-
angle contained by UC and IB; and ſince, by ſuppoſition, EB is greater
than AB, and UB leſs than IB, the rectangle EC, AB, or its equal, the
rectangle UC, IB will be leſs than the rectangle EC, EB, and greater than
the rectangle UC, UB; and conſequently the ſquare on HO will be leſs
than the ſquare on EF and greater than the ſquare on UG; and therefore
HO leſs than EF and greater than UG: but this could not be the Caſe
unleſs O fell between E and U, as was was required.
Case VI. If the order of the points be retained; but the given ratio be
of a greater to a leſs, B muſt then fall beyond I, and C beyond E; and
DH is drawn as in the preceding Caſe (See Fig. 37.) moreover, that O
will fall between E and U may be made to appear, by reaſonings ſimilar to
thoſe there made uſe on the like occaſion.
Case VII. Is conſtructed exactly in the ſame manner as Caſe V. and,
Case VIII. As Caſe VI: the truth of which will appear by. barely in-
ſpecting Fig. 38 and 39.
Epitagma II. In this Epitagma there are alſo eight Caſes, viz. when the
order of the given points is A, U, E, I; A, U, I, E; U, A, E, I; or
U, A, I, E; and the given ratio of a leſs to a greater: and there are
four others wherein the order of the points are the ſame as in thoſe,
111[34] the ratio of a greater to a leſs; but theſe, I ſhall ſhew, may be reduced
to four.
Case I. The order of the given points being A, U, E, I; and the given
ratio a leſs to a greater, the conſtruction will be eſſected by Fig. 40,
wherein B is made to fall beyond A with reſpect to I and C beyond U, and
DH is drawn through the center of the circle on BC. And O will fall
between U and E for reaſons ſimilar to thoſe urged in the ſirſt Caſe of Epi-
tagma I. It is moreover obvious that the conſtruction will not be eſſentially
different ſhould the points E and I change places, and therefore need not
here be made a new Caſe.
Case II. The order of the points being the ſame as in the laſt Caſe, let
the given ratio be of a greater to a leſs; then, as in Fig. 41, B muſt fall
beyond I, and C beyond E; but DC muſt ſtill be drawn through the
center of the circle on BC. It is manifeſt that this conſtruction will ſerve
for that Caſe wherein the points A and U change ſituations, if the ratio be,
as here, of a greater to a leſs.
Case III. Here, let the order of the points be U, A, I, E, and the
given ratio of a leſs to a greater, and the Conſtruction will be aſſected by
Fig. 42, in which B falls beyond A, and C beyond U with reſpect to I and
E: and the ſame conſtruction will ſerve if I and E change places, but the
ratio remain the ſame.
Case IV. If the poſition of the points be retained, but the ratio be
made of a leſs to a greater; then muſt B fall beyond I (Fig. 43.) and C
beyond E; but DH drawn as before. That O muſt fall as was required,
in theſe three laſt caſes, is obvious enough from what has been ſaid be-
fore on the like occaſion: and it is alſo plain that the conſtruction will
not be materially diſſerent though A and U change places.
Scholium. That none of the Caſes of theſe two Epitagmas are ſubject
to Limitations, might be proved with the utmoſt rigour of geometrical
reaſoning was it not ſuſſiciently manifeſt from conſidering that as the point
O approaches points A, or U, the ratio of the rectangle AO, OU to the
rectangle EO, OI will become very ſmall, and as it approaches the points
E, or I the ſaid ratio will become very great: and nothing hinders that
the ſaid point may ſall any where between thoſe.
112[35]
Epitagma III. There are here but four Caſes, viz. when the order of the
given points is A, E, I, U; A, I, E, U; E, A, U, I; or E, U, A, I;
the two ſirſt of theſe are not poſſible unleſs the given ratio be the ratio of a
greater to a leſs; nor the two latter, unleſs it be of a leſs to a greater, and
as theſe are reduced to the ſirſt two by reading every where E for A, I
for U, and the contrary, I ſhall omit ſpecifying them.
Case I. If the order of the given points be A, E, I, U, the conſtruc-
tion will be effected by Fig. 44, wherein B is made to fall beyond I, and
C beyond E, and DH is drawn parallel to BC. That O, when this con-
ſtruction is uſed, will fall between E and I, is eaſily made appear by rea-
ſoning in a manner ſimilar to what was done in Caſe V. of Epitagma I.
Case II. The conſtruction of this Caſe, where the order of the points
is A, I, E, U, is not materially different from that above exhibited as ap-
pears by Fig. 45, and that O will fall between I and E is manifeſt without
farther illuſtration.
Limitation. In theſe two Caſes the given ratio of R to S cannot be
leſs than that which the ſquare on AU bears to the ſquare on a line which
is the difference of two mean proportionals between AI and EU, AE and
IU. For by Lemma I. the leaſt ratio which the rectangle contained by AO
and UO can have to the rectangle contained by EO and IO; or, which
is the ſame thing, that R can have to S, will be when the point O is
the interſection of the diameter AU, of a circle AYUV, with a ſtraight
line YV. joining the tops of two perpendiculars EV, IY to the indeſi-
nite line, on contrary ſides thereof, and terminating in the periphery of
the circle. Produce VE (Fig. 27.) to meet the circle again in K, and
draw the diameter KL; join LY and KY, on which, produced, let fall
the perpendicular UF. Now, ſince by Lemma III. KF is a mean propor-
tional between AI and EU, and YF a mean proportional between AE
and IU: it remains only to prove that the ratio of the rectangle con-
tained by AO and OU to the rectangle contained by EO and OI is the
ſame with the ratio which the ſquare on AU bears to the ſquare on KY,
which is the diſſerence between KF and YF. Becauſe the angles E and
KYL are both right, and the angles EVO and KYL equal (Eu. III. 21.)
the triangles EVO and YLK are ſimilar; and ſo VO is to EO as AU (LK)
is to KY; or the ſquare on VO is to the ſquare on EO as the ſquare on AU
is to the ſquare on KY. Now the triangles EVO, IYO being alſo
113[36] OY will be to OV as OI is to OE, and (Eu. V. 15. 16.) the rectangle
contained by VO and YO or its equal, the rectangle contained by AO and
OU, is to the rectangle contained by EO and OI as the ſquare on OV is
to the ſquare on EO, as the ſquaree on AU is to the ſquare on KY.
Q. E. D.
Scholium. It might be obſerved that in the two Caſes of this Epitagma
where the points A and U are means, the limiting ratio will be a maxi-
mum inſtead of a minimum; and that ratio will be the ſame with that which
the ſquare on KY bears to the ſquare on EI, as is plain from what hath
been advanced above.
PROBLEM II. (Fig. 46 to 57.)
Where O is ſought between a mean and an extream point: and here,
as in the firſt Problem, there are three Epitagmas. Firſt when the points
A and U, which bound the ſegments containing the antecedent rectangle,
are one an extreme, and the other an alternate mean; ſecondly when they
are both means, or both extremes; thirdly when they are one an extreme,
and the other an adjacent mean.
Epitagma I. There are here eight Caſes, but they are conſtructed at
four times, becauſe it is indifferent whether the given ratio be of a leſs to a
greater, or of a greater to a leſs.
Case I. The order of the given points being A, E, U, I, as in Fig. 46,
make B to fall between A and I, C between U and E, and draw DH
through the center of the circle on BC; and O will fall between A and
E, becauſe AB is to BO as CO is to CE, and therefore, if AB be greater
than BO, CO muſt be greater than CE, and if leſs, leſs; but this cannot
be the caſe if O falls either beyond A or E: and the like abſurdity fol.
lows if o be ſuppoſed to fall otherwiſe than between I and U.
Case II. Wherein the order of the points is U, I, A, E; and it is con-
ſtructed in the very ſame manner that Caſe I. is, as appears by barely
inſpecting Fig. 47.
Case III. If the order of the given points be A, I, U, E (Fig. 48.) the
points B and C muſt be made to fall as in the two preceding Caſes; but
DH muſt be drawn parallel to BC, and O will fall as required. For
114[37] at I, the perpendicular IG: by Eu. VI. 13. 17. the ſquare on IG is equal
to the rectangle contained by IB and IC; and the ſquare on HO is equal
to the rectangle contained by IB and UC. Now IC is by ſuppoſition
greater than UC, and therefore the rectangle IB, IC is greater than the
rectangle IB, UC: conſequently the ſquare on IG is greater than the
ſquare on HO, and IG than HO; whence O muſt fall between I and B,
much more between I and A. And in the ſame manner it may be proved
that the point o falls between U and E.
Case IV. In which the order of the given points is U, E, A, I; it is
conſtructed exactly in the ſame manner as Caſe III, and is exhibited by
Fig. 49.
Epitagma II. There are here only four Caſes, becauſe, as in Epi-
tagma I. it is indifferent whether the given ratio be of a leſs to a greater,
or of a greater to a leſs; and the two laſt of thoſe, viz. where the order
of the given points is E, A, U, I; or E, U, A, I, being reducible to the
two former by reading every where I for A, E for U, and the contrary,
I ſhall omit ſaying any thing of their conſtructions, except that they are
exhibited by Fig. 52 and 53.
Case I. The order of the given points, being A,E,I,U, make B to
fall between A and I, C between E and U, and draw DH through the
center of the circle on BC, as is done in Fig. 50; and O will fall as re-
quired for reaſons ſimilar to thoſe urged in Caſe I. of the firſt Epitagma
of this Problem.
Case II. If the order of the given points be A, I, E, U, the conſtruc-
tion will be as in Fig. 51, where B and C are made to fall, and DH is
drawn as in Caſe I.
Epitagma III. Here there are eight Caſes, viz. four where in the
order of the given points is A, U, E, I; A, U, I, E; U, A, E, I; and
U, A, I, E, and the given ratio of a greater to a leſs, when O will fall
between the two given points, which bound the conſequent rectangle;
and four others@ wherein the order of the given points is the ſame as
here, but the given ratio of a leſs to a greater, and in which the point
O will fall between the points that bound the antecedent rectangle; but
as theſe laſt are reducible to the former by the ſame means which have been
uſed on former ſimilar occaſions, I ſhall not ſtop to ſpecify them.
115[38]
The former four are all conſtructed by making B to fall between A and
I, C between U and E, and drawing DH parallel to BC; and it will ap-
pear by reaſonings ſimilar to thoſe uſed for the like purpoſe in Caſe III. of
Epitagma I. that O muſt fall between E and I as was propoſed. See Fig.
54, 55, 56 and 57.
Limitation. In theſe four Caſes, the given ratio of R to S muſt not
be leſs than that which the ſquare on the ſum of two mean proportionals
between AE and IU, AI and EU bears to the ſquare on EI. For it has
been proved (Lem. II.) that when the ratio of the rectangle contained by
AO and UO to that contained by EO and IO; or, which is the ſame thing,
the given ratio of R to S is the leaſt poſſible, the ſquare on EO will be to
the ſquare on IO as the rectangle contained by AE and UE is to that con-
tained by AI and UI; and (Lem. V. Fig. 31.) that FG will then be the
fum of two mean proportionals between AE and UI, AI and UE: it
therefore only remains to prove that the rectangle contained by AO and
UO is to that contained by EO and IO as the ſquare on FG is to the
fquare on EI. Now it has been proved in demonſtrating Lem. V. that the
triangles EOG and IOF are ſimilar, and that the angle at V is right,
whence it follows that the triangles AOG and FOU are alſo ſimilar, and
conſequently that AO is to OG as OF is to UO; therefore the rectangle
contained by AO and UO is equal to that contained by GO and OF. More-
over GO is to OF as EO is to IO, and ſo by compoſition and permutation,
FG is to EI as OG is to EO, and as OF is to IO: hence by compound
ratio the ſquare on FG is to the ſquare on EI as the rectangle contained
by (OG and OF) AO and UO is to that contained by EO and IO.
Q. E. D.
Scholium. In the four Caſes, wherein the given ratio is of a leſs to a
greater, and wherein the point O muſt fall between thoſe given ones which
bound the antecedent rectangle, the limiting ratio will be a maximum, and
the ſame with that which the ſquare on AU bears to the ſquare on FG.
116[39]
PROBLEM III.
In this, the point O is ſought without all the given ones, and the three
Epitagmas are as in Problem I.
Epitagma I. There are here eight Caſes, viz. four when the order of
the given points is the ſame as ſpecified in Epitagma I. of Problem I, and
O ſought beyond the given point which bounds the antecedent rectangle;
and four others when O is ſought beyond that which bounds the conſe-
quent one: the Conſtructions of the four firſt are ſhewn by the ſmall
letters b and o in Fig. 32, 34, 36 and 38; and the four latter ones by
the ſame letters in Fig. 33, 35, 37 and 39; and the demonſtrations that
o will fall as required by the Problem are exactly the ſame as thoſe made
uſe of in the laſt mentioned Epitagma. It is farther obſervable, that the
four firſt Caſes are not poſſible, unleſs the given ratio be of a leſs to a
greater; nor the four latter, unleſs it be of a greater to a leſs, as is ma-
nifeſt without farther illuſtration.
Epitagma II. Here, as in the ſecond Epitagma of Problem I, the points
A and U are one an extreme, and the other an adjacent mean, and there
are eight Caſes; but it will be ſufficient to exhibit the conſtructions of
four of them, the others being not eſſentially different; and theſe are ſhewn
by the ſmall b and o in Fig. 40, 41, 42 and 43; the demonſtrations that o
will fall as required need not be pointed out here; but it may be neceſſary
to remark that the firſt and third are not poſſible unleſs the given ratio be
of a leſs to a greater, nor the ſecond and fourth unleſs it be of a greater to
a leſs, as is obvious enough.
Epitagma III. In which the points A and U are both means, or both
extremes; and there are here eight Caſes, viz. four wherem theſe points
are extremes, and four others wherein they are means: but theſe laſt being
reducible to the former by the ſame method that was uſed in the third Epi-
tagmas of the two preceding Problems, I ſhall omit them.
All the Caſes of this Epitagma are conſtructed by making B fall beyond
I, and C beyond E, with reſpect to A and U; and drawing DH parallel to
BC. That O will fall beyond A in Fig. 58 and 60, and beyond U in Fig.
59 and 61 appears hence. Draw AG perpendicular to BC, meeting the
circle on BC in G: by Eu. VI. 13. 17, the ſquare on AG is equal to
117[40] rectangle contained by AB and AC; but the ſquare on HO is equal to
the rectangle contained by AB and EC: now EC is, by ſuppoſition,
greater than AC, therefore the rectangle AB, EC is greater than the
rectangle AB, AC, and the ſquare on HO greater than the ſquare on AG,
conſequently HO is itſelf greater than AG; but this could not be the
Caſe unleſs O fell beyond A. In the ſame manner my it be proved that O
will fall beyond U in Fig. 59 and 60.
Limitation. In the above four Caſes the given ratio of R to S muſt
not exceed that which the ſquare on AU bears to the ſquare on the ſum of
two mean proportionals between AI and UE, AE and UI. For (Fig. 30.)
demit from A, on KO produced, the perpendicular AH. Now it has been
proved (Lem. III.) that the ratio of the rectangle continued by AO and UO
to that contained by EO and IO, or which is the ſame thing, the given
ratio of R to S is the greateſt poſſible; and (Lem. IV.) that KF is a mean
proportional between AI and UE, alſo that YF is a mean proportional
between AE and UI: but HK is equal to YF, therefore HF is equal to
the ſum of two mean proportionals between AI and UE, AE and UI;
it only then remains to prove, that the rectangle contained by AO and UO
is to that contained by EO and IO as the ſquare on AU is to the ſquare
on HF. The triangles OEK, OHA, OIY and OUF are all ſimilar; con-
ſequently OK is to OE as OA is to OH, as OY is to OI, and therefore
by compound ratio, the rectangle contained by AO and UO (OK and
OY) is to that contained by EO and IO as the ſquare on AO is to the
ſquare on OH; but alſo AO is to UO as HO is to EO, and by compoſi-
tion and permutation, AU is to HF as AO is to HO, or (Eu. VI. 22.)
the ſquare on AU is to the ſquare on HF as the ſquare on AO is to the
ſquare on HO, and ſo by equality of ratios, the rectangle contained by AO
and UO is to that contained by EO and IO as the ſquare on AU is to the
ſquare on HF.
Q. E. D.
Scholium. In the four Caſes wherein the points A and U are means,
the limiting ratio will be a minimum, and the ſame with that which the
ſquare on HF bears to the ſquare on EI.
THE END.
118 15[Figure 15]
119 16[Figure 16]
120
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121 17[Figure 17]
122 18[Figure 18]
123
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124 19[Figure 19]
125 20[Figure 20]
126
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127 21[Figure 21]
128 22[Figure 22]
129
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130 23[Figure 23]
131 24[Figure 24]
132
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133
A
SYNOPSIS
OF ALL THE DATA FOR THE
Conſtruction of Triangles,
FROM WHICH
GEOMETRICAL SOLUTIONS
Have hitherto been in Print.
With References to the Authors, where thoſe Solutions are to be found:
By JOHN LAWSON, B. D.
Rector of Swanscombe, in KENT.
ROCHESTER:
Printed by T. Fisher; and Sold by J. Nourse, B. White, T. Payne, and J. Wilkie, in London@
MDCCLXXIII.
[Price One Shilling.]
134
[Empty page]
135
ADVERTISEMENT.
IT is but few years ago ſince the Compiler of this Synopſis
conceived his firſt idea of the uſeſulneſs of ſuch an undertaking,
and he exhibited a ſmall ſpecimen thereof in a periodical Work
then publiſhing under the Title of The British Oracle. Here-
upon he received ſeveral letters from Mathematical friends, ex-
preſſing their ſenſe of the great propriety of ſuch a collection, and
ſtrongly encouraging him to purſue the undertaking. Since that
time it has been a growing work, and would continue ſo, were
the publication delayed ever ſo long, as freſh Problems are con-
tinually propoſed to the public. He has therefore now determined
to ſend it abroad, as complete as he can make it to the preſent
period, and leave additions to be made by future collectors.
136
[Empty page]
137
AN
EXPLANATION
OF THE
SYMBOLS made uſe of in this SYNOPSIS.
11
H. # repreſents the Hypothenuſe of a right-angled triangle.
V. # Vertical angle.
B. # Baſe or ſide oppoſite V.
P. # Perpendicular from V on B.
S&s. # Sides about V; S the greater, s the leſs.
A & a. # Angles at B; A the greater, a the leſs.
m & n. # Segments of B by P; m the greater, n the leſs.
Ar. # Area.
Per. # Perimeter.
L. # Line from V to B, biſecting V.
λ. # Line from V to B, cutting B in a given ratio.
l. # Any other line ſpecified how drawn.
R. # Radius of inſcribed circle.
☉(sun). # Circle.
◻. # Square.
: # Ratio: thus, S: s ſignifies the ratio of the ſides.
138[ii]
N. B. Between each of the Data a full ſtop is placed. Moreover, m and n are
ſometimes uſed for different ſegments than thoſe of B by P, but then it is ſignified in
words: - the ſame alſo is to be obſeved of P.
Obſerve likewiſe, for the more ready finding any propoſed Problem in the Synop-
ſis, that the data are ranged in the ſame order as the Symbols here recited; viz.
all data whereof V is one are placed firſt; next thoſe where B is given, either ſim-
ply by itſelf, or combined with any other datum; next P, & c.
Moreover, when in the references you find this mark*, it ſignifies that ſuch Authors
have only conſtructed the Problem partially, and not generally as propoſed in the Synop-
ſis, e. g. for a right-angled triangle, when it is propoſed for a triangle in general;
and again, for a line biſecting another, when it is propoſed to cut it ſo that the ſeg-
ments may be in any given ratio.
139[i]
INDEX
OF THE
Authors refered to in the SYNOPSIS.
11
ANDERSONI Var. Prob. Practice, cum Supplemento Apollonii
# Redivivi, 4to. # Pariſiis 1612
Anderſoni Exercitationum Math. Deas ima. 4to. # ibid. 1619
Aſhby’s Algebra, 2d Edit. 12mo. # Lond. 1741
Britiſh Oracle (Vol. I. being all that was publiſhed) 12mo. # ibid. 1769
Caſtillioneus inNewtoni Arith. Univerſalem, 4to. # Amſtelod. 1761
Clavius in Euclidem, var. Ed.
Court Magazine.
D’Omerique (Hugonis) Analyſis Geometrica, 4to. # Gadibus 1699
Diarian Repoſitory, Periodical Work, printed for Robinſon, 4to. # Lond. 1770, &c.
Foſter’s Miſcellanies, or Math. Lugubrations, fo. # ibid. 1659
General Magazine.
Gentleman’s Magazine.
Gentleman’s Diary.
Ghetaldi (Marini) Var. Prob. Collectio, 4to. # Venetiis 1607
Ghetaldus de Reſolutione & Compoſitione Math. fo. # Romæ 1640
Gregorius a Sancto Vincentio, fo. # Antverp. 1647
Herigoni Curſus Math. Lat. & Gallicè, 8vo. 5 Tom. # Paris 1644
Hutton’s Ladies Diaries.
# Mathematical Miſcellany.
Imperial Magazine.
Ladies Diaries.
Martin’s Math. Correſpondence, in his Magazine.
Mathematical Magazine.
Mathematician, Periodical Work. 8vo. # Lond.
11140[ii] Miſcellanea Curioſæ, Periodical Work. 8vo. 6 Nos. # rork 1734-5
Miſcellanea Cur. Math. Period Work, by Holliday, 4to. 2 vols. # Lond. 1745
Miſcellanea Scientiſica Curioſa, Period. Work, 4to. # ibid. 1766
Oughtred’s Clavis, var. Editions Lat. & Eng.
Palladium, Periodical Work.
Pappus Alexandrinus Commandini, fo. # Bononiæ 1660
Regiomontanus de Triangulis, fo. # Baſiliæ 1561
Renaldinus (Carolus) de Res. & Comp. Math. fo. # Patavii 1668
Ronayne’s Algebra, 2d Edit. 8vo. # Lond. 1727
Rudd’s Practical Geometry in 2 Parts, 4to. # ibid. 1650
Saunderſon’s Algebra, 2 vols. 4to. # Camb. 1740
Schooten’s (Franciſcus à) Exercitationes Math. 4to. # Lug. Bat. 1657
Simpſon’s Algebra, 8vo. 3d Edit. \\ 1ſt Edit. # Lond. 1767 \\ ibid. 1745
# Select Exerciſes, 8vo. # ibid. 1752
# Geometry. 8vo. 2d Edit. # ibid. 1760
Supplement to Gentleman’s Diary, 3 Nos. 12mo. # ibid. 1743, &c.
Town and Country Magazine.
Turner’s Mathematical Exerciſes, Periodical Work, 8vo. # ibid. 1750
Univerſal Magazine.
Vietæ Opera, fo. # Lug. Bat. 1646
Weſt’s Mathematics, 2d Edit. 8vo. # Lond. 1763
Wolfius’s Algebra, tranſlated by Hanna, 8vo. # ibid. 1739
25[Figure 25]
Lately was publiſhed by the ſame Author;
[Price Six Shillings in Boards.]
APOLLONIUS concerning Tangencies, as reſtored by Vieta & Ghetaldus,
with a Supplement; the 2d Edit. To which is now added a Second Supplement,
being Fermat’s Treatiſe on Spherical Tangencies. Likewiſe Apollonius con-
cerning Determinate Section, as reſtored by Willebrordus Snellius; to which
is added an entire new Work, being the ſame reſtored by Mr. W. Wales.
141[1]
SYNOPSIS.
11
1. V. B. P. # SIMPSON’s Alg. pr. 5. - Mis. Cur. Math. Vol. I. \\ page 31. - Vieta Iſt. Ap. to Apollonius Gallus, pr. 5. \\ - *Aſhby’s Alg. pag. 111. - *Rudd’s Pract Geo. part 2d, qu. \\ 4.-L. Diary, qu. 160.
2. V. B. P: m. # Town and Country Mag. Nov. and Dec. 1772.
3. V. B. P±m. # Mathematician, pr. 77. - Univerſal Mag. Mar. 1749.
4. V. B. S: s. # Simpſon’s Alg. pr. 3. - Simp. Geom. pr. 13. - Pappus Lib. \\ VII. pr. 155. - Herigon App. Geometriæ planæ, pr. 13. - \\ D’Omerique Lib. III. pr. 35. - Court Mag. July, 1762.
5. V. B. S + s. # Simpſon’s Alg. pr. 1. - Ghetaldus var. prob. 13 & *7. - Ghetaldus \\ deRes. & Comp. Math. Lib. V. c. 4, pr. 4, pag. 337, & *Lib. II. \\ pr. 9, pag. 93. - Renaldinus pag. 318, 326, 524, 79. - \\ Saunderſon’s Alg. art. 332. - *Rudd’s Prac. Geom. part \\ 2d, qu. 38. - *Aſhby’s Alg. pag. 102.
6. V. B. S - s. # Simpſon’s Alg. pr. 2. - Ghetaldus var. prob. 12 & *6. - \\ Ghetaldus de Res. & Comp. Lib. V. c. 4, pr. 3, and *Lib. II. pr. 8. \\ - Renaldinus pag. 317, 326, 527, 79. - *Simpſon’s Sel. Ex. pr. \\ 2. - *Wolfius’s Alg. pr. 12
8. - *Aſhby’s Alg. pag. 106.
7. V. B. S + s x S. # *Town and Country Mag. Jan. and Feb. 1769.
8. V. B. S + s + P. # Arith. Univ. Caſtillionei, pr. 5.
9. V. B. Ar. # *Simpſon’s Alg. pr. 33. - Simp. Geom. pr. 5. - *Saunder - \\ ſon’s Alg. art. 329. - *D’Omerique L. III. pr. 36. - *Oughtred’s
142[2]11
# Clavis, ch. 19, pr. 24. - *Vieta Geo. Eff. pr. 20. - *Herigon \\ App. Geo. planæ, pr. 10. - *Rudd’s Pract. Geom. part 2d, \\ qu. 47. - *Palladium, 1754, pa. 22.
10. V. B. Per. # Reducible to V. B. S + s.
11. V. B. L. # Simpſon’s Alg. pr. 72. - Simp. Geom. pr. 21.
12. V. B. λ. # *Simpſon’s Alg. pr. 58. - *Ghetaldus var. prob. 3. - Regio - \\ montanus de triangulis, Lib. II. pr. 29.
13. V. B. Direction \\ of l thro’ V. # }Simpſon’s Sel. Ex. pr. 48.
14. V. B. R. # *Simpſon’s Sel. Ex. pr. 29. - Br. Oracle, qu. 67, Cor. - \\ *Weſt’s Mathematics, 2d. Ed. pag. 45.
15. V. B. Side of \\ ins. ◻. # }Mathematician, pr. 25. - Br. Oracle, qu. 19.
16. V. B: P. S±s. # Brit. Oracle, qu. 31.
17. V. B±P. S±s. # Arith. Un. Caſtillionei, pr. 7. - *Simpſon’s Sel. Ex. pr. 31. - \\ *Simpſon’s Alg. Iſt. Ed. pr. 81.
18. V. B±P. Ar. # *Turner’s Math. Ex. pr. 18.
19. V. B±S. s. # Ghetaldus var. prob. 16, 17. - *Idem, 10, II. - Idem de \\ Res. &. Comp. Lib. V. cap. 4, pr. 7, 8. - *Anderſon var. prob. 2; \\ - *Simp. Sel. Ex. pr. 1. - Twyſden in Foſter’s Miſcell. pr. I. \\ - Renaldinus pag. 526, 529, *437. - *Mathematician, pr. \\ 42. - *Aſhby’s Alg. pr. 31, 32. - and in other places.
20. V. B±S. S±s. # Br. Oracle, qu. 102. - *Court Mag. Nov. 1761. - L. Diary, \\ qu. 661. - *Hutton’s L. Diary, qu. 147.
21. V. B±S. B±s. # *Clavius’s Euclid at end of B. II. - *Ghetaldus var. prob. 23. \\ - *Schooten pr. 37. - *Oughtred ch. 19, pr. 17. - *Saun - \\ derſon, art. 327. - *Anderſon var. prob. 15, 16. - *L. Diary, \\ 1770, p. 35.
22. V. B2: m2. P. # *Mis. Scient. Cur. qu. 54.
143[3]11
23. V. Point in B. \\ : \\ {S±s.\x} # }Simpſon’s Sel. Ex. pr. 43. - Simp. Geom. pr. 19, 20.
24. V. S + s - B. P. # Hutton’s Miscellany, qu. 5.
25. V. S + s - B. Ar. # Math. Mag. No. III. pr. 4.
26. V. B + S - s. \\ S + s + m - n. # }Gent. Mag. 1768, pag. 428, 519.
27. V. B±S. Per. # *Anderſon var. prob. 1, 3. - See V. B±S. s.
28. V. B ‖ to a given \\ line. Ar. # }Simpſon’s Sel. Ex. pr. 41. - Simp. Geom. pr. 4.
29. V. Pointin B. Ar. # Simpſon’s Sel. Ex. pr. 42.
30. V. ∠ of B with λ. λ. # *Mathematician, pr. II. - *York Mis. Cur. qu. 15.
31. V. P. S: s. # Simp. Alg. pr. 3. - Court Mag. Octo. 1762.
32. V. P. S±s. # Simp. Alg. pr. 80, 78. - *Oughtred ch. 19, pr. 9, 10. - \\ *Ghetaldus var. prob. 20, 21. - *De Res. & Comp. Lib. III. \\ pr. 3, 4. - Diarian Repoſitory, pag. 24. - *Hutton’s L. Diaries, \\ qu. 22. - *Wolfius’s Alg. pr. 127. - *Ronayne’s Alg. B. II. c. 2, \\ pr. 2.
33. V. P. S x s. # D’Omerique Lib. I. pr. 31. - Gent. D. qu. 149.
34. V. P. m: n. # Simp. Alg. pr. 11.
35. V. P. m - n. # Simp. Alg. pr. 10. - Foſter’s Lug. pr. 2. - *Rudd’s Pr. \\ Geom. part 2d, qu. 5.
36. V. P. Per. # Simp. Alg. pr. 60. - Ronayne’s Alg. B. II. ch. 1, pr. 2. - \\ Arith. Univ. Caſtillionei, pr. 4. - Mis. Cur. Math. qu. 61.
37. V. P. λ. # *Town and C. Mag. 1769, pag. 296, 381.
38. V. P. Ifrom A or a \\ to bis. S or s. # }*Br. Oracle, qu. 74.
39. V. P. R. # *Br. Oracle, qu. 51.
40. V. P. Ra. of cir- \\ cum. ☉(sun). # }Gent. Diary, 1767, qu. 300.
41. V. P + s. S - s: \\ m - n. # }Gent. Diary, 1751, qu. 108.
144[4]11
42. V. S or s. S x s: \\ m x n, ſegts. by L. # }Brit. Oracle, qu. 87. - Renaldinus, pa. 337.
43. V. S x s. m x n. # Brit. Oracle, qu. 60.
44. V. S: s. m - n. # Simpſon’s Alg. pr. 3.
# ☞ When V and S: s are given, the triangle is given in \\ ſpecies, and therefore may be conſtructed with any other \\ datum which does not affect the angles.
45. V. S: s. R. # Br. Oracle, qu. 21.
46. V. S±s. m: n. # Simpſon’s Alg. pr. 7, 9. - *York Mis. Cur. qu. 46.
47. V. S±s. m - n. # Simpſon’s Alg. pr. 6, 8. - *Oughtred ch. 19, pr. 12, 13. - \\ *Ghetaldus var. pr. 8, 9. - Idem, pr. 14, 15. - Renaldinus \\ pag. 319, 529.
48. V. S±s. Ar. # Anderſon var. prob. 22. - *Simp. Alg. pr. 34.
49. V. S±s. L. # Deducible from Simpſon’s Geo. pr. 19, 20.
50. V. S - s. R. # *Br. Oracle, qu. 20.
51. V. S2 + s2. λ. # Hutton’s Mis. qu. 24.
52. V. m. n. # Simpſon’s Alg. pr. 4. - York Mis. Cur. qu. 17, Caſe 3d. - \\ *Rudd’s Pract. Geo. part 2d, qu. 2. - Court Mag. Feb. 1763, \\ m and n being ſegts. by L.
53. V. Ar. Per. # *D’Omerique L. III. pr. 34. - *Simpſon’s Alg. pr. 35. - \\ *Simpſon’s Sel. Ex. pr. 30. - Arith. Uni. Caſtillionei, pr. 8, \\ *3. - *Rudd’s Prac. Geo. part 2d, qu. 7. - L. Diary, 1761, \\ qu. 480. - Mathematician, qu. 49. - Gent. Diary, 1744, qu. \\ 40. - *Wolfius’s Alg. pr. 113.
54. V. Ar. Side of \\ ins. ◻. # }L. Diary, 1763, qu. 507. - Court Mag. Dec. 1762.
55. V. Per. L. # Reducible to V. P. Per.
56. V. Per. R. # *Rudd’s Prac. Geo. part Iſt, qu. 19. - Reducible to \\ V. B. S + s.
57. V. L. m: n, \\ ſegts. by L. # }Mis. Cur. Math. V. I. qu. 26.
58. V. L. m - n, \\ ſegts. by L. # }L. Diary, 1773, qu. 662.
145[5]11
59. V. L. Side of \\ ins. ◻. # }T. and Country Mag. Nov. and Dec. 1772.
60. V. λ. Ra. of \\ circ. ☉(sun). # }*Gent. Diary, 1766, qu. 282. - Reducible to V. B. λ.
61. B. P. S: s. # D’Omerique L. 1. 49. - Vieta 1ſt. App. Apoll. Galli, pr. 2. - \\ Ghetaldus de Res. & Comp. L. II. pag. 48, 49, &c - Simp. Alg. \\ pr. 23. - Simp. Sel. Ex. pr. 19. - Simp. Geom. pr. 13. - \\ Schooten, pr. 22. - Turner’s Math. Exer. prob. 57. - Hutton’s \\ Miſcel. qu. 58.
62. B. P. S + s. # D’Omerique L. III. 25. - Vieta ib. pr. 3. - Gregorius a S. \\ Vinc. pr. 82, pag. 48. - Anderſon var. prob. 20, Cor. - \\ Simp. Alg. pr. 77. - Simp. Geom. pr. 15. - Arith. Un. \\ Caſtillionei, pr. 9.
63. B. P. S - s. # D’Omerique L. III. 26. - Vieta ib. pr. 4. - Simp. Alg. pr. \\ 76. - Simp. Sel. Ex. pr. 20. - Simp. Geom. pr. 15.
64. B. P. S x s. # D’Omerique L. I. 31. - Vieta ib. pr. 1. - Ghetaldus de Res. \\ & Comp. pag. 52. - Simpſ. Sel. Ex. pr. 21. - Simp. Geom. \\ pr. 16.
65. B. P. A - a. # Simpſ. Alg. pr. 15.
66. B. P. L. # Anderſon var. prob. 21.
67. B. P. Supp. of \\ A=Comp. of a. # }Math. Mag. No. I. pr. 1.
68. B. P: S. S + s. # D’Omerique L. III. 30.
69. B. S - P. s - P. # Ghetaldus de Res. & Comp. pag. 264.
70. B. ∠ of P with S. s. # Imperial Mag. Sep. 1760.
71. B. S±s. A or a. # Reducible to V. B±S. s.
72. B. S: s. A. # Gent. Diary, 1749, qu. 81.
73. B. S: s. A=2a. # Hutton’s Miſcel. qu. 16.
74. B. S: s. A - a. # D’Omerique L. I. 50. - Simp. Alg. pr. 14.
75. B. S±s. A - a # Simp. Alg. pr. 12, 13.
76. B. S + s. m: n. # D’Omerique L. III. 32. - Renaldinus, pag. 331.
77. B. S - s. Ar. # York Mis. Cur. qu. 32. - See B. P. S-s.
146[6]11
78. B. S±s. Side \\ of ins. ◻. # }Reducible to B. P. S±s.
79. B. A. Ar. # Gent. Diary, 1741, qu. 5.
80. B. A - a. λ. # *Simpſ. Algebra, pr. 59.
81. B. A or a. R. # Imperial Mag. Nov. 1760.
82. B. R. Ra. of \\ circums. ☉(sun). # }Mathematician, pr. 66.
83. B±P. A - a. m - n. # L. Diary, qu. 646.
84. B±S. All the \\ angles. # }*Britiſh Oracle, qu. 50.
85. B + s. S. n. # D’Omerique L. III. 29.
86. B - S. S±. m - n. # Oughtred ch. 19, pr. 14. - Ghetaldus var. prob. 5. - Ghe- \\ taldus de Res. & Comp. pag. 66.
87. B + S - s. Ar. Per. # Gent. Diary, 1749, qu. 89.
88. ∠ of B with L. \\ S: s. # }Town and C. Mag. 1769, pag. 606, 662.
89. ∠ of B with L. \\ S: s. λ. # }*Palladium, 1752, qu. 47.
90. ∠ of B with L. \\ m. n, \\ ſegts. by L. # }Ghetaldus var. prob. 4. - Regiomontanus de triang. L. II. \\ 33. - Miſ. Scient. Cur. pr. 27.
91. B x La max. S. s. # L. Diary, 1773, qu. 656.
92. B x λ a max. S. s. # *L. Diary, 1762, qu. 495.
93. P. P. P. # Mis. Cur. Math. Vol. I. pag. 30. - Rudd’s Pract. Geom. part \\ 2d, qu. 43.
94. P. S: s. A - a. # Simp. Alg. pr. 14.
95. P. S±s. A - a. # Simp. Alg. pr. 79.
96. P. S: s. m: n. # D’Omerique L. III. 33. - Simp. Alg. pr. 25.
97. P. S: s. m - n. # Simpſ. Alg. pr. 24.
98. P. S + s. m: n. # D’Omerique L. III. 31.
147[7]11
99. P. S + s. m - n. # Oughtred ch. 19, pr. 15. - Ghetaldus var. prob. 2. - Ghe- \\ taldus de Res. & Comp. pag. 56. - D’Omerique L. III. 27. \\ - Renaldinus, pag. 455, 456.
100. P. S - s. m - n. # Oughtred ch. 19, pr. 16. - Ghetaldus var. prob. 1. - Ghetal- \\ dus de Res. & Comp. pag. 36. - D’Omerique L. III. 28. - \\ Renaldinus, pag. 460.
101. P. A or a. Per. # Reducible to B. S + s. A or a.
102. P. A - a. m: n. # Simp. Alg. pr. 17.
103. P. A - a. m - n. # Simp. Alg. pr. 19.
104. P. m - n. Ra. \\ of circum. ☉(sun). # } Martin’s Mag. qu. 395.
105. P. L. λ. # *Mathematician, pr. 10. - *L. Diary, qu. 270.
106. P: S. P - n. m - n. # Mathematician, pr. 64.
107. P: L. S - s. \\ Ra. of cir@um. ☉(sun). # }Gent. Diary, qu. 363.
108. S. s. Ar. # Mis. Cur. Math. qu. 121. - Saunderſon, art. 333.
109. S. s. λ. # *Mathematician, pr. 9. - *Gent. Diary, 1759, qu. 186. - \\ Simp. Sel. Ex. pr. 33. - Schooten, pr. 23. - *Rudd’s Prac. \\ Geometry, part 2d, qu. 14, 16.
110. S. L. n. \\ or s. L. m, \\ ſegts. by L. # }Br. Oracle, qu. 81.
111. S: s. A. L. # Town and C. Mag. Aug. and Sep. 1770.
112. S: s. A - a. m - n. # Simp. Alg. pr. 14.
113. S±s. A - a. m: n. # Simp. Alg. pr. 17.
114. S±s. A - a. m - n. # Simp. Alg. pr. 16, 18.
115. S: s. m. n. # Simp. Alg. pr. 22.
116. S±s. m. n. # Simp. Alg. pr. 20, 21.
148[8]11
117. S - s. m - n. L. \\ Segts. by L. # }Gent. Mag. 1768, pag. 471, 570.
118. Sxm. s x n. L. \\ Segts. by L. # }L. Diary, qu. 622.
119. S - s. m - n. R. # Br. Oracle, qu. 61.
120. A or a. R. Side \\ of ins. ◻. # } *Simpſ. Sel. Ex. pr. 27.
121. A - a. Per. L. # Simp. Alg. pr. 61.
122. m. n. L. # Simp. Alg. pr. 57.
123. Per. All the \\ angles. # }Mathematician, pr. 44.
# ☞ When all the angles are given, the triangle is given in \\ ſpecies, and therefore may be conſtructed, by ſimilar triangles, \\ with any other datum.
149[9]
Continuation of the Synopsis,
Containing ſuch Data as cannot readily be expreſſed by the
Symbols before uſed without more words at length.
124. IT is required to conſtruct an iſoceles triangle ſuch, as to have its equal legs
to a given line, and moreover ſuppoſing a circle inſcribed therein, and a di-
ameter thereof drawn parallel to the baſe, and continued to meet the equal legs;
ſuch line ſhall divide the area in a given ratio.
Rudd’s Prac. Geom. part 2d, qu. 46. - Supp. to Gent. Diary, 1741, 1742, qu. II.
125. In a plane triangle, ABC, there is given the angle
26[Figure 26] at C, and the parts or ſegments of the baſe AD, AE;
to conſtruct the triangle ſo, that if BD be drawn, the
angle ABD may be a maximum, and BC to EC in
a given ratio.
L. Diary, 1773, qu. 659.
126. V. B. Line from A or a to divide S or s in a given ratio.
Town and Coun. Mag. Dec. 1772.
127. V. B. Difference of two lines from the angles at the baſe to the centre of the
inſcribed ☉(sun).
*Simpſon’s Sel. Ex. pr. 18.
128. V. B. Two lines from A and a meeting in a point O given in poſition, as alſo
the angle of a line from V to O with either of the ſides about the vertical angle.
Mis. Cur. Math. qu. 107.
129. V. A point in B. Rect. of the ſegments of B made by that point.
Simp. Sel. Ex. pr. 44.
130. Poſition of a line through V. B. S - s.
Simp. Sel. Ex. pr. 49.
131. Poſition of a line through V. B. S, line biſect. B, & s in geometrical progreſſion.
Simp. Sel. Ex. pr. 51.
132. V. S. The angle of a line from extreme of S with the baſe, and ſegment of
s cut off thereby adjacent to the baſe.
*Mis. Cur. Math. Vol. II. qu. 30.
150[10]
133. V. The ſquare of the baſe equal to the rectangle of one ſide and a given line.
General Mag. qu. 61.
134. V. P. The angle of two lines from extremes of B to middle of P.
Mathematician, pr. 65.
135. V. B. The angle of two lines from extremes of B to the middle of P.
Brit. Oracle, qu. 91.
136. V. S. The ratio of the ſquare of a line, drawn in a given direction from V to B,
to the rectangle of the ſegments of B made thereby.
Anderſoni Exercitationes Math. No. 8.
137. V. S or s. m: n, theſe being ſegts. by a line from V to B dividing V into
given angles.
Hutton’s La. Diary, qu. 139.
138. V. A line from V to B dividing V in a given ratio. Area a minimum.
L. Diary, 1761, qu. 479.
139. V. Line from A biſecting S. Line from a biſecting s.
*Simpſon’s Sel. Ex. pr. 15.
140. V. Per. Line parallel to B biſecting the area.
*Gent. Di. 1750, qu. 98. - Reducible to V. B. S + s.
141. S. s. Line from V to centre of ins. ☉(sun).
L. Diary, 1771, qu. 635.
142. Baſe. One ſide. Ratio of a line from V, making a given angle with ſaid ſide,
to alternate part of B.
Hutton’s Mis. pag. 63, Cor.
143. B. Point of contact therein of ins. ☉(sun). mxn.
Hutton’s L. Diaries, 1722, qu. 94.
144. L. A perpendicular thereto from one of the angles at the baſe. The other
angle at the baſe.
L. Diary, 1769, 1770, qu. 604.
145. L. A perpendicular thereto from A. Another from a.
L. Diary 1768, 1769, qu. 588.
146. One of the angles at the baſe. Perpendicular therefrom to oppoſite ſide a max-
imum. A line from the other angle at the baſe biſecting its oppoſite ſide.
*L. Diary, 1769, 1770, qu. 607.
147. S. s. m: n, theſe being ſegts. by a line from V to B, and the ratio of this line to
m or n.
Rudd’s Prac. Geom. part 2d, qu. 40.
151[11]
148. S. s. Line from V making an angle with S=A.
Diarian Repoſitory, qu. 197.
149. S. m. n, theſe being ſegments of B by l from V making a given angle with S.
T. and Country Mag. 1769, pag 662.
150. One angle at the baſe. The ſi le adjacent. The ratio of the other ſide to a line
drawn from V to an unknown point in B, and the length of a line drawn from
the ſaid point parallel to the given ſide to terminate in the unknown ſide.
Hutton’s Miſc. qu. 23.
151. A - a. R. Line from centre of ins. circle to middle of B.
Gent. Diary, 1771 - 2, qu. 349.
152. Three lines from the angles biſecting the oppoſite ſides.
Mathematician, pr. 48. - Simp. Sel. Ex. pr. 22. \\ - Palladium, 1752. qu. 43.
153. Sxs: mxn, ſegts by L. m - n. Angle made by L & I biſecting the baſe.
Brit. Oracle, qu. 93.
154. The baſe of an iſoceles triangle, and the diſtance of the vertical angle from
the foot of a perpendicular from one of the equal angles upon the oppoſite ſide.
Brit. Oracle, qu. 8.
155. P - n. m of an iſoceles triangle, m and n being ſegts. by a perpendicular from
one of the angles at the baſe on one of the equal ſides.
Court Mag. Sep. 1761.
156. V. Line biſecting A or a. Neareſt diſtance from V to periphery of ins. ☉(sun).
*Gent. Diary, qu. 129.
157. V. The ſegments of S made by a line drawn from A to make a given angle
with B.
*Mis. Cur. Math. Vol. I. qu. 79.
158. V. B. Line from A or a to the centre of inſcribed circle.
Mis. Sci. Cur. qu. 53.
159. B. L. Line from extremity of L parallel to S or s.
Math. Mag. No. III. prob. 7.
160. V. The ſegments of B made by a line dividing V into given angles.
Mis. Cur. Math. Vol. II. qu. 48.
161. V. One ſide. Ratio of the ſegments of the baſe made by a line dividing V
into given angles.
Palladium, 1756, pa: 43.
152[12]
SYNOPSIS
Of Data for Right-angled Triangles which have not yet
been conſtructed in general, the vertical angle being
ſuppoſed acute or obtuſe.
☞ The much greater part of theſe problems are purpoſely leſt without any reference.
The Compiler has ſeen an Author from whom Conſtructions to them all may
be derived, but he forbears to name him, in order to leave them as Exerciſes
for young Geometricians.
11
1. H. H x P + P2.
2. H. P2 - n2.
3. H. m2 - P2.
4. H. S±m, or H. s±n.
5. H. S2 + m2, or H. s2 + n2.
6. H. S2 + n2.
7. H. m2 - n2.
8. H. l biſecting A or a. # Univer. Muſeum, July, 1767. - Ladies \\ Diary, 1772, qu. 633, Cor. - Mis. \\ Scient. Curioſa, pag. 196, Cor. II.
153[13]11
9. H2: S x s. Any other datum.
10. H2: S2 + m2, or H2: s2 + n2. any other.
11. H + P. H x P.
12. H + P. H x P + P2.
13. H + P. S2 + n2.
14. H + P. m2 + n2.
15. H x P + P2. P.
16. H x P: S2 + n2. any other.
17. H x P + P2: S2 + n2. any other.
18. {1/2}H - P. S - s. # Turner’s Math. Ex. pr. 37.
19. H±S. m, or H±s. n.
20. H + S. n, or H + s. m.
21. H2 + S2: P2, or H2 + s2: P2. any other.
22. H2 + S2: S2 + m2, or H2 + s2: s2 + n2. any other.
23. H2 + S2. m, or H2 + s2. n.
24. H2 + S2. n, or H2 + s2. m.
25. H2 + S2: m2, or H2 + s2: n2. any other.
26. H x S. s. or Hxs. S. # D’Omerique L. III. pr. 37. - Vieta Geo. \\ Eff. pr. 18. - Oughtred, ch. 19, pr. \\ 25. - Herigon Geo. planæ, pr. 9. - \\ Schooten, pr. 38.
27. H x S: s2, or H x s: S2. any other
28. H + S + m. H2 + S2 + m2.
29. H + S + m. H x S + S x m + S2.
30. H + S + m. H x S + S x m.
31. H + S + m. H2 + m2.
32. H2 + S2 + m2: H x S + S x m. any other.
33. H2 + S2 + m2: H x S + S x m + S2. any other.
154[14]11
34. H2 + S2 + m2. H + m.
35. H2 + S2 + m2: H2 - m2. any other.
36. H2 + S2 + m2: H2 + 2m2. any other.
37. H2 + S2 + m2: P2. any other.
38. H2 + S2 + m2. S.
39. H2 + S2 + m2: s2. any other.
40. H2 + S2 + m2: 2S2 + m2. any other.
41. H2 + S2 + m2.: S2 + 2m2. any other.
42. H2 + S2 + m2: Sxn. any other.
43. H2 + S2 + m2. n.
44. H x S + S x m + S2. H + m.
45. H x S + S x m + S2. S.
46. H + m. H2 - m2.
47. H + m. P.
48. H + m. s.
49. H2 + 2m2: P2. any other.
50. H2±m2: S2. any other.
51. H2 - m2: S2 + 2m2. any other.
52. H2 - m2. n.
53. P. m2 + n2.
54. P2: m2±n2. any other.
55. P + m. n, or P + n. m.
56. P - n. m.
57. m - P. n.
58. P + m. m - n.
59. P + n. m - n.
60. P2 + s2: m2 - P2. any other.
61. P2 + s2: S2 + n2. any other.
155[15]11
62. P2 - n2. m2 - P2.
63. P2 - n2: S2. any other.
64. m2 - P2: s2. any other.
65. m2 - P2: S2 + n2. any other.
66. P2 - n2: S2 + n2. any other.
67. P2 - n2. m.
68. m2 - P2. n.
69. P2 - n2: m2. any other.
70. m2 - P2: n2. any other.
71. P2 - n2. m - n.
72. m2 - P2. m - n.
73. P2 - n2: m2 + 2n2. any other.
74. Pxm - P x n: S2 + n2. any other.
75. S. n, or s. m # Oughtred ch. 19, pr. II. - Ronayne’s Alg. \\ B. II. ch. 2, pr. I - Ghetaldus var. prob. 19. \\ - Idem de Res. & Comp. L. III. pr. 2. - \\ Renaldinus. pag. 518.
76. S. m - n, or s. m - n. # Oughtred ch. 19, pr. 8. - D’Omerique, L. \\ III. pr. 24. - Renaldinus, pag. 412. - Ghe- \\ taldus var. prob. 18. - Idem de Res. & Comp. \\ L. III. pr. I. - Foſter’s Math. Lug. pr. 18.
77. S2: n2, or s2: m2. any other.
78. S2: s2 + n2, or s2: S2 + m2. any other.
79. S2: 2s2 + n2, or s2: 2S2 + m2. any other.
80. S + m. n, or s + n. m.
81. S2 + m2. n, or s2 + n2. m.
82. S2 + n2. m, or S2 + n2. n.
83. S2 + n2: s2 + n2. any other.
156[16]11
84. S2 + n2. m - n.
85. S2 + n2: m2 - n2. any other.
86. S2 + n2: m2 + 2n2. any other.
87. s2 + n2: m2 - n2. any other.
88. m - n. m2 - n2.
89. Ar. Sides in arithmetical progreſſion.
Simp. Alg. pr. 36. - Simp. Sel. Ex. pr. 45.
90. Ar. Sides in geom. progreffion.
Wolfiu’s Alg. pr. 114. - Simps. Sel. Ex. pr.
46. - Vieta Iſt. App. to Apoll. Gallus, pr. 7.
91. Per. Sides in geom. progreſſion.
Simps. Alg. pr. 39.
92. I biſecting an acute angle. I from right angle biſecting the foregoing given line.
Gent. Diary, qu. 266.
93. H. Part of S adjacent to the right angle intercepted by a perpendicular to H
from middle of H.
L. Diary, qu. 633.
94. One leg and a line parallel thereto intercepted by the hypothenuſe and
the other leg being given; to determine the triangle ſuch, that the rectangle un-
der the hypothenuſe and a line from the acute angle, adjacent to the given leg,
to the point of interſection of the parallel and the other leg may be of a given
magnitude.
L. Diary. qu. 648.
N. B. In all the Nos. from 28 to 52 incluſive, S may be changed for s, and m
for n, though this be not expreſſed as in others.
FINIS.
157
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158
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159 27[Figure 27]
160
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