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A SECOND
SUPPLEMENT,
BEING
Mon$. DE FERMAT’S Treati$e on
Spherical Tangencies.
PROBLEM I.
HAVING four points N, O, M, F, given, to de$cribe a $phere which
$hall pa$s 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 circum$cribed, which will be given in
magnitude and po$ition. That this circle is in the $urface of the $phere
$ought appears from hence; becau$e if a $phere be cut by any plane, the
$ection will be a circle; but only one circle can be drawn to pa$s through the
three given points N, O, M; therefore this circle mu$t be in the $urface of
the $phere. 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 $phere
$ought will be in this line CB. From the fourth given point F let FB be
drawn perpendicular to CB, which FB will be al$o given in magnitude and
po$ition. Through C draw ACD parallel to FB, and this line will be a
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diameter of the circle given in po$ition, and therefore the points A and D
will al$o be given.
LETus now $uppo$e the thing done, and that the center of the $phere
$ought is E, which, as ob$erved before, mu$t be in the line CB. Drawing
EF, EA, ED, the$e lines mu$t be equal, $ince the points F, A, D, have
been $hewn to be in the $urface of the $phere. But the$e lines EF, EA, ED,
are in the $ame plane, $ince FB and AD are parallel, and BC perpendicular
to each of them. If therefore a circle be de$cribed to pa$s through the three
points F, A, D, who$e center is E, it will be in the line CB, and will be
the center of the $phere required.
PROBLEM II.
HAVING three points N, O, M, given, and a plane AD, to de$cribe a-
$phere which $hall pa$s through the three given points; and al$o touch the
given plane.
LET a circle ENOM be de$cribed to pa$s through the three given points,
it will be in the $urface of the $phere $ought, from what was $aid under the
former Problem. From it’s center I let a perpendicular to it’s plane IA be
erected; the center of the $phere $ought will be in this line IA; let the line
IA meet the given plane in the point A, which point will be therefore given.
From the center of the given circle ENOM, let a perpendicular to the given
plane ID be drawn, the point D will then be given, and therefore the line
AD both in po$ition and magnitude, as likewi$e the lines ID, IA, and the
plane of the triangle ADI. But the plane of the circle NOM is al$o given
in po$ition, and therefore al$o their common $ection EIF, and hence the
points E and F.
Suppo$e now the thing done, and that the center of the $phere $ought is B.
Draw BE, BF, and BC parallel to ID. Since the triangle ADI, and the
line EIF are in the $ame plane, therefore BE, BF, BC, will be in the $ame
plane. But the line ID is perpendicular to the given plane, therefore the
line BC parallel to it, will al$o be perpendicular to the given plane. Since
then a $phere is to be de$cribed 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 po$ition, in
which plane is al$o the right line AD. The que$tion is then reduced to this,
Having two points E and F given, as al$o a right line AD in the $ame plane,
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to find the center of a circle which will pa$s through the two points, and like-
wi$e touch the right line, which is the VIIth of the preceeding Problems.
PROBLEM III.
HAVING three points N, O, M, given, as likewi$e a $phere IG, to de-
$cribe a $phere which will pa$s through the three given points, and likewi$e
touch the given $phere.
The circle NOM in the $urface of the $phere $ought is given, and a per-
pendicular to its plane from it’s center FA being drawn, the center of the
$phere required will be in this line. From I the center of the given $phere
let IB be drawn perpendicular to FA, and through F, ED parallel to IB,
which, from what has been before proved, will be in the plane of the circle
NOM, and the points E and D will be given.
Suppo$e now the thing done, and that the center of the $phere required is
C. Then the lines CI, CE, CD, will be in the $ame plane, which is given, as
the points I, E, and D are given. But the point of contact of two $pheres is
in the line joining their centers; therefore the $phere $ought will touch the
$phere given in the point G, and the line IC will exceed the lines EC, ED, by
IG the radius of the given $phere: with center I therefore and this di$tance
IG let a circle be de$cribed in the plane of the lines CI, CE, CD, and it
will pa$s through the point G and be given in magnitude and po$ition; but
the points D and E are al$o in the $ame plane; and therefore the que$tion is
reduced to this, Having two points E and D given, as likewi$e a circle
IGH, to find the center of a circle which will pa$s through the two points
and likewi$e 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-
$cribe a $phere which $hall touch them all four.
IF two planes touch a $phere, the center of that $phere will be in a plane
be$ect<007>ng the inclination of the other two. And if the planes be parallel, it
will be in a parallel plane be$ecting their interval. This being allowed,
which is too evident to need further proof; the center of the $phere $ought
will be in a plane bi$ecting the inclination of two planes CB and BA; it will
likewi$e be in another plane bi$ecting the inclination of the two planes BA and
AH; and therefore in a right line, which is the common $ection of the$e two
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bi$ecting planes; let this right line be EF. Moreover, the center of the
$phere $ought will al$o be in a plane bi$ecting the inclination of the two planes
AH and GH, and the inter$ection of this la$t bi$ecting plane with the right
line EF will give a point D, which will be the center of the $phere required.
PROBLEM V.
HAVING three planes AB, BC, CD, given, and al$o a point H; to find
a $phere which $hall pa$s through the given point, and likewi$e touch the
three given planes.
SUPPOSE it done. The three planes, by what was $aid under the la$t pro-
po$ition, will give a right line in po$ition, in which will be the center of the
$phere required. Let this right line be GE, perpendicular to which from H
the given point let HI be drawn, which therefore will be given in magnitude
and po$ition. Let HI be produced, and FI taken equal to HI; the point
F will then be given. Now $ince the center of the $phere required is in the
line GE, and FH is perpendicular thereto and bi$ected thereby, and one
extreme H is _by hypothe$is_ in the $urface of the $aid $phere, the other extreme
F will be $o too. Nay even a circle de$cribed with I center and IH radius
in a plane perpendicular to GE will be in the $aid $pherical $urface. Here
then we have a circle given in magnitude and po$ition, and taking any one
of the given planes AB, by an evident corollary from Problem II. of this
Supplement, a $phere may be de$cribed which will touch the given plane,
and likewi$e have the given circle in it’s $urface; and $uch a $phere will
an$wer every thing here required.
PROBLEM VI.
HAVING three planes ED, DB, BC, given, and al$o a $phere RM, to
con$truct a $phere which $hall touch the given one, and likewi$e the three
given planes.
SUPPOSE it done, and that the $phere ERCA is the required one, viz.
touches the $phere in R, and the planes in E, A, C. Let the center of this
$phere be O; then drawing RO, EO, AO, CO, they will all be equal, and
RO will pa$s through M the center of the given $phere; 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 $uppo$ed drawn parallel to the given ones ED, DB, BC,
re$pectively. Since OR is equal to OE, and OM equal to OU, RM will be
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equal to UE: but RM is given in magnitude, being the radius of the given
$phere, therefore UE is al$o given in magnitude. And $ince OE is perpen-
dicular to the plane DE, it will be al$o 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 po$ition _by hypo_-
_the$is_, the other PU will al$o be given in po$ition. In the $ame manner it
may be proved that the planes GH, IN, are given in po$ition, and that the
lines OG, OI, are perpendicular thereto re$pectively, and each al$o equal to
OM. A $phere therefore de$cribed with center O and OM radius will touch
the three planes PU, GH, IN, given in po$ition: but the point M is given,
being the center of the given $phere. The que$tion is then reduced to this,
Having three planes given PU, GH, IN, and a point M, to find the radius
of a $phere which $hall touch the given planes, and pa$s through the given
point; which is the $ame as the preceeding Problem. [And this radius be-
ing increa$ed or dimini$hed by MR, according as R is taken in the further or
nearer $urface of the given $phere, will give the radius of a $phere which will
touch the three given planes DE, DB, BC, and likewi$e the given
$phere.]
BY a like method, when among the _Data_ there are no points, but only
planes and $pheres, we $hall always be able to $ub$titute a given point in the
place of a given $phere.
PROBLEM VII.
HAVING two points H, M, as al$o two planes AB, BC, given, to find a
$phere which $hall pa$s through the given points, and touch the given
planes.
DRAW HM and bi$ect it in I, the point I will be given, through the
point I let a plane be erected perpendicular to the right line HM, this plane
will be given in po$ition, and the center of the $phere required will be in this
plane. But becau$e it is al$o to touch the planes AB, BC, its center will be
al$o in another plane given in po$ition (by what has been proved, Prob. IV.)
and therefore in a right line which is their inter$ection, given in po$ition,
which let be GE; to which line GE from one of the given points M demit-
ting a perpendicular MF, it will be given in magnitude and po$ition, and
being continued to D $o that FD equals MF, the point D will be given;
and, from what has been proved before, will be in the $pherical $urface.
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Thercfore there are given three points H, M, D, as likewi$e a plane AB,
or AC, through which points the $phere is to pa$s, and al$o touch the given
plane. Hence it appears that this Problem is reduced to the IId of this
Supplement.
BEFORE we proceed, the following ea$y Lemmas mu$t be premi$ed.
LEMMA I.
LET there be a circle BCD, and a point E taken without it, and i$ from
E a line EDOB be drawn to pa$s 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 $uppo$e a $phere who$e center is O,
and one of who$e great circles is ACDB; if from the $ame point E a line
ECA be any-how drawn to meet the $pherical $urface in the points C and A,
I $ay the rectangle AEC will $till be equal to the rectangle BED. For if we
$uppo$e the circle and right line ECA to revolve upon EDB as an immove-
able axis, the lines EC and EA will not be changed, becau$e the points C
and A de$cribe circles who$e planes are perpendicular to that axis; and
therefore the rectangle AEC will in any plane be $till equal to the rectangle
BED.
LEMMA II.
BY the $ame method of rea$oning, the Vth Lemma immediately preceed-
ing Problem XIII, in the Treati$e of Circular Tangencies, may be extended
al$o to $pheres, viz. that in any plane ($ee the Figures belonging to that
Lemma) MG X MB = MH X MA. And al$o that MF X MC = ME X MI.
LEMMA III.
LET there be two $pheres YN, XM, through who$e centers let the right
line RYNXMU pa$s, and let it be as the radius YN to the radius XM, $o
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 $phere OTS be de$cribed to pa$s through the points T and S, and to
touch one of the given $pheres XM as in O, it will al$o touch the other
given $phere YN. For joining UO, and producing it to meet the $urface of
the $phere OTS in Q; the rectangle QUO = the rectangle SUT, by
Lemma I. but the rectangle SUT = the rectangle RUM. by _con$truction_,
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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 $urface of the $phere YN.
Therefore the point Q is in the $urface of the $phere YN; it is therefore
common to the $pheres YN and OTS; and I $ay that the$e $pheres touch in
the $aid point Q. For from the point U let a line UZ be drawn in any
plane of the $phere OTS, and being produced let it cut the three
$pheres in the points Z, D, H, K, P, B. The rectangle ZUB in the
$phere OTS is by Lemma I. and II. equal to the rectangle DUP terminated
by the $pheres XM and YN. But DU is greater than ZU, becau$e the
$pheres XM and OTS touch in the point O, and therefore any other line from
U but UO mu$t meet the $urface of OTS before it meets the $urface XM.
Since then ZUB = DUP, and DU is greater than ZU, UP mu$t be le$s
than UB, and the point B will fall without the $phere YN; and by the
fame rea$on, all other points in the $urface of the $phere OTS, except the
point Q.
THE Demon$tration is $imilar and equally ea$y in all ca$es, whether the
$pheres touch _exlernally_ or _internally_.
LEMMA IV.
LET there be a plane AC, and a $phere FGD through who$e center O let
FODB be drawn perpendicular to the plane, and from F any right line
FGA cutting the $phere in G and the plane in A; I $ay that the rectangle
AFG = the rectangle BFD. For let the given $phere and plane be cut by
the plane of the triangle ABF, the $ection of the one will be the circle GDF,
and of the other the right line ABC. Since the line FB is perpendicular
to the plane AC, it will be al$o to the right line AC. Having then a circle
FDB, and a right line AC in the $ame plane; and a line FDB pa$$ing thro'
the center perpendicular to AC, join D and G, and in the quadrilateral
figure ABDG the angles at B and G being both right ones, it will be in a
circle, and the rectangle AFG = the rectangle BFD; and the $ame may be
proved in any other $ection of the $phere.
LEMMA V.
LET there be a plane ABD and a $phere EGF, through who$e center O
let FOEC be drawn perpendicular to the plane, and in any other plane let
FHI be drawn $o that the rectangle IFH = the rectangle CFE: if through
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the points I and H a $phere be de$cribed which touches the plane AC, I $ay
it will al$o touch the $phere EGF. From F draw FB to the point of contact
of the $phere and plane, and make the rectangle BFN = the rectangle CFE,
and the point N will be in the $urface of the $phere EGF, by Lemma IV.
But the rectangle CFE, by _con$truction_, = the rectangle IFH; therefore
IFH = BFN, and the point N will be al$o in the $ur$ace of the $phere IHB.
It remains then to be proved that the$e $pheres touch in N, which is very ea$y
to be done. For from the point F through any point R in the $pherical $ur-
face EGF let the line FR be drawn, which may cut the $pherical $urface
IBH in L and P, and the plane AC in K. The rectangle KFR = the rect-
angle CFE, by Lemma IV. = the rectangle IFH, by _con$truction_, = the
rectangle PFL. Since then KFR = PFL, and KF is greater than PF, be-
cau$e the $phere IHB touches the plane AC in B, therefore FR is le$s than
FL, and the point R is without the $phere IHB, and the $ame may be $hewn
of every other point in the $pherical $urface EGF, except the point N.
The$e Lemmas, though they be very ea$y, are very elegant and valuable,
e$pecially the IIId and Vth. In the IIId. though there be an in$inite num-
ber of $pheres which, pa$$ing through the points T and S, may touch the
$phere XM, yet they will all al$o touch the $phere YN, by what is there
proved. In the Vth, though there be an infinite number of $pheres which,
pa$$ing through the points I and H, may touch the plane AC, yet they will
all al$o touch the $phere EGF, by what is there proved.
We $hall now be able to go through the remaining Problems with ea$e.
PROBLEM VIII.
Let there be given a plane ABC, and two points H and M, and al$o a
$phere DFE; to find a $phere which $hall pa$s through the given points, and
touch the given plane, and likewi$e the given $phere.
Through the center O of the given $phere let EODB be demitted perpen-
dicular to the g<007>ven plane ABC, and let HE be drawn, and make the rect-
angle HEG equal to the rectangle BED, and G will then be given. Find
then a $phere, by Problem II. which $hall pa$s through the three points M,
H, G, and touch the plane ABC, and it will be the $phere here required.
For it pa$$es through the points M and H, and touches the plane ABC,
by _con$truction_; it likewi$e touches the $phere DFE, by Lemma V. For
$ince the rectangle HEG = the rectangle BED, every $phere which pa$$es
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through the points H and G, and touches the plane ABC, touches likewi$e
the $phere DFE.
PROBLEM IX.
LET there be given two $pheres AB, DE, as al$o two points H and M;
to find a $phere which $hall pa$s through the two given points, and likewi$e
touch the two given $pheres.
LET the right line AF be drawn pa$$ing through the centers of the
$pheres, and as the radius AB is to the radius DE, $o make BF to EF, and
the point F will be given. Make the rectangle HFG = the rectangle NFA,
and the point G will be given. Now having given three points M, H, G,
as al$o a $phere DE; find a $phere by Problem III, which $hall pa$s through
the given points, and touch the given $phere; and, by Lemma III, it will be
the $phere here required.
PROBLEM X.
LET there be given two planes AB, BD, a point H, and a $phere
EGF; to find a $phere which $hall pa$s through the given point, and touch
the given $phere, as al$o the two given planes.
THROUGH the center O of the given $phere let a perpendicular to either of
the given planes CEOF be demitted, and make the rectangle HFI = the
rectangle CFE. Then having given the two points H and I, as al$o the
two planes AB, BD; find a $phere, by Problem VII, which $hall pa$s
through the two given points, and likewi$e touch the two given planes; and,
by Lemma V, it will be the $phere required.
PROBLEM XI.
LET there be given a point, a plane, and two $pheres; to find a $phere
which $hall pa$s through the point, touch the plane, and al$o the two
$pheres.
This Problem, by a like method of rea$oning, is immediately reduced to
the VIIIth, where two points, a plane, and a $phere are given, and that by
means of the Vth Lemma. But if you chu$e to u$e the IIId Lemma, it will
be reduced to the $ame Problem by a different method, and a different
con$truction.
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PROBLEM XII.
LET there be given a point and three $pheres, to $ind a $phere which $hall
pa$s through the point, and touch all the three $pheres.
WE a$$ign no figure to this Problem al$o, becau$e, by help of Lemma III,
it may immediately be reduced to Problem IX, where two points and two
$pheres are given.
PROBLEM XIII.
LET there be two planes, and al$o two $pheres given; to find a $phere
which $hall touch the planes, as al$o the $pheres.
Suppo$e the thing done. If therefore we imagine another $pherical $urface
parallel to that which is required, and which we now $uppo$e found, and
who$e radius is le$s than it's by the radius of the le$$er of the two given
$pheres; this new $pherical $urface will touch two planes parallel to the two
given ones, and who$e di$tance therefrom will be equal to the radius of the
le$$er of the given $pheres; it will al$o touch a $phere concentric to the
greater given one who$e radius is le$s than it's by the radius of the le$$er given
one; and it will likewife pa$s through the center of the le$$er given one.
The Que$tion is then reduced to Problem X, where a point, two planes and
a $phere are given.
PROBLEM XIV.
HAVING three $pheres and a plane given; to find a $phere which $hall
touch them all.
BY a like method to what is u$ed in the preceeding, and in the VIth Pro-
blem, this is reduced to Problem XI, where a point, a plane, and two
$pheres are given.
PROBLEM XV.
HAVING four $pheres given; to $ind a $phere which $hall touch them all.
SUPPOSE the thing done. As, in the treati$e of Circular Tangencies, the
la$t Problem, where it is required, having three circles given, to find a fourth
which $hall touch them all, is reduced to another, where a point and two
circles are given; $o al$o this, by a like method, and $imilar to what has been
u$ed in the preceding Problems, is reduced to Problem XII, where three
$pheres and a point are given.
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THE various _Ca$es_, _Determinations_ and other _Minuliæ_ we have taken no
notice of: for if we had, this Treati$e would have very much exceeded that
to which it was intended as a Supplement.
Synop$is of the PROBLEMS.
N. B. A point is repre$ented by., a plane by 1, and a $phere by 0.
1. .... # 4. 1111 # 15. 0000
2. ...1 # 5. 111. # 12. 000.
3. ...0 # 6. 1110 # 14. 0001
7. ..11 # 13. 1100 # 9. 00..
8. ..10 # 10. 11.0 # 11. 00.1
|
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THE
TWO BOOKS
OF
APOLLONIUS PERGÆUS,
CONCERNING
DETERMINATE SECTION,
As they have been Re$tored by
WILLEBRORDUS SNELLIUS.
By JOHN LAWSON, B. D. Rector of Swan$combe, Kent.
TO WHICH ARE ADDED,
THE SAME TWO BOOKS, BY WILLIAM WALES,
BEING
AN ENTIRE NEW WORK.
LONDON:
Printed by G. BIGG, Succe$$or to D. LEACH.
And $old by B. WHITE, in Fleet-Street; L. DAVIS, in Holborne; J. NOURSE, in the
Strand; and T. PAYNE, near the Mews-Gate.
MDCC LXXII.
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[0059]
ADVERTISEMENT.
SINCE the publication of the preceding Tract on
TANGENCIES, the Tran$lator thereof has ob$erved,
that tho$e pieces of WILLEBRORDUS SNELLIUS, which he
mentioned in his Preface thereto, are exceeding $carce
in England. His _Re$u$citata Geometria de $ectione rationis_
_& $patii_, 1607, he has never once had an opportunity
of $eeing; but $uppo$ing this $hould in a $hort time be
lo$t, more than ample amends is made by what
Dr. HALLEY has done on the $ame $ubject. Le$t the
other Tract, _De Sectione Determinatâ_, $hould undergo
the $ame fate with the original APOLLONIUS, he was
determined to re$cue it therefrom, or re$pite it at lea$t
for $ome time, by putting it into an Engli$h dre$s.
While he was doing this, he happened to communicate
the piece to $ome friends; one of whom has ventured,
after SNELLIUS, on this $ubject, and he pre$umes with $ome
$ucce$s, as every Reader will allow, when he peru$es the
Propo$itions here printed after tho$e of SNELLIUS. Yet,
notwith$tanding this, the Editor per$i$ted in his re$olu-
tion of printing his tran$lation of SNELLIUS, as the work
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has much merit, and was in danger of being lo$t, and as he
was the fir$t that con$tructed _Quadratic Equations_ after
this particular manner, as Dr. SIMSON ob$erves in his
Note on EUC. VI. 28 and 29.
The Editor leaves his Friend to $peak for him$elf
in relation to what he has done, and tru$ts that the
candid Reader will not think more meanly of his
performance from the mode$t manner in which he
$peaks of it him$elf.
[0061]
EXTRACT from PAPPUS's Preface to his Seventh Book
in Dr. HALLEY's Tran$lation.
DE SECTIONE DETERMINATA II.
HIS $ubjiciuntur libri duo de _Sectione Determinatâ,_
quas etiam ad modum præcedentium unam pro-
po$itionem dicere liceat, $ed disjunctam: quæ huju$-
modi e$t. “ Datam rectam infinitam in uno puncto $e-
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 $ub _aliâ_ unâ interceptâ & datá quâdum;
vel etiam ad contentum $ub duabus _aliis_ interceptis:
idque ad quam partem velis punctorum datorum.”
Hujus autem, qua$i bis disjunctæ, & intricatos Dio-
ri$mos habentis, per plura nece$$ario facta e$t demon-
$tratio. Hanc autem dedit _Apollonius_ communi methodo
tentamen faciens, ac $olis rectis lineis u$us, ad exemplum
$ecundi libri Elementorum primorum _Euclidis:_ ac rur-
$us idem demon$travit ingenio$e quidem, & magis ad
in$titutionem accomodate, per $emicirculos. Habet
From hence it appears that EUCLID'S were called the _fir$t_ Elements,
and that the other Analytical Tracts, recited by PAPPUS, were called the
_$econd_ Elements.
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autem primus liber Problemata $ex, Epitagmata, _$ive_
_Di$po$itiones punctorum,_ $edecim; Diori$mos quinque:
quorum quatuor quidem Maximi $unt, Minimus vero
unus. Sunt autem maximi, ad $ecundum Epitagma $e-
cundi problematis; item ad tertium quarti problematis;
ad tertium quinti & ad tertium $exti. Minimus vero
e$t ad tertium Epitagma tertii problematis.-Secundus
liber de Sectione Determinatâ tria habet Problemata,
Di$po$itiones novem, Determinationes tres; e quibus
Minima $unt ad tertium primi, ut & ad tertium $ecun-
di; Maximum autem e$t ad tertium tertii problematis.
-Lemmata habit liber primus XXVII, $ecundus vero
XXIV. In$unt autem in utroque libro de Sectione
determinatâ Theorenata octoginta tria.
[0063]
THE
PREFACE.
HAD not a motive more prevalent than Cu$tom induced me to
$ay $omething by way of Preface to the Performance which
I herewith offer to the Public, the difficulty I find in doing it with
propriety, would have determined me to remain entirely $ilent.
The $ubject has employed the Pen of one of the able$t Geometers
of the la$t Century; it may therefore $eem very pre$umptuous, in
me at lea$t, who am but young in the$e matters, to attempt it after
him. To obviate, if po$$ible, this Cen$ure is my only intention
here; and I hope I $hall not be deemed impertinent, if I attempt to
$hew wherein I apprehend I have come nearer to the great original
than he hath done.
[0064]
PAPPUS, in his preface to the $eventh Book of _Matbematical_
_Collections_, tells us that this Tract of APOLLONIUS was divided into
two Books; that the fir$t Book contained $ix Problems, and the
$econd three: now SNELLIUS' whole work contains but four; and
it $eemed to me difficult to $hew how tho$e could contain the $ub-
$tance of nine, and yet the $ix fir$t have $ixteen _Epitagmas,_ or ge-
neral Ca$es, and the three la$t nine. I fir$t, therefore began with
inquiring whether, or no, other Problems could not be found,
wherein the $ection of an indefinite $traight line is propo$ed to be
effected, “So, that of the $egments contained between the point of
$ection $ought, and given points in the $aid line, either the $quare
on one of them, or the rectangle contained by two of them, may
have a given ratio to the rectangle contained by one of them and a
given external line, or to the rectangle contained by two of them;“
as is de$cribed by PAPPUS.
In this inquiry it $oon occurred to me, that the three problems
which make my fir$t, $econd 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 $hould. Hence I
concluded, that there were in the$e, $ome, more general than tho$e
of APOLLONIUS, which ought therefore to be divided.
My next bu$ine$s was, if po$$ible, to find out the order in which
APOLLONIUS had arranged them: and here, with re$pect to the
fir$t Book, I had no other information to guide me, but what is to
[0065]
be met with in the above mentioned Preface of PAPPUS; where he
tells us that in the $ix _Problems_ of _Book_ I. there were “Sixteen
_Epitagmas,_ or general Ca$es, five _Determinations;_ and of the$e,
four were _Maxima,_ and one a _minimum_: That the _maxima_ are at the
$econd _Epitagma_ of the $econd _Problem_, at the third of the fourth,
_the third of the fifth, and the third of the $ixth; but that the minimum_
_was at the third Epitagma of the third problem.”_ It moreover $eem-
ed rea$onable to me, that the$e Problems wherein the fewe$t points
are given, would be antecedent to tho$e wherein there were more;
and of the$e wherein the number of given points are the $ame, that
tho$e would be prior to the others, wherein there was a given ex-
ternal line concerned: and la$tly, that when the number of given
points were two, the $econd Ca$e, or Epitagma, would naturally
be when the required point O is $ought between the two given
ones.
Now the three new Problems, together with the three fir$t of
SNELLIUS, making exactly $ixteen Epitagmas, _viz_. one in the fir$t,
and three in each of the others; it $eemed highly probable, that
the$e compo$ed the fir$t book. Al$o that the _Problem_, wherein
only one point was given, would be the fir$t; and it $eemed ea$y
to a$$ign the $econd, becau$e it is the only one wherein the _limita_-
The words which are in Italics were entirely omitted in SNELLIUS'S Extract
from Pappus, both in the Greek and Latin, and are read with $ome variation in
COMMANDINE'S tran$lation; 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 ri$k in $ub$cribing to the opinion
of $o excellent a Geometer.
[0066]
_tion_ is at the $econd _Epitagma_; and farther, the Limiting Ratio is
therein a maximum, as it ought. Again, the Problem, wherein
it is propo$ed to make the $quare 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 $ought beyond (E) that of the given ones
which bounds the $egment concerned in the con$equent term of
the ratio; which, therefore, I apprehend mu$t have been the third
_Epitagma_, and if $o, this of cour$e mu$t have been the third _Pro_-
_blem_: and as there remains only one wherein the number of given
pointsare two, I make that the fourth. With re$pect to the fifth and
$ixth Problems, in which three points are given, it $hould $eem
that that would be the fir$t in order, wherein there is a given ex-
ternal line concerned.
But it $hould, by no means, be di$$embled that objections may be
brought again$t the identity, and arrangment of $ome of the$e Pro-
blems. For fir$t, PAPPUS no where expre$sly $ays that APOLLO-
NIUS compared together two $quares, wherefore, if this cannot be
implied, the identity of the fourth _Problem_ is deeply $truck at:
and moreover, this fourth _Problem_ perhaps cannot with propriety,
be $aid to have its limiting ratio either _maximum_ or _minimum_, un-
le$s the ratio of equality, can be admitted as $uch. La$tly, in the
fifth _Problem_, the $aid limiting ratio is a _minimum_, and not a _maxi_-
_mum_ as it is $aid to have been by PAPPUS: either, therefore, a
[0067]
mi$take mu$t be admitted in this Author, or the fifth Problem is
wrong placed. I am not prepared farther to obviate the$e objec-
tions, and only mention them to $hew that although I $aw them
in their full force, I could by no means agree, that they are pow-
erful enough to overturn tho$e already advanced for what I have
done.
I come now to _Book_ II, which if I am not much mi$taken, was
entirely employed about what SNELLIUS makes his fourth _Problem._
In this I am confirmed not only by the account which PAPPUS gives
in his Preface, but much more by the _Lemmas_ of APOLLONIUS
which he hath left us. For we there find that LEMMA 21, where-
in is a$$igned the lea$t ratio which the rectangle contained by AO
and UO can bear to that contained by EO and IO, when O is $ought
between the two mean points of the four given ones, is $aid to be
concerned in determining the μοναχὴ, or _$ingle Ca$e_ , of _Problem_ 1.
This _Problem_ therefore of APOLLONIUS contained only tho$e
Ca$es of the general one, where O is $ought between the two mean
points. In like manner, we gather from _Lemma_ 22, that his $e-
cond _Problem_ was concerned in determining the point O when $ought
between a mean point, and an extreme one. And la$tly, from
_Lemma_ 24, that the third _Problem_ of _Book_ II. determined
the point O when required without all the given ones.
So called, I conceive, becau$e in every other Ca$e of the third _Epitagma_,
except this extreme, or limiting one, there are two points which will $atis$y
the _Problem._
[0068]
The Limitations of the two former are $aid by PAPPUS to have
been _minimums_, and that of the third a _maximum_, in conformity,
to which, I have here made them $o; although I cannot $ee with
what propriety: each of them admitting, in $ome Ca$es, of a _maxi_-
_mum_ and in others of _minimum_, as I have intimated in a _$cbolium_ at
the end of each _Problem_. But notwith$tanding I have conformed to
the manner of APOLLONIUS in dividing this Problem into three,
which it mu$t be confe$$ed contributes much to order in enumera-
ting $uch a multitude of Ca$es, yet have I previou$ly $hewn how
the whole may be generally con$tructed at once; and that by a me-
thod, which I flatter my$elf will not be found inferior to any that
hath heretofore been given of this very intricate and general
_Problem_.
Such are the things that I have attempted, and $uch the rea$ons
for what I have done in the following little Tract. The merit due
to each of them I chearfully $ubmit (where every one ought) to
the deci$ion of the impartial Reader. In the Con$tructions, my
chief Aim was novelty and uniformity: I could have given more
$imple con$tructions to one or two of them; in particular the $ixth
of _Book_ I: but it was not my intention to give any thing that I
knew had been done before. I know of many imperfections, but
no fal$e rea$onings, and hope none will be found; but if there
$hould, I hope the candid Geometer will be more inclined to ex-
cu$e than exult, when I a$$ure him the greate$t part of the work
has been executed at different times, amid$t the hurry and perplexi-
ties which it may ea$ily be conceived attend the fitting out for a
three years Voyage to the $outh $eas.
[0069]
I cannot conclude without acknowledging, in the warme$t man-
ner, the obligations I am under to my truly worthy and ingenious
friend, the Tran$lator of SNELLIUS; for the great pains and trou-
ble he hath taken to furni$h me with tran$lations from various Au-
thors, which my utter want of the Greek, and little acquaintance
with the Latin Language made ab$olutely nece$$ary to me: And
after all, had it not been for his kindne$s, this attempt might $till
have remained in as great ob$curity as its Author.
[0070]
[0071]
PROBLEMS
CONCERNING
DETERMINATE SECTION.
PROBLEM I.
TO cut a given indefinite right line in one point, $o that of the $egments
intercepted between that point and two other points given in the inde-
finite right line, the $quare of one of them may be to the rectangle under the
other and a given external right line, in a given ratio.
In the given indefinite right line let be a$$igned the points A and E, it is then
required to cut it in the point O, $o that
AO
<_>2 may be to OE into a given
line AU in the ratio of R to S; which ratio let be expre$$ed by AI to AU,
$etting off AI from A either way, either towards E or the contrary; and
then from A and I erect two perpendiculars AY equal to AE, and IR
equal to AI, and the$e on the $ame $ide of the given indefinite line, if AI
was $et off towards E; but on oppo$ite $ides, if AI was $et off the other way.
The former con$truction I will beg leave to call _Homotactical_, and the latter
_Antitactical_. Let now the extremities of the$e perpendiculars Y and R be
joined, and upon YR as a diameter let a circle be de$cribed, I $ay that the
inter$ection of this circle with the given indefinite line $olves the Problem.
If it inter$ects the line in two places, the Problem admits of two Solutions;
[0072]
but if it only touches, then only of one; if it neither touches nor cuts, it is
then impo$$ible.
DEMONSTRATION. Let a point of inter$ection 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 $imilar.
Hence AY = AE: AO:: OI: IR = AI
And by div . or comp . EO: AO:: AO: AI
And
AO
<_>2 = EO x AI
Therefore AO
<_>2 (= EO x AI): EO x AU:: AI: AU:: R: S
Q. E. D.
This Problem admits of two Ca$es. The 1$t 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 $ide of the required $quare [fig. 1. 2.] and here the LLMITATION is,
that AI mu$t not be given le$s than four times AE, as appears from fig. 2.
for AE: AO:: OI: AI; and here OI being the half of AI, AE will be
the half of AO, or the fourth part of AI. In this Ca$e the _Homotactical Con_-
_$truction_ is u$ed.
CASE II. Is when AO the $ide of the required $quare is part of OE the co-
efficient of the given external line AU, [fig. 3.] and this is unlimited, for here
the _Anlitactical Con$truction_ is u$ed. Or if O be required between A and E,
this is effected by the $ame _Con$truction_.
LEMMA I.
If from the extremes of any diameter perpendiculars be let fall upon any
Chord, I $ay that the $egments of the$e perpendiculars intercepted by like Arcs
are equal, and moreover al$o the $egments of the Chords them$elves.
That YO is equal to IU may be thus $hewn. Having joined YI, the
angle IYE is a right one, being in a $emicircle, and the angle at O is right by
_hypotbe$is;_ hence YI is parallel to the Chord, and YOUI is a parallelogram,
and the oppo$ite $ides YO and IU will be equal. In the $ame manner OE is
proved equal to UL. And as to the $egments of the Chord, it is thus $hewn.
By Euc. III. 35. and 36, the rect. EOY = rect. SOR, and rect. LUI = rect.
SUR. But, by what has been ju$t proved, rect. EOY = rect. LUI; hence
rect. SOR = rect. SUR, and the $egments SO and OR are re$pectively equal
to the $egments UR and SU.
[0073]
LEMMA II.
If of four proportionals the $um of two, being either extremes or means, be
greater than the $um of the other two; then I $ay the$e will be greate$t and
lea$t of all.
This is the _conver$e_ of Euc. V. 25. and may be thus demon$trated. Draw
a circle who$e diameter may be equal to the greater $um; and in it in$cribe the
le$$er $um IO, which will therefore not pa$s through the center, and let the
parts be IU and UO; then through U draw a diameter AUE, and the other
two terms will be AU and EU, of which AU is greate$t of all and EU lea$t of
all, and IU and UO of intermediate magnitude, by Euc. III. 7.
LEMMA III.
If of four proportionals the difference of two, being either extremes or
means, be greater than the difference of the other two, then I $ay the$e will be
the greate$t and lea$t of all.
This is demon$trated in the $ame manner as the preceding by Euc. III. 8.
PROBLEM II.
To cut a given indefinite right line in one point, $o that, of the three $eg-
ments intercepted between the $aid point and three points given in the $ame in-
definite right line, the rectangle under one of them and a given external right
line may be to the rectangle under the other two in a given ratio.
In the given indefinite line let the a$$igned points be A, E, I. It is then
required to cut it again in the point O, $o 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 $et off IU = AE the contrary way from A; but if A be the
middle point, then $et it off towards A. Then from U $et off UN = S the
con$equent of the given ratio, either towards A, or the contrary way; for as
the Ca$es vary, it’s po$ition will vary. From A and N erect perpendiculars
AY and NM to the given indefinite right line equal to AE and AI re-
$pectively, and the$e _bomotactical_ if A be an extreme point, but _antitactical_ if
A be the middle point of the three given ones. Join the extremes of the$e
perpendiculars Y and M, and upon YM as a Diameter de$cribe a circle. I $ay
that the inter$ection of this circle with the given indefinite line $olves the
Problem. If it inter$ects the line in two points, then the Problem admits of
[0074]
two $olutions; if it only touches, then but of one; if it neither cuts nor
touches, it is then impo$$ible.
DEMONSTRATION. Let the point of inter$ection then be O or _o_. We $hall
have, by Lemma I. AO x ON = MN x NK = MN x AY = AI x AE, by
_con$truction._ Let now from N be $et off NL = AI in the $ame direction as A
is from I; then by what has been demon$trated NL: ON:: AO: AE.
And by Divi$ion or Compo$ition OL: ON:: OE: AE
And by Permutation OL: OE:: ON: AE
But by what has been proved ON: AE:: AI: AO
Therefore by Equality OL: OE:: AI: AO
And by Divi$ion or Compo$ition LE: OE:: OI: AO
And LE x AO = OE x OI
But LE = NU. for NL was put = AI, and IU = AE
Hence NU x AO = S x AO = OE x OI
And R: S:: R x AO: S x AO or OE x OI
Q. E. D.
This Problem may be con$idered as having two EPITACMAS, the fir$t,
when the $egment a$$igned for the coefficient of the given external line R is
terminated by an extreme point of the three given ones and the point $ought;
and this again admits of three Ca$es. The other is when the afore$aid $eg-
ment is terminated by the middle point of the three given ones and the
point $ought.
EPITAGMA I. CASE I. Let the a$$igned points be A, E, I. A an extreme
and E the middle one. And let the point O $ought ($uch that AO x R: OE
x OI:: R: S) be required to lie between A and E, or el$e beyond I, which
will ari$e from the $ame _con$truction._
Here the _Homotactical con$truction_ is u$ed, and IU as likewi$e UN is $et off
in the $ame direction as AI. And $ince AO: AE:: AI: ON, and AO + ON
is greater than AE + AI or AU, by LEMMA II. AO and ON will be the
lea$t and greate$t of all; and AO will therefore be le$s than AE, as likewi$e
A_o_ (being equal ON by LEMMA I.) greater than AI. This Ca$e is
unlimited.
CASE II. Let the a$$igned points be in the $ame po$ition as before, and let
the point O $ought be required between E and I.
Here the _con$truction_ is _Homotactical_, and UN is $et off the contraty way, viz.
in the direction IA. And $ince AO: AE:: AI: ON, and AO + ON is le$s
than AE + AI or AU, by LEMMA II. AE and AI will be the lea$t and
[0075]
greate$t of all, and AE will therefore be le$s than AO, and AI greater. And
the $ame will hold with regard to A_o_.
Here is a LIMITATION, which is this; that UN or S the con$equent of the
given ratio, $et off from R, mu$t not be given greater than the difference of
the $um of AE and AI and of a line who$e $quare is equal to four times their
rectangle [i. e. to expre$s it in the modern manner, UN mu$t not exceed AI +
AE - 4 AI x AE.] This appears by Fig. 2. to this Ca$e, the circle there
touching the given indefinite line, and pointing out the Limit.
CASE III. Let the a$$igned points be $till in the $ame po$ition, and let the
point $ought be now required on the contrary $ide of A.
Here the _con$truction_ is $till _Homotactical_, and UN is $et off the $ame way as
in the la$t Ca$e; and the LIMITATION is, that UN mu$t not be given le$s than
the $um of AI, AE, and a line who$e $quare is equal to four times their rect-
angle [or expre$$ing it Algebraically, UN mu$t not be le$s 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 $ought be required either between A and one of the
extremes, or beyond either of the extremes.
Here having $et off IU = AE toward A, you may $et off UN either way,
and u$ing the _Antitactical con$truction,_ the $olution will be unlimited. The
only difference is, that if UN be in the direction UI, two $olutions will ari$e,
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 $olutions will ari$e, 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 u$ed, as LEMMA II. was in Ca$e I. II.
COROLLARY I. If then the given ratio be that of AT to TI, or of AE to
EP $et off from A the other way, $o that EP be le$s than AE, I $ay then
that O will fall between E and P, as likewi$e ο between T and I, provided _o_
falls beyond I.
For by _con$truction_ IU = AE, and UN = PE. therefore IN = AP. But by
LEMMA I. _o_N = AO. therefore (_o_ falling beyond I by _hypotbe$is_) O will fall
beyond P; but by _hypotbe$is_ it falls $hort of E; therefore O falls between
P and E.
Next to $hew that ο will fall between T and I, we have AT: TI:: AE: EP
And by Divi$ion AT: AI:: AE: AP
Hence AT x AP = IAE or _o_ AO
[0076]
Therefore AT x AO is greater than _o_ AO
Or AT greater than A_o_.
COROLLARY II. If the three given points be I, A, E; and O falls between
A and I, $o as to make AO x PE: IOE:: AL: LI, I $ay then O will fall
beyond L.
For let us $uppo$e that O and L coincide; then by _hypotbe$is_ AL: LI::
AL x PE: IL x LE
And by the next following LEMMA IV. AL x IL: IL x PE:: AL: LE
i. e. AL: PE:: AL: LE
Hence PE is equal to LE, a part to the whole, which is manife$tly ab$urd.
LEMMA IV.
If it be as a line to a line $o a rectangle to a rectangle; then I $ay it will be
as the flr$t line into the breadth of the $econd rectangle to the $econd line into
the breadth of the fir$t rectangle, $o the length of the fir$t rectangle to the
length of the $econd.
Suppo$ition. AE: IO:: UYN: SRL.
Conclu$ion. 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
$egments is equal to that under the whole and the middle $egment, together
with the rectangle under the extremes.
DEM. AI x IE + IO x IE = AO x IE.
Hence {AI x IE + IO x IE + AE x IO \\ i. e. AI x IE + AI x IO \\ i. e. AI x EO} = AO x IE + AE x IO.
Q. E. D.
N. B. The$e two LEMMAS $ave much Circumlocution and Tautology in
the two following Propo$itions, and indeed are highly u$eful in all ca$es where
compound ratios are concerned.
[0077]
PROBLEM III.
To cut a given indefinite right line in one point, $o that of the three $eg-
ments intercepted between the $ame, and three points given, the rectangle
under two of them may be to the $quare 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, $o that OA x OE may be to
OI
<_>2 (let the $ituation of I be
what it may) in a given ratio, which ratio let be expre$$ed by EL to LI.
[And here I cannot but ob$erve with HUGO D'OMERIQUE, page 113. that
this Problem, viz. _‘To exhibit two lines in a given ratio who$e $um, or who$e_
_difference is given,’_ ought to have had a place in the Elements as a Propo$ition;
or at lea$t to have been annext as a Scholium to the 9th or 10th of the VIth
Book.] And be the $ituation of L al$o 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 $uch, that AI x OE: OI x OL:: EI: IL. And let $uch a Ca$e be
cho$en of PROBLEM II, that, according as AO is greater or le$s than AI, $o of
the three rectangles, de$cribed in LEMMA V, made by the four points E, O,
I, L, that of IO x EL may accordingly be greater or le$s than that of EI x OL.
DEMONSTRATION. On $uppo$ition then that $uch a Ca$e of PROBLEM II. is
made u$e 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 Divi$ion or Compo$ition EL x OI: OE x IL:: AO: OI
This appears from LEMMA V.
Then again by LEMMA IV, AO x OE:
OI
<_>2:: EL: IL.
Q. E. D.
This Problem has two EPITAGMAS. The fir$t wherein OI, who$e $quare is
$ought, is bounded by I an extreme point of the three given ones. And this
again admits of three Ca$es. The $econd is when the point I is the middle
point. And this again has three ca$es. And there remain two Anomalous
Ca$es, wherein Problem II. is of no u$e, which mu$t therefore be con$tructed
by them$elves.
EPITAGMA I. CASE I. Let the ratio given, EL to LI, be _inequalitatis_
_majoris_, i. e. of a greater to a le$s; and the point O $ought be required to lie
between I and the next point to it E, or el$e to lie beyond I the other way;
for the $ame _con$truction_ $erves for both. Here CASE I. of PROBLEM II. is to
[0078]
be u$ed, 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
le$s to a greater, and the point O $ought be required to lie between I and the
next point to it E; or el$e to fall beyond A the other extreme. For the $ame
_con$truction_ $erves for both. Here CASE IV. of PROBLEM II. is to be u$ed, and
the point O will fall between E and I, and _o_ beyond A, if we u$e one of the
_con$tructions_ there recited: but if we u$e the other, the points will $hift places,
as was ob$erved under that Ca$e, viz. O will fall beyond I the other way, and
_o_ between L and E.
CASE III. Let now the point O be $ought between A and E. Here $et off
the given ratio in $uch a manner that EI may be the $um of the terms, and
make u$e 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 mu$t not be le$s than IE + EL + 4 IEL.
EPITAGMA II. CASE IV. Here OI the line who$e $quare 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 $ought between I and either extreine A or E. the $ame
_con$truction_ $erving for both. The given ratio mu$t here be $et off in $uch a
manner that EI may be the _$um_ of the terms of it; and make u$e
of I$t CASE of the IId PROBLEM; with this caution, that of the two $egments
AI, IE, you cho$e the le$$er IE whereon to exhibit the given ratio; for then
it will appear by the work it$elf that O falling between E and L, _o_ will al$o
fall between A and I: otherwi$e, if AI was le$s than IE, there would want
$ome proof of this. Therefore of the two extreme given points call that E
which bounds the le$$er $egment, and then the general Demon$tration will fit
this Ca$e as well as the re$t.
CASE V. Let the given ratio of EL to LI be _inæqualitatis minoris;_ and let
the point $ought be required to lie beyond either extreme. The $ame _con-_
_$truction_ $erves for both. Here we mu$t u$e 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 $egment. I have
in the Figure made AI = IE on purpo$e to $hew that in that ca$e the point N
will coincide with A. But if IE be greater than AI, the point N will
always fall beyond A, and con$equently the point _o_ more $o.
[0079]
CASE VI. Let the given ratio of EL to LI be _inæqualitatis majoris_, and
let the point $ought be required to lie beyond either extreme. Here we mu$t
u$e the IIId CASE of the IId PROBLEM; and the DETERMINATION is that
UN (found in the $ame ratio to AI as IL is to IE) mu$t not be le$s than
IE + EL + 4 IEL.
CASE VII. Let the $ituation of O be required the $ame as in the two la$t
Ca$es, but let the given ratio be that of equality, which was there $uppo$ed
of inequality. Here the IId PROBLEM will be of no u$e, and this Ca$e
requires a particular con$truction.
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, $uch that AO x OE may equal
OI
<_>2.
CONSTRUCTION. Upon AE diameter de$cribe 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
$ay O is the point required.
DEMONSTRATION. Joining YR, the triangles SUI and SYR will be $imi-
lar, and the angle UIS or RIO = SYR. But the angle IRO made by the
tangent and $ecant = SYR in the alternate $egment. Therefore RIO = IRO,
and OR = OI. But by the property of the circle AOE =
OR
<_>2. And
therefore AOE = OI
<_>2.
Q. E. D.
The DETERMINATION is that AI mu$t be greater than IE.
CASE VIII. Whereas in the I$t and IId CASES the given ratio was that of
inequality, let us now $uppo$e it that of equality; and let the three points be
A, E, I, and E the middle one; and let a fourth O be $ought between E and
I, $uch that AOE may equal
OI
<_>2.
The CONSTRUCTION and DEMONSTRATION of this Ca$e is in every re$pect
the $ame as that of the preceeding, as will appear by comparing the figures.
PROBLEM IV.
To cut a given indefinite right line $o in one point that, of the four $eg-
ments intercepted between the $ame and four points given in the indefinite
line, the rectangle under any two a$$igned ones may be to the rectangle under
the two remaining ones in a given ratio.
[0080]
In the indefinite line let the four points be A, E, I, U. It is then required
to be cut again in O $o that OA x OU may be to OE x OI (be the Po$ition
of the four given points what they may) in the ratio of AL to LE, let the
point L fall al$o 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, $o that UI x OE may be to
OU x OL as AE to AL. And let $uch a CASE be cho$en of the IId PROBLEM
that, according as UO is required greater or le$s than UI, or their $um $hall
con$titute OI, $o of the three rectangles de$cribed in the Vth LEMMA made
by the four points E, O, A, L, that of OE x AL may accordingly be greater
or le$s than OL x AE, or their $um con$titute that of OA x EL.
N. B. U being now u$ed to repre$ent one of the given points, in all the
following Diagrams I have $ub$tituted V in the place where U was u$ed
before.
DEMONSTRATION. On $uppo$ition therefore that $uch a CASE of the
IId PROBLEM is made u$e of,
We have UI x OE: OU x OL:: AE: AL
And by Inver$ion OU x OL: UI x OE:: AL: AE
And by LEMMA IV. AL x OE: AE x OL:: OU: UI
Hence by Compo$ition or Divi$ion, &c. AL x OE: OA x LE:: OU
: OI as appears by LEMMA V.
Then again by LEMMA IV. OU x OA: OI x OE:: AL: LE
Q. E. D.
This PROBLEM has three EPITAGMAS. The I$t whereof is when of the
two a$$igned points A and U, the one of them is an extreme, and the other an
alternate mean; and this admits of three CASES. The IId is when A and U
are both of them extremes; and this has four CASES. The IIId is when of
A and U one of them is an extreme, and the other is the point next to it; and
this has three Ca$es. And there remain three more _Anomalous Ca$es_, wherein
the IId PROBLEM is of no u$e, but which may be reduced to one, as $hall be
$hewn in it's proper place.
EPITAGMA I. CASE I. Let A the fir$t a$$igned point be an extreme, and
U the $econd a$$igned point be an alternate mean; and let the point O be
$ought between the fir$t a$$igned A and the next point to it E; or between
the $econd a$$igned U and the la$t I. For the $ame _Con$truction_ $erves
for both.
[0081]
Here AE is to be made the $um of the terms of the given ratio, and we are
to u$e the IVth CASE of the IId PROBLEM, whereby O falling between
L and E, _o_ will fall beyond U; and that it will fall $hort of I appears from
the I$t COROLLARY from the IVth CASE of the IId PROBLEM.
CASE II. The given ratio being _inæqualitatis minoris_, let the point $ought
be required between the $econd a$$igned U and the $econd in order E, or be-
yond the fir$t A, which ari$es from the $ame _Con$truction_. Here AE is to be
made the _difference_ of the terms of the given ratio, and we are to u$e 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 $ought
be required between the $econd a$$igned U and the $econd in order E, or be-
yond the la$t I, which ari$es from the $ame _Con$truction_. Here AE is to be
made the _difference_ of the terms of the given ratio, and L is to be $et off the
contrary way to what it was in the la$t CASE; and we are to u$e the I$t CASE
of the IId PROBLEM, whereby O being made to fall between E and L, or
between E and U, according as L or U is neare$t to the point E, _o_ will fall
beyond I, as any one will $ee who con$iders the _Con$truction_ of that CASE with
due attention.
EPITAGMA II. CASE IV. Let the a$$igned points now be the extremes
A and U, and let O the point $ought be required now between the fir$t
a$$igned A and the next to it E, or, which is effected by the $ame _Con$truction_,
between the $econd a$$igned U and the next to it I. Here AE is to be made
the _$um_ of the terms of the given ratio, and the IVth CASE of the IId PRO-
BLEM is to be u$ed, $o that of the three points L, E, U, O being made to fall
beyond L one of the extremes, and _o_ within U the other extreme, it will appear
from the I$t COROLLARY from the IVth CASE of the $aid 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 $ought be
required between the $econd and third in order, _viz_. between E and I. Here
AE mu$t be the _difference_ of the terms of the given ratio, and L $et off to-
wards I, and the IId CASE of the IId PROBLEM u$ed, and then O, as like-
wi$e _o_, will fall between E and I, if the PROBLEM be po$$ible.
As to the DETERMINATION, $ee LEMMA VII. following.
CASE VI. The given ratio being _inæqualitatis majoris_, let the point $ought
be required beyond the la$t a$$igned, that is the la$t in order, U. Here AE
mu$t be the _difference_ of the terms of the given ratio, [and L mu$t evidently
[0082]
fall beyond U, but for a more particular DETERMINATION $ee LEMMA VII.
following] and the IId CASE of the IId PROBLEM is to be u$ed, and then O,
as likewi$e _o_, will fall beyond U.
CASE VII. The given ratio being _inæqualitatis minoris_, let the point $ought
be required to lie beyond either of the a$$igned ones, i. e. beyond either ex-
treme, the $ame _Con$truction_ $erving for both. Here AE is to be the _di$-_
_ference_ of the terms of the given ratio, and L to be $et off backwards beyond
A; and the IVth CASE of the IId PROBLEM u$ed, that $o O being made to
fall beyond U, it will appear, by the IId COROLLARY from the IVth CASE of
the $aid PROBLEM, that _o_ will al$o fall beyond A.
EPITAGMA III. CASE VIII. Let the a$$igned points A and U be now
one an extreme, and the other the point next it: and let the point $ought be re-
quired to fall between the two a$$igned ones. Here AE mu$t be the _$um_ of
the terms of the given ratio, and the IId CASE of the IId PROBLEM u$ed.
And $o O, as likewi$e _o_, being made to fall between L and U, they will
much more fall between A and U.
The LIMITATION is, that VN (found in the $ame ratio to UI as AL to AE)
mu$t not exceed LE + EU - 4 LEU.
CASE IX. The given ratio being _inæqualitatis minoris_, let the point $ought
be required between the $econd a$$igned U and the third in order E, or el$e
beyond the fir$t a$$igned A, the $ame _Con$truction_ $erving for both. Here AE
is to be the _difference_ of the terms of the given ratio, and L to be $et off be-
yond A, and the I$t CASE of the IId PROBLEM u$ed: and $o O being made
to fall between E and U, _o_ will fall beyond L, and much more be-
yond A.
CASE X. The given ratio being _inæqualitatis majoris_, let the point $ought
be required between the $econd a$$igned U and the third in order E, or el$e
beyond the la$t in order I, the $ame con$truction $erving for both. Here AE
is to be the _difference_ of the terms of the given ratio, and L to be $et off be-
yond E, and the IVth CASE of the IId PROBLEM u$ed, $o that O being
made to fall between U and E, _o_ will fall beyond I, as any one will $ee who
con$iders the _con$truction_ of that CASE with due attention.
CASE XI. As to the three _Anomalous Ca$es_, in which the IId PROBLEM is
of no u$e, and which I $aid before might be reduced to one, they are the$e:
whereas in the IId and IIId CASES, as al$o in the VIth and VIIth, and like-
wi$e in the IXth and Xth the ratio given was that of inequality, let us now
[0083]
$uppo$e it that of equality, and they may all be $olved by one _con$truction_, 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 divi$. UO: OI:: EO: AO
Hence AOU = EOI.
LEMMA VI. Let there be two $imilar triangles IAE, UAO, having their
ba$es IE and UO parallel; I $ay I$t when they are _right-angled_, that the ex-
ce$s of the rectangle EAO, under the greater $ides of each, above the rect-
angle IAU, under the le$$er $ides of each, will be equal to the rectangle
IE x OU, under their ba$es. IIdly, When they are _obtu$e-angled_, that the
$aid exce$s will be equal to the rectangle under the ba$e of one and the $um
of the di$tances of the angles at the ba$e of the other from the perpendicular,
viz. EI x
OS + US
. IIIdly, When they are _acute-angled_, that then the $aid
exce$s will be equal to the rectangle under the ba$e of one and the difference
of the $egments of the ba$e 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 $imilar, and when I$t 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 ca$e
they be _obtu$e-angled_, EAO = YAS + EY x OS by part I$t; and IAU =
YAS + IY x US by the $ame. 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 $et off YL = YI, and draw LAR
which will be equal and $imilarly 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 oppo$ite $ides of the diameter, cut the $aid diameter in O, and
A and U be the extremes of the $aid diameter, I $ay 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 $ame $ide of the diameter, and con$equently meets the diameter _produced_
in O, that then the ratio of AOU to EOI is a _Maximum_.
[0084]
DEMONSTRATION. Through S, any other point taken at plea$ure, 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 $rmilar 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
la$t rectangle we may $ub$titute MY x NF: for the triangles VLS and NRF
are $imilar, being each of them $imilar 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 $ame holds good with regard to any other point S taken between E
and I, $o that the ratio of AOU to EOI is a _Minimum_, and _$ingular_, or
what the Antients called μοναχ@.
Let now VY join the tops of two perpendiculars drawn on the $ame $ide of
the diameter, and meet the diameter produced in O; I $ay that the ratio of
AOU to EOI is a _Maximum_.
For u$ing the $ame _con$truction_ as before, it will appear that the rectangle
LSM: ESI:: VOY or AOU: EOI. And it may be proved in the $ame
manner that VSR or ASU is le$s 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 al$o equal to
VS x SN. Now VS x NR is equal to MY x NF or LV x NF from the $imi-
larity of the triangles LVS, NRF. Therefore now al$o MY x NF together
with VSR is proved equal to VS x SN. But as NK is le$s than NF, VSR
will be a le$s 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 $ame 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.
[0085]
DETERMINATE SECTION.
BOOK I.
PROBLEM I. (Fig. 1.)
In any indefinite $traight line, let the Point A be a$$igned; it is required
to cut it in $ome other point O, $o that the $quare on the $egment AO
may be to the $quare on a given line, P, in the ratio of two given $traight
lines R and S.
ANALYSIS. Since, by Hypothe$is, the $quare on AO mu$t be to the
$quare on P as R is to S, the $quare on AO will be to the Square on P as
the $quare 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 $quare on P as the $quare on R is
to the $quare on AD, or (EU. VI. 22) AO is to P as R to AD; con$e-
quently, AO is given by EU. VI. 12.
SYNTHESIS. Make AB equal to R, AC equal to S, and de$cribe on
BC a $emi-circle; erect at A the indefinite perpendicular AF, meeting the
circle in D, and take AF equal to P; draw DB, and parallel thereto FO,
meeting the indefinite line in O, the point required.
For, by rea$on of the $imilar triangles ADB, AFO, AO is to AF (P) as
AB (R) is to AD; therefore (EU. VI. 22.) the $quare on AO is to the
$quare on P as the $quare on R is to the $quare on AD; but the $quare on
AD is equal to the rectangle contained by AB (R) and AC (S) by EU. VI.
13. 17; and $o the $quare on AO is to the $quare on P as the $quare on R
is to the rectangle contained by R and S; that is (EU. V. 15.) as R is to S.
Q. E. D.
SCHOLIUM. Here are no _limitations_, nor any precautions whatever to be
ob$erved, except that AB (R) mu$t be $et off from A that way which O
is required to fall.
[0086]
PROBLEM II. (Fig. 2 and 3.)
In any indefinite $traight line, let there be a$$igned the points A and E;
it is required to cut it in another point O, $o that the rectangle contained
by the $egments AO, EO may be to the $quare on a given line P, in the
ratio of two given $traight lines R and S.
ANALYSIS. Conceive the thing done, and O the point $ought: then
would the rectangle AO, EO be to the $quare on the given line P as R is
to S, by hypothe$is. Make AQ to P as R is to S: then the rectangle AO,
EO will be to the $quare 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 $quare on P is
to the rectangle EO, P; and therefore AO is to AQ as P is to EO; con-
$equently (EU. VI. 16.) the rectangle AO, EO is equal to the rectangle
AQ, P; and hence, as the $um, or difference of AO and EO is al$o given,
the$e lines them$elves are given by the 85th or 86th of the _Data._
SYNTHESIS. On AE de$cribe 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; la$tly, draw HO perpendicular to DH, meeting the inde-
finite line in O, the point required.
For it is manife$t from the con$truction that AD and HO are equal;
hence, and EU. VI. 17, the rectangle AQ, P is equal to the $quare on HO;
con$equently equal to the rectangle AO, EO (EU. III. 35. 36) and $o
(EU. VI. 16.) AO is to P as AQ is to EO; but by con$truction AQ is to
P as R to S, therefore by compound ratio, the rectangle AO, AQ is to the
$quare 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 $quare 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_. Fir$t, when O is $ought
beyond A; $econdly, when it is $ought between A and E, and la$tly, when
it is $ought beyond E. The fir$t and la$t of the$e are con$tructed by
_Fig_. 2, and have no limitations; but in the $econd, (_Fig_. 3.) the given
ratio of R to S mu$t not be greater than that which the $quare on half
AE bears to the $quare on P: $ince if it be, a third proportional to S
[0087]
and P will be greater than one to R and half AE, and of cour$e, AQ
(a fourth proportional to S, R and P) greater than a third proportional
to P and half AE; in which ca$e the rectangle AQ, P will be greater
than the $quare on half AE, and $o 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 impo$$ible.
PROBLEM III. (Fig. 4. and 5.)
In any indefinite $traight line let there be a$$igned the points A and E;
it is required to cut it in another point O, $o that the $quare on the $egment
AO may be to the rectangle contained by the $egment EO and a given line
P, in the ratio of two given $traight lines R and S.
ANALYSIS. Suppo$e the thing done, and that O is the point $ought:
then will the $quare on AO be to the rectangle EO, P as R to S. Make
AQ to P as R is to S; then will the $quare on AO be to the rectangle EO,
P as AQ is to P; or (EU. V. 15.) the $quare 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; con$equently by compo$ition, or divi$ion, AO is to AE
as AQ is to OQ, and $o (EU. VI. 16.) the rectangle AO, OQ is equal to
the rectangle AE, AQ; and hence, as the $um or difference of AO and OQ
<007>s al$o given, the$e lines them$elves are given by the 85th or 86th of the
_Data_.
SYNTHESIS. Take AQ a fourth proportional to S, P and R, and
de$cribe 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 $ought beyond A, or between A and
E, cutting the $aid circle in H: La$tly, from H draw HO perpendicular
to DH, which will cut the indefinite line in O, the point required.
For it is plain from the Con$truction, that AD and HO are equal; and
(EU. VI. 17) the rectangle AE, AQ is equal to the $quare on AD, and
therefore equal to the $quare on HO; but the $quare on HO is equal to the
rectangle AO, OQ, (EU. III. 35. 36) con$equently the rectangle AO, OQ
[0088]
is equal to the rectangle AE, AQ; and hence (EU. VI. 16.) OQ is to AQ
as AE is to AO; therefore, by compo$ition or divi$ion, AO is to AQ as
EO is to AO; but by con$truction, AQ is to R as P is to S, and $o, 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 $quare
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_ al$o, which I $till enu-
merate as before. The fir$t and $econd are con$tructed by _Fig_. 4, where DH
is drawn through F, the center of the circle on AQ: and the$e have no limi-
tations. The third is con$tructed as in _Fig_. 5, where DH is drawn parallel
to AQ; and here the given ratio of R to S mu$t not be le$s 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
ca$e the rectangle contained by AE and AQ will be greater than the $quare
on half AQ, and con$equently AD (a mean proportional between AE and
AQ) greater than half AQ; but it is plain when this is the ca$e, that DH
will neither cut nor touch the circle on AQ, and therefore the problem is
impo$$ible.
PROBLEM IV. (Fig. 6. 7. and 8.)
In any indefinite $traight line, let there be a$$igned the points A and E;
it is required to cut it in another point O, $o that the two $quares on the
$egments AO, EO, may obtain the Ratio of two given $traight lines,
R and S.
ANALYSIS. Imagine the thing to be effected, and that O is really the
point required: then will the $quare on AO be to the $quare on EO as
R to S; or (EU. V. 15.) the $quare on AO will be to the $quare on EO
as the $quare on R is to the rectangle contained by R and S. Let DE be
made a mean proportional between EB (R) and EC (S). Then
(EU. VI. 17.) the $quare on AO will be to the $quare on EO as the
$quare on R to the $quare on DE; and $o (EU. VI. 22.) AO to EO as R to
DE; and hence both AO and EO will be given by the conver$e of _Prop_. 38.
of _Eu. Data_.
[0089]
SYNTHESIS. Make EB equal to R, EC equal to S, and de$cribe on
BC a circle; erect at E the perpendicular ED, meeting the periphery of the
circle in D; al$o at A erect the perpendicular AF equal to R; draw AD,
which produce, if nece$$ary, to cut the indefinite line, as in O, which will
be the point required.
For becau$e of the $imilar triangles AOF, EOD, AO is to EO as AF
(R) is to DE; therefore the $quare on AO is to the $quare on EO as the
$quare on R is to the $quare on DE (EU. VI. 22); but the $quare on DE
is equal to the rectangle contained by R and S; therefore the $quare on AO is
to the $quare on EO as the $quare on R is to the rectangle R, S; that is as
R is to S, by EU. V. 15.
Q. E. D.
SCHOLIUM. This Problem al$o hath three _Epitagmas,_ which I enumerate
as in the la$t. The fir$t is con$tructed by _Fig_. 6, wherein the perpendicu-
lars DE and AF are $et off on the $ame $ide of the indefinite line; the $econd
by _Fig_. 7, where they are $et off on contrary $ides, and the third by _Fig._ 8,
in which they are again $et off on the $ame $ide. The $econd has no limits;
but in the fir$t R mu$t be le$s, and in the third greater than S, for rea$ons
too obvious to be in$i$ted on; and hence, both the$e ca$es are impo$$ible
when the given ratio is that of equality.
PROBLEM V. (Fig. 9. 10. 11. 12. 13. 14. 15. 16.)
In any indefinite $traight line let there be a$$igned the points A, E and I;
it is required to cut it in another point, O, $o that the rectangle contained
by the $egment AO and a given $traight line P may be to the rectangle
contained by the $egments EO, IO in the ratio of two given $traight lines
R and S.
ANALYSIS. Conceive the thing done, and O the point $ought: then
would the rectangle AO, P be to the rectangle EO, IO as R to S. Make
IQ to P as S is to R; then will the rectangle AO, P be to the rectangle
EO, IO as P is to IQ; or (EU. V. 15.) the rectangle AO, P be to the
rectangle EO, IO as the rectangle IO, P is to the rectangle IO, IQ; and
hence (EU. V. 15. 16) AO is to EO as IO is to IQ; whence, by compo$ition
or divi$ion, AE is to EO as OQ is to IQ: therefore (EU. VI. 16.) the
[0090]
rectangle EO, OQ is equal to the rectangle AE, IQ; con$equently, as the
$um or difference of EO and OQ is al$o given, tho$e lines them$elves are
given by the 85th or 86th of the _Data._
SYNTHESIS. Take IQ a fourth proportional to R, S and P, and
de$cribe on EQ a circle; erect at E the indefinite perpendicular EK, and
take therein ED a mean proportional between AE and IQ; from D draw
DH, parallel to EQ, if O mu$t lie any where between the points E and Q;
but through F, the center of the circle on EQ if it mu$t fall without them,
cutting the $aid circle in H: La$tly, draw HO perpendicular to DH, which
will meet the inde$inite line in O, the point required.
For it is manife$t from the con$truction that ED and HO are equal; and
(EU. VI. 17.) the rectangle AE, IQ is equal to the $quare on ED, and
therefore equal to the $quare on HO; but the $quare on HO is equal to the
rectangle EO, OQ (EU. III. 35. 36.): therefore the rectangle AE, IQ is
equal to the rectangle EO, OQ; and hence (EU. VI. 16.) AE is to OE as OQ
to IQ, whence, by compo$ition or divi$ion, AO is to EO as OI to IQ; but
IQ is to P as S to R, or inver$ely, P is to IQ as R to S; and $o, by compound
ratio, the rectangle AO, P is to the rectangle EO, IQ as the rectangle IO,
R is to the rectangle IQ, S; that is (EU. V. 15 and 16.) the rectangle AO,
P is to the rectangle IO, R as EO is to S; or the rectangle AO, P is to the
rectangle IO, R as the rectangle IO, EO is to the rectangle IO, S
(EU. V. 16.) the rectangle AO, P is to the rectangle EO, IO as the rectangle
IO, R is to the rectangle IO, S; that is (EU. V. 15.) as R is to S. Q. E. D.
SCHOLIUM. This Problem may be con$idered as having three _Epitagmas_,
or general Ca$es, _viz_. when A, the point which bounds the $egment a$$igned
for the co efficient of the given line P being an extreme, O is $ought be-
tween it and the next thereto, or beyond all the points with re$pect to A;
$econdly, where A is the middle point; and thirdly, when A being again
an extreme, O is $ought beyond it, or between the other two points E and
I: and each of the$e is $ubdivi$ible into four more particular ones.
EPITAGMA I. Here the four _Ca$es_ are when E being the middle point,
O is required between A and E, or beyond I; and the$e are both con-
$tructed at once by _Fig_. 9: when I is the middle point and O $ought between
A and I or beyond E; and the$e are both con$tructed at once by _Fig_. 10.
and in both of the$e IQ is $et off from I contrary to that direction which A
[0091]
bears therefrom, and DH drawn through F, the center of the circle on EQ:
none of the$e Ca$es are $ubject to any _Limitations_.
EPITAGMA II. Wherein A is the middle point, and the _Ca$es_, when O
is $ought beyond E, between E and A, between A and I or beyond I. The
fir$t and third of which are con$tructed at once by _Fig_. 11, wherein IQ is
$et off from I towards A and DH drawn through F, the center of the circle
on EQ. The $econd and fourth are con$tructed at once, al$o, by _Fig_. 12.
where IQ is $et off from I the contrary way to that which A lies, and DH
drawn parallel to EQ. There are no _Limitations_ to any of the$e Ca$es.
EPITAGMA III. Here, E being the middle point, the Ca$es are, when O
mu$t lie beyond A, or between E and I; and the $ame Ca$es occur when
I is made the middle point. The fir$t is con$tructed by _Fig_. 13, the $econd
by _Fig_. 14, the third by _Fig_. 15, and the fourth by _Fig_. 16: in every one
of which IQ is $et off from I towards A, and DH drawn parallel to EQ.
The Limits are that the given ratio of R to S, mu$t not be le$s than the ratio
which the rectangle AE, P bears to the $quare on half the Sum, or half the
difference of AE, and a fourth propor tional to R, S and P; that is, to the
$quare on half EQ: $ince if it $hould, the rectangle contained by AE and
the $aid fourth proportional will be greater than the $quare on half EQ;
and of cour$e ED (a mean proportional between them) greater than half
EQ, in which Ca$e DH can neither cut nor touch the circle on EQ, and
$o the problem be impo$$ible. It is farther ob$ervable in the two la$t ca$es,
that to have the former of them po$$ible, AE mu$t be le$s, and to have
the latter po$$ible, EI mu$t be greater than the above-mentioned half
$um; for if this latter part of the Limitation be not ob$erved, the$e ca$es
are changed into one another.
PROBLEM VI.
(Fig. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.)
In any indefinite $traight line let there be a$$igned the points A, E and I;
it is required to cut it in another point O, $o that the rectangle contained
by the $egments AO, EO may be to the $quare on IO in the ratio of two
given $traight lines, R and S.
[0092]
ANALYSIS. Let us conceive the thing effected, and that O is really the
point $ought. Then, by $uppo$ition, the rectangle AO, EO is to the
$quare on IO as R to S. Make EC to IC as R is to S; and the rectangle
AO, EO is to the $quare 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 $quare on IO as the rectangle EC, OB is to the rectangle
IC, OB; and $o by permutation, the rectangle AO, EO is to the rectangle
EC, OB as the $quare on IO is to the rectangle IC, OB; and becau$e EO is
to EC as IO to BO, AO will be to OB as IO to IC, and $o by compo$ition,
or divi$ion 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 $um or difference of CO and BO,
as al$o the rectangle contained by them, equal to the rectangle AB, IC,
wherefore the$e lines them$elves are given by the 85th or 86th of the _Data._
SYNTHESIS. Make AB to IB and EC to IC in the given ratio, and
de$cribe 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 mu$t fall between B and C;
but through F, the center of the circle on BC, if it mu$t fall without them,
cutting the $@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 con$truction that BD and HO are equal, and
(EU. IV. 17.) the rectangle AB, IC, or the rectangle IB, EC is equal to the
$quare on BD, and therefore equal to the $quare on HO, which (EU. III.
35. 36.) is equal to the rectangle BO, CO: con$equently (EU. VI. 16.) AB
is to BO as CO to IC; al$o EC is to CO as BO is to IB; wherefore, by
compo$ition or divi$ion, AO is to BO as IO to IC, and EO to EC as IO
to BO: con$equently by compound ratio, the rectangle contained by AO and
EO is to the rectangle contained by BO and EC, as the $quare on IO is to
the rectangle contained by BO and IC; by permutation, the rectangle
contained by AO and EO is to the $quare 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.
[0093]
SCHOLIUM. This _Problem_ al$o hath three _Epitagmas_; $ir$t, when O is
$ought between I, the point which bounds the $egment who$e $quare is
concerned, and either of the other given ones; $econdly, the $aid point
being an extreme one, when O is $ought beyond it, or beyond both the
other given points with re$pect to it; and thirdly, when O is required be-
yond the next in order to the abovementioned point I: the$e are each of
them $ubdivi$ible into other more particular _ca$es_.
EPITAGMA I. Here O is $ought between I, the point which bounds the
$egment who$e $quare is concerned, and the next in order to it: and there
are four ca$es, _viz_. when I is an extreme point; and the given ratio of R
to S the ratio of a greater to a le$s; when I, remaining as before, the given
ratio is of a le$s to a greater; again when I is the middle point, and O
$ought 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; becau$e by
con$truction, 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 le$s, le$s; but it is plain
that if O $hould fall either beyond E or I, this could not be the Ca$e. It
is farther manife$t that $hould the points A and E change places, the con-
$truction would be no otherwi$e altered than that AB would then be greater
than IC.
CASE II. If the given points retain their po$ition, but the ratio be made
of a le$s to a greater, the con$truction will then be by _Fig._ 18, where B
mu$t be made to fall beyond A, and C beyond E with re$pect to I; but
DH is $till drawn through the center of the circle on BC: and that O will
fall as required may be made appear by rea$onings $imilar to tho$e u$ed in
_Ca$e_ I. Moreover no change will en$ue in the con$truction when the Points
A and E change places, except that B and C will change $ituations al$o.
CASES III and IV, Are con$tructed at once by _Fig_. 19, when B mu$t
fall between A and I, C between I and E, and DH be drawn as before:
and it is here evident that the con$truction will be the $ame let the given
ratio be what it will. None of tho$e Ca$es admit of any _Limitations_.
EPITAGMA II. There are here only two Ca$es, _viz_. when O is required
beyond I, the point which bounds the $egment who$e $quare is concerned;
[0094]
and $econdly when it is $ought beyond both the other given points: in the
fir$t, the ratio of R to S mu$t be of a greater to a le$s, and in the latter of
a le$s to a greater. The former is con$tructed in _Fig_. 17. at the $ame time
with _Ca$e_ I of _Epitagma_ I. and is repre$ented by the $mall letters _h_ and _o_:
the latter in _Fig_. 18, and pointed out by the $ame letters. That O will
fall in both as is required needs not in$i$ting on.
EPITAGMA III. In which there are $ix Ca$es, _viz_. I being the middle
point, when O is $ought beyond A, or beyond E; and that whether the
given ratio be of a le$s to a greater, or of a greater to a le$s; and again, I
being an extreme point, when O is $ought 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
le$s to a greater; and the$e are both con$tructed at once by _Fig_. 20,
wherein B is made to fall beyond A, and C beyond E with re$pect 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 le$s; and the con$truction 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
$ought between A and E: in con$tructing this Ca$e, B mu$t 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 Con$tructing Ca$e VI. are exactly
the $ame, as will appear by barely in$pecting _Fig_. 24.
LIMITATION. It is plain that in the four la$t Ca$es, the ratio which the
rectangle contained by AO and EO bears to the $quare on IO, or which is
the $ame thing, the given ratio of R to S cannot exceed a certain limit;
and it is farther obvious that the $aid limit will be when the $traight line
DH becomes a tangent to the circle on BC, as in _Fig_. 25. 26, for after
that the problem is manife$tly impo$$ible. Now when DH is a tangent to
the circle on BC, HO will be equal to half BC; but the $quare on HO
is equal to the rectangle contained by IB and EC, wherefore the $quare on
half BC will then be equal to the rectangle contained by IB and EC.
Moreover, by the con$truction, R is to S as AB is to IB, and as EC is to
IC; therefore by compo$ition or divi$ion, the $um or difference of R and
S is to R as EI to EC, and the $aid $um or difference is to S as AI is to
[0095]
IB, as EI is to IC: and hence, by compound ratio, the $quare on the
abovementioned $um or difference is to the rectangle contained by R and S
as the rectangle contained by AI and EI is to the rectangle contained by
IB and EC, al$o by permutation, AI is to EI as IB is to IC; wherefore,
by compo$ition or divi$ion, AE is to AI as BC is to IB, by permutation,
AE is to BC as AI is to IB, therefore by equality, the $um or difference
of R and S is to S as AE is to BC; or (EU. V. 15.) as half AE is to
half BC; con$equently (EU. VI. 22.) the $quare on the above mentioned
$um or di$$erence is to the $quare on S as the $quare on half AE is to the
$quare on half BC, or to the rectangle contained by IB and EC. Hence
_exæquo perturbaté_, the rectangle contained by R and S is to the $quare on
S as the $quare on half AE is to the rectangle contained by AI and EI,
or (EU. V. 15.) R is to S as the $quare on half AE is to the rectangle
contained by AI and EI; and which is therefore the greate$t ratio which
R can have to S in tho$e _Ca$es_.
It ought farther to be remarked, that to have _Ca$e_ III po$$ible, where
O is $ought beyond A, and the ratio of a greater to a le$s, it is nece$$ary
that AI be le$s than IE, and to have _Ca$e_ IV. po$$ible, that it be greater.
For it is plain from the Con$truction, that IB mu$t in the former ca$e be
le$s, and in the latter greater than I C; but as R is to S $o is AB to
IB, and $o is EC to IC, wherefore by divi$ion, the exce$s of R above S
is to S as AI is to IB, and as EI is to IC; and $o by permutation AI is
to EI as IB is to IC: con$equently when IB is greater than IC, AI will
be greater than EI; and when le$s, le$s.
With re$pect to tho$e ca$es wherein the given ratio is that of equality,
it may be $ufficient to remark, that none of the Ca$es of _Epitagma_ II.
are po$$ible under that ratio: that one of _Ca$es_ III. and IV. _Epitagma_ III.
is always impo$$ible when the given ratio of R to S is the ratio of equality;
and both are $o if AI be at the $ame time equal to IE. La$tly _Ca$es_ V.
and VI. are never po$$ible under the ratio of equality, unle$s the $quare on
half AE be equal to, or exceed the rectangle contained by AI and EI;
all which naturally follows from what has been delivered above.
See Prop. A. Book V. of Dr. Sim$on’s Euclid.
THE END OF BOOK I.
[0096]
DETERMINATE SECTION.
BOOK II.
LEMMA I.
If from two points E and I in the diameter AU of a circle AYUV (_Fig_.
27.) two perpendiculars EV, IY be drawn contrary ways to terminate in
the Circumference; and if their extremes V and Y be joined by a $traight
line VY, cutting the faid diameter in O; then will the ratio which the
rectangle contained by AO and UO bears to the rectangle contained by
EO and IO be the lea$t po$$ible.
*** This being demon$trated in the preceeding Tract of SNELLIUS, I
$hall not attempt it here.
LEMMA II.
If to a circle de$cribed 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 $traight line YVO be drawn to cut the line
AI in O; then will the ratio which the rectangle contained by AO and
UO bears to that contained by EO and IO be the lea$t po$$ible: and
moreover, the $quare on EO will be the $quare on IO as the rectangle con-
tained by AE and UE is to the rectangle contained by AI and UI.
DEMONSTRATION. If the $aid ratio be not then a minimum, let it be
when the $egments are bounded by $ome other point S, through which and
the point V, let the $traight line SV be drawn, meeting the circle again in
R; draw SM parallel to OY, meeting the tangents EV and IY in L and
M, and through R and Y draw the $traight line RY meeting SM produced
in N: the triangles ESL and EOV, ISM and IOY are $imilar; wherefore
LS is to SE as VO is to EO, and SM is to SI as YO is to OI; con$e-
[0097]
quently, by compound ratio, the rectangle contained by LS and SM is to
that contained by SE and SI as the rectangle contained by VO and OY,
or its equal, the rectangle contained by AO and OU is to that contained
by EO and IO. Now the triangles VSL and NSR having the angles at
R and L equal, and the angle at S common, are $imilar; and therefore SR
is to SN as SL is to SV; con$equently, the rectangle contained by SR and
SV, or its equal, the rectangle contained by SA and SU is equal to that
contained by SN and SL: but SN is nece$$arily greater than SM, in con-
$equence whereof the rectangle contained by SN and SL, or its equal, the
rectangle contained by AS and SU is greater than that contained by SM
and SL; wherefore the ratio which the rectangle AS, SU bears to the rect-
angle ES, SI is greater than that which the rectangle SM, SL bears to it,
and of cour$e, greater than the ratio which the rectangle AO, UO bears to
the rectangle EO, IO; and that, on which $ide $oever 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 $imilar, becau$e the angles EVO
and IYO are equal, and $o EO is to IO as EB is to IC, al$o 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 $quare on EO is to the $quare on IO as the $quare
on EV is to the $quare 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 $ame $ide thereof to
terminate in the periphery, and if their extremes V and Y be joined by a
$traight line VY, cutting the $aid diameter, produced, in O; then will the
ratio which the rectangle contained by AO and UO bears to the rectangle
contained by EO and IO be the greate$t po$$ible.
*** This, like the $ir$t, is demon$trated by SNELLIUS, and needs not be
repeated.
[0098]
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, al$o YF a mean proportional
between AE and IU.
DEMONSTRATION. Draw UY, UK, KA and AY, the angles I and F
being right by con$truction, and the angles IDU, and FKV equal, being
both equal to the angle UAY, the triangles IYU and FKU are $imilar,
and con$equently IY is to UY as KF is to UK; or (EU. VI. 22.) the
$quare on IY is to the $quare on UY as the $quare on KF is to the $quare
on UK: now the $quare on IY is (EU. VI. 8. 17.) equal to the rectangle
contained by AI and IU, the $quare on UY to the rectangle contained by
AU and IU, and the $quare 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 $quare on KF is to the rectangle contained by AU
and UE, whence (EU. V. 15.) AI is to AU as the $quare 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 $quare on KF is to the
rectangle contained by AU and UE; $eeing then that the con$equents are
here the $ame, the antecedents mu$t 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, becau$e AYU is a
right angle; and therefore as the angles F and E are both right, the tri-
angles AKE and YUF are $imilar, and AK is to AE as YU is to YF,
wherefore the $quare on AK is to the $quare on AE as the $quare on YU
is to the $quare on YF: but the $quare on AK is equal to the rectangle
contained by AU and AE, and the $quare on YU is equal to the rectangle
contained by AU and IU, con$equently the rectangle contained by AU
and AE is to the $quare on AE as the rectangle contained by AU and IU
is to the $quare on YF; whence (EU. V. 15.) AU is to AE as the rect-
angle contained by AU and IU is to the $quare on YF, or the rectangle
contained by AU and IU is to that contained by AE and IU as the rect-
[0099]
angle contained by AU and IU is to the $quare on YF; hence the rect-
angle AE, IU is equal to the $quare on YF, and AE is to YF as YF is
to IU.
Q. E. D.
LEMMA V.
If in any $traight line four points A, U, E and I (_Fig_. 31.) be a$$igned,
and if the point O be $o taken by LEMMA II, that the ratio of the rect-
angle contained by AO and UO to that contained by EO and IO may be
the lea$t po$$ible; al$o if through O the indefinite perpendicular FG be
drawn; and la$tly, 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 $hall FG be equal to the $um 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 the$e two la$t will be re$pectively perpendiculars to IF and UG, be-
cau$e the three perpendiculars of every plane triangle meet in a point.
Since by con$truction and EU. VI. 17, the $quare on EG is equal to the
rectangle contained by AE and UE, and the $quare on IF to that con-
tained by AI and UI, and becau$e (LEM. II.) the $quare on EO is to the
$quare on IO as the rectangle AE, UE is to the rectangle AI, UI; the
$quare on EO is to the $quare on IO as the $quare on EG is to the $quare
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 $imilar, and HG parallel to
IF, and the angle UHE equal to the angle UFI. Again, becau$e AI is
to IF as IF is to UI, the triangles AIF, and UFI are $imilar, and $o, for
like rea$ons, are the triangles AEG and GEU, wherefore the angles UFI
and FAE are equal, and al$o 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 con$equently, the angles VAY and
YUV are together equal to two right angles; wherefore the points AYUV
are in a circle: hence, and becau$e AO is perpendicular to FG, GY will
See PAP. _Math. Collect._ B. vii. prop. 60.
[0100]
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 $imilar, and UN is to UI as UE is to UN, con$equently the $quare on
UN is equal to the rectangle contained by UI and UE. Moreover, $ince
IX pa$$es through N, and is perpendicular to FU, by EU. I. 47, the dif-
ference of the $quares on IF and IU is equal to the difference of the
$quares on NF and NU: now the $quare 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 $quare on UI, the difference of
the $quares on IF and UI, and con$equently the difference of the $quares on
NF and NU is equal to the rectangle contained by AU and UI; but the
$quare on NU has been proved equal to the rectangle contained by UI
and UE, therefore the $quare on NF is equal to the rectangle contained
by EU and IU together with that contained by AU and IU, that is (EU.
II. 1.) to the rectangle contained by AE and UI; wherefore AE is to NF
as NF is to UI.
La$tly, for the like rea$ons which were urged above, the difference of
the $quares on NU and NG is equal to the difference of tho$e on GE and
UE: now the $quare 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 $quare on UE; therefore the difference of the $quares on GE
and UE, or the difference of tho$e on NU and NG, is equal to the rect-
angle contained by AU and UE; but the $quare on NU is equal to the
rectangle contained by UI and UE, therefore the $quare 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 $o AI is to NG as NG is to UE. Now FG is equal to the $um
of NF and NG; therefore FG is equal to the $um of two mean propor-
tionals between AE and UI, AI and UE.
Q. E. D.
[0101]
PROBLEM VII. (Fig. 32, 33, 34, &c.)
In any inde$inite $traight line let there be a$$igned the points A, E, I
and U; it is required to cut it in another point, O, $o that the rectangle
contained by the $egments AO, UO may be to that contained by the $eg-
ments EO, IO in the ratio of two given $traight lines, R and S.
ANALYSIS. Imagine the thing done, and O the point $ought: 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 $ince UO is to UC
as IO to OB, by con$truction, AO will be to BO as EO to EC; and $o by
compo$ition or divi$ion, 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 $um or dif-
ference of CO and BO, as al$o the rectangle contained by them, equal to
the rectangle CU, IB, whence tho$e lines them$elves 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-
$cribe on BC a circle; erect, at B the inde$inite 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
$ought any where without them, cutting the circle on BC in H. La$tly,
draw HO perpendicular to DH, which will cut the inde$inite line in O,
the point required.
For it is plain from the con$truction that HO and BD are equal, and
(EU. VI. 17.) the rectangle AB, EC, or the rectangle IB, UC is equal to
the $quare on BD, and therefore equal to the $quare 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 compo$ition or divi$ion, AO is to BO as EO is to CE, and UO is to
[0102]
CU as IO is to BO; and $o, 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 con$truction as R to S.
Q. E. D.
SCHOLIUM. In enumerating the $everal Ca$es of this _Problem_ I $hall en-
deavour to follow the method which I conceive _Apollonius_ did: and there-
fore, notwith$tanding the preceding _Analy$is_ and Con$truction are general
for the whole, divide it into three _Problems_, each Problem into three _Epi-_
_tagmas_, or general _Ca$es_, and the$e again into their $everal particular ones.
PROBLEM I. (Fig. 32 to 45.)
Here O is $ought between the two mean points of the four given ones:
and the three Epitagmas are, fir$t, when A and U, the points which bound
the $egments containing the antecedent rectangle, are one an extreme, and
the other an alternate mean; $econdly, when tho$e points are one an ex-
treme and the other an adjacent mean; thirdly, when they are both means,
or both extremes.
EPITAGMA I, Con$i$ts of eight _Ca$es, 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 le$s to a greater, and four others wherein the order of the
points is the $ame as in tho$e, but the ratio of R to S, the ratio of a greater
to a le$s.
CASE I. Let the order of the given points be A, I, U, E, and the given
ratio of a le$s to a greater; and the Con$truction will be as in _Fig_. 32, where
B is made to fall beyond A, with re$pect to I, and C beyond U with re-
$pect to E, and DH is drawn through F, the center of the circle on BC.
That O, when this con$truction is u$ed, will fall between I and U is
plain, becau$e 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 le$s, le$s; but this,
it is manife$t, cannot be the Ca$e if O falls either beyond I or U, and
therefore it falls between them.
[0103]
CASE II. If the order o$ the given points be retained, but the ratio be of
a greater to a le$s, B mu$t $all beyond I with re$pect to A, and C beyond
E, as in _Fig_. 33, and DH mu$t be drawn as before: and it may be proved
by rea$onings $imilar to tho$e u$ed in the $ir$t Ca$e that O will fall between
I and U, as was required.
CASES III and IV. The$e ca$es are con$tructed exactly in the $ame man-
ner as _Ca$es_ I and II re$pectively; and the rea$onings to prove that O will
fall as it ought arethe $ame with tho$e made u$e of in _Ca$e_ I, as will appear
by in$pecting _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 le$s to a greater. Then mu$t 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 $quare on
EF is equal to the rectangle contained by EC and EB, the $quare on UG
to the rectangle contained by UC and UB; and, by con$truction, the
$quare on HO to the rectangle contained by EC and AB, or to the rect-
angle contained by UC and IB; and $ince, by $uppo$ition, EB is greater
than AB, and UB le$s than IB, the rectangle EC, AB, or its equal, the
rectangle UC, IB will be le$s than the rectangle EC, EB, and greater than
the rectangle UC, UB; and con$equently the $quare on HO will be le$s
than the $quare on EF and greater than the $quare on UG; and therefore
HO le$s than EF and greater than UG: but this could not be the Ca$e
unle$s 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 le$s, B mu$t then fall beyond I, and C beyond E; and
DH is drawn as in the preceding Ca$e (See _Fig_. 37.) moreover, that O
will fall between E and U may be made to appear, by rea$onings $imilar to
tho$e there made u$e on the like occa$ion.
CASE VII. Is con$tructed exactly in the $ame manner as Ca$e V. and,
CASE VIII. As Ca$e VI: the truth of which will appear by. barely in-
$pecting _Fig_. 38 and 39.
EPITAGMA II. In this Epitagma there are al$o eight _Ca$es, 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 le$s to a greater: and there are
four others wherein the order of the points are the $ame as in tho$e, but
[0104]
the ratio of a greater to a le$s; but the$e, I $hall $hew, may be reduced
to four.
CASE I. The order of the given points being A, U, E, I; and the given
ratio a le$s to a greater, the con$truction will be e$$ected by _Fig_. 40,
wherein B is made to fall beyond A with re$pect to I and C beyond U, and
DH is drawn through the center of the circle on BC. And O will fall
between U and E for rea$ons $imilar to tho$e urged in the $ir$t Ca$e of _Epi-_
_tagma_ I. It is moreover obvious that the con$truction will not be e$$entially
different $hould the points E and I change places, and therefore need not
here be made a new Ca$e.
CASE II. The order of the points being the $ame as in the la$t Ca$e, let
the given ratio be of a greater to a le$s; then, as in _Fig_. 41, B mu$t fall
beyond I, and C beyond E; but DC mu$t $till be drawn through the
center of the circle on BC. It is manife$t that this con$truction will $erve
for that Ca$e wherein the points A and U change $ituations, if the ratio be,
as here, of a greater to a le$s.
CASE III. Here, let the order of the points be U, A, I, E, and the
given ratio of a le$s to a greater, and the Con$truction will be a$$ected by
_Fig_. 42, in which B falls beyond A, and C beyond U with re$pect to I and
E: and the $ame con$truction will $erve if I and E change places, but the
ratio remain the $ame.
CASE IV. If the po$ition of the points be retained, but the ratio be
made of a le$s to a greater; then mu$t B fall beyond I (_Fig_. 43.) and C
beyond E; but DH drawn as before. That O mu$t fall as was required,
in the$e three la$t ca$es, is obvious enough from what has been $aid be-
fore on the like occa$ion: and it is al$o plain that the con$truction will
not be materially di$$erent though A and U change places.
SCHOLIUM. That none of the Ca$es of the$e two _Epitagmas_ are $ubject
to _Limitations_, might be proved with the utmo$t rigour of geometrical
rea$oning was it not $u$$iciently manife$t from con$idering 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 $mall, and as it approaches the points
E, or I the $aid ratio will become very great: and nothing hinders that
the $aid point may $all any where between tho$e.
[0105]
EPITAGMA III. There are here but four _Ca$es, 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 $ir$t of the$e are not po$$ible unle$s the given ratio be the ratio of a
greater to a le$s; nor the two latter, unle$s it be of a le$s to a greater, and
as the$e are reduced to the $ir$t two by reading every where E for A, I
for U, and the contrary, I $hall omit $pecifying them.
Case I. If the order of the given points be A, E, I, U, the con$truc-
tion will be effected by _Fig_. 44, wherein B is made to fall beyond I, and
C beyond E, and DH is drawn parallel to BC. That O, when this con-
$truction is u$ed, will fall between E and I, is ea$ily made appear by rea-
$oning in a manner $imilar to what was done in _Ca$e_ V. of _Epitagma_ I.
CASE II. The con$truction of this Ca$e, where the order of the points
is A, I, E, U, is not materially different from that above exhibited as ap-
pears by _Fig_. 45, and that O will fall between I and E is manife$t without
farther illu$tration.
LIMITATION. In the$e two Ca$es the given ratio of R to S cannot be
le$s than that which the $quare on AU bears to the $quare on a line which
is the difference of two mean proportionals between AI and EU, AE and
IU. For by _Lemma_ I. the lea$t ratio which the rectangle contained by AO
and UO can have to the rectangle contained by EO and IO; or, which
is the $ame thing, that R can have to S, will be when the point O is
the inter$ection of the diameter AU, of a circle AYUV, with a $traight
line YV. joining the tops of two perpendiculars EV, IY to the inde$i-
nite line, on contrary $ides thereof, and terminating in the periphery of
the circle. Produce VE (_Fig_. 27.) to meet the circle again in K, and
draw the diameter KL; join LY and KY, on which, produced, let fall
the perpendicular UF. Now, $ince by _Lemma_ III. KF is a mean propor-
tional between AI and EU, and YF a mean proportional between AE
and IU: it remains only to prove that the ratio of the rectangle con-
tained by AO and OU to the rectangle contained by EO and OI is the
$ame with the ratio which the $quare on AU bears to the $quare on KY,
which is the di$$erence between KF and YF. Becau$e the angles E and
KYL are both right, and the angles EVO and KYL equal (EU. III. 21.)
the triangles EVO and YLK are $imilar; and $o VO is to EO as AU (LK)
is to KY; or the $quare on VO is to the $quare on EO as the $quare on AU
is to the $quare on KY. Now the triangles EVO, IYO being al$o $imilar,
[0106]
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 $quare on OV is
to the $quare on EO, as the $quaree on AU is to the $quare on KY.
Q. E. D.
SCHOLIUM. It might be ob$erved that in the two Ca$es of this _Epitagma_
where the points A and U are means, the _limiting ratio_ will be a _maxi_-
_mum_ in$tead of a _minimum_; and that ratio will be the $ame with that which
the $quare on KY bears to the $quare on EI, as is plain from what hath
been advanced above.
PROBLEM II. (Fig. 46 to 57.)
Where O is $ought between a mean and an extream point: and here,
as in the fir$t _Problem_, there are three _Epitagmas_. Fir$t when the points
A and U, which bound the $egments containing the antecedent rectangle,
are one an extreme, and the other an alternate mean; $econdly when they
are both means, or both extremes; thirdly when they are one an extreme,
and the other an adjacent mean.
EPITAGMA I. There are here eight Ca$es, but they are con$tructed at
four times, becau$e it is indifferent whether the given ratio be of a le$s to a
greater, or of a greater to a le$s.
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, becau$e AB is to BO as CO is to CE, and therefore, if AB be greater
than BO, CO mu$t be greater than CE, and if le$s, le$s; but this cannot
be the ca$e if O falls either beyond A or E: and the like ab$urdity fol.
lows if _o_ be $uppo$ed to fall otherwi$e than between I and U.
CASE II. Wherein the order of the points is U, I, A, E; and it is con-
$tructed in the very $ame manner that _Ca$e_ I. is, as appears by barely
in$pecting _Fig_. 47.
CASE III. If the order of the given points be A, I, U, E (_Fig_. 48.) the
points B and C mu$t be made to fall as in the two preceding Ca$es; but
DH mu$t be drawn parallel to BC, and O will fall as required. For erect
[0107]
at I, the perpendicular IG: by Eu. VI. 13. 17. the $quare on IG is equal
to the rectangle contained by IB and IC; and the $quare on HO is equal
to the rectangle contained by IB and UC. Now IC is by $uppo$ition
greater than UC, and therefore the rectangle IB, IC is greater than the
rectangle IB, UC: con$equently the $quare on IG is greater than the
$quare on HO, and IG than HO; whence O mu$t fall between I and B,
much more between I and A. And in the $ame manner it may be proved
that the point _o_ falls between U and E.
CASE IV. In which the order of the given points is U, E, A, I; it is
con$tructed exactly in the $ame manner as _Ca$e_ III, and is exhibited by
_Fig_. 49.
EPITAGMA II. There are here only four _Ca$es_, becau$e, as in _Epi_-
_tagma_ I. it is indifferent whether the given ratio be of a le$s to a greater,
or of a greater to a le$s; and the two la$t of tho$e, _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 $hall omit $aying any thing of their con$tructions, 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 rea$ons $imilar to tho$e urged in _Ca$e_ I. of the fir$t Epitagma
of this Problem.
CASE II. If the order of the given points be A, I, E, U, the con$truc-
tion will be as in _Fig_. 51, where B and C are made to fall, and DH is
drawn as in _Ca$e_ I.
EPITAGMA III. Here there are eight Ca$es, _viz_. four where in the
order of the given points is A, U, E, I; A, U, I, E; U, A, E, I; and
U, A, I, E, and the given ratio of a greater to a le$s, when O will fall
between the two given points, which bound the con$equent rectangle;
and four others@ wherein the order of the given points is the $ame as
here, but the given ratio of a le$s to a greater, and in which the point
O will fall between the points that bound the antecedent rectangle; but
as the$e la$t are reducible to the former by the $ame means which have been
u$ed on former $imilar occa$ions, I $hall not $top to $pecify them.
[0108]
The former four are all con$tructed 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 rea$onings $imilar to tho$e u$ed for the like purpo$e in Ca$e III. of
_Epitagma_ I. that O mu$t fall between E and I as was propo$ed. _See Fig_.
54, 55, 56 and 57.
LIMITATION. In the$e four Ca$es, the given ratio of R to S mu$t not
be le$s than that which the $quare on the $um of two mean proportionals
between AE and IU, AI and EU bears to the $quare 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 $ame thing,
the given ratio of R to S is the lea$t po$$ible, the $quare on EO will be to
the $quare 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 $quare on FG is to the
fquare on EI. Now it has been proved in demon$trating _Lem_. V. that the
triangles EOG and IOF are $imilar, and that the angle at V is right,
whence it follows that the triangles AOG and FOU are al$o $imilar, and
con$equently 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 $o by compo$ition and permutation,
FG is to EI as OG is to EO, and as OF is to IO: hence by compound
ratio the $quare on FG is to the $quare 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 Ca$es, wherein the given ratio is of a le$s to a
greater, and wherein the point O mu$t fall between tho$e given ones which
bound the antecedent rectangle, the _limiting ratio_ will be a _maximum_, and
the $ame with that which the $quare on AU bears to the $quare on FG.
[0109]
PROBLEM III.
In this, the point O is $ought without all the given ones, and the three
_Epitagmas_ are as in _Problem_ I.
EPITAGMA I. There are here eight _Ca$es, viz_. four when the order of
the given points is the $ame as $pecified in _Epitagma_ I. of _Problem_ I, and
O $ought beyond the given point which bounds the antecedent rectangle;
and four others when O is $ought beyond that which bounds the con$e-
quent one: the Con$tructions of the four fir$t are $hewn by the $mall
letters _b_ and _o_ in _Fig_. 32, 34, 36 and 38; and the four latter ones by
the $ame letters in _Fig_. 33, 35, 37 and 39; and the demon$trations that
_o_ will fall as required by the _Problem_ are exactly the $ame as tho$e made
u$e of in the la$t mentioned _Epitagma_. It is farther ob$ervable, that the
four fir$t _Ca$es_ are not po$$ible, unle$s the given ratio be of a le$s to a
greater; nor the four latter, unle$s it be of a greater to a le$s, as is ma-
nife$t without farther illu$tration.
EPITAGMA II. Here, as in the $econd _Epitagma_ of _Problem_ I, the points
A and U are one an extreme, and the other an adjacent mean, and there
are eight _Ca$es_; but it will be $ufficient to exhibit the con$tructions of
four of them, the others being not e$$entially different; and the$e are $hewn
by the $mall _b_ and _o_ in _Fig_. 40, 41, 42 and 43; the demon$trations that _o_
will fall as required need not be pointed out here; but it may be nece$$ary
to remark that the fir$t and third are not po$$ible unle$s the given ratio be
of a le$s to a greater, nor the $econd and fourth unle$s it be of a greater to
a le$s, as is obvious enough.
EPITAGMA III. In which the points A and U are both means, or both
extremes; and there are here eight _Ca$es, viz_. four wherem the$e points
are extremes, and four others wherein they are means: but the$e la$t being
reducible to the former by the $ame method that was u$ed in the third _Epi_-
_tagmas_ of the two preceding _Problems_, I $hall omit them.
All the Ca$es of this Epitagma are con$tructed by making B fall beyond
I, and C beyond E, with re$pect to A and U; and drawing DH parallel to
BC. That O will fall beyond A in _Fig_. 58 and 60, and beyond U in _Fig_.
59 and 61 appears hence. Draw AG perpendicular to BC, meeting the
circle on BC in G: by Eu. VI. 13. 17, the $quare on AG is equal to the
[0110]
rectangle contained by AB and AC; but the $quare on HO is equal to
the rectangle contained by AB and EC: now EC is, by $uppo$ition,
greater than AC, therefore the rectangle AB, EC is greater than the
rectangle AB, AC, and the $quare on HO greater than the $quare on AG,
con$equently HO is it$elf greater than AG; but this could not be the
Ca$e unle$s O fell beyond A. In the $ame manner my it be proved that O
will fall beyond U in _Fig_. 59 and 60.
LIMITATION. In the above four _Ca$es_ the given ratio of R to S mu$t
not exceed that which the $quare on AU bears to the $quare on the $um of
two mean proportionals between AI and UE, AE and UI. For (_Fig_. 30.)
demit from A, on KO produced, the perpendicular AH. Now it has been
proved (_Lem_. III.) that the ratio of the rectangle continued by AO and UO
to that contained by EO and IO, or which is the $ame thing, the given
ratio of R to S is the greate$t po$$ible; and (_Lem_. IV.) that KF is a mean
proportional between AI and UE, al$o that YF is a mean proportional
between AE and UI: but HK is equal to YF, therefore HF is equal to
the $um 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 $quare on AU is to the $quare
on HF. The triangles OEK, OHA, OIY and OUF are all $imilar; con-
$equently 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 $quare on AO is to the
$quare on OH; but al$o AO is to UO as HO is to EO, and by compo$i-
tion and permutation, AU is to HF as AO is to HO, or (Eu. VI. 22.)
the $quare on AU is to the $quare on HF as the $quare on AO is to the
$quare on HO, and $o by equality of ratios, the rectangle contained by AO
and UO is to that contained by EO and IO as the $quare on AU is to the
$quare on HF.
Q.E.D.
SCHOLIUM. In the four Ca$es wherein the points A and U are means,
the limiting ratio will be a _minimum_, and the $ame with that which the
$quare on HF bears to the $quare on EI.
THE END.
[0111]
[0111a]
[0112]
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[0114]
[0115]
[0115a]
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[0117a]
[0118]
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[0121]
A
SYNOPSIS
OF ALL THE DATA FOR THE
Con$truction of TRIANGLES,
FROM WHICH
GEOMETRICAL SOLUTIONS
HAVE HITHERTO BEEN IN PRINT.
With References to the Authors, where tho$e 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.]
[0122]
[0123]
ADVERTISEMENT.
IT is but few years ago $ince the Compiler of this Synop$is
conceived his fir$t idea of the u$e$ulne$s of $uch an undertaking,
and he exhibited a $mall $pecimen thereof in a periodical Work
then publi$hing under the Title of THE BRITISH ORACLE. Here-
upon he received $everal letters from Mathematical friends, ex-
pre$$ing their $en$e of the great propriety of $uch a collection, and
$trongly encouraging him to pur$ue the undertaking. Since that
time it has been a growing work, and would continue $o, were
the publication delayed ever $o long, as fre$h Problems are con-
tinually propo$ed to the public. He has therefore now determined
to $end it abroad, as complete as he can make it to the pre$ent
period, and leave additions to be made by future collectors.
[0124]
[0125]
AN
EXPLANATION
OF THE
SYMBOLS made u$e of in this SYNOPSIS.
H. # repre$ents the Hypothenu$e of a right-angled triangle.
V. # Vertical angle.
B. # Ba$e or $ide oppo$ite V.
P. # Perpendicular from V on B.
S&s. # Sides about V; S the greater, s the le$s.
A & a. # Angles at B; A the greater, a the le$s.
m & n. # Segments of B by P; m the greater, n the le$s.
Ar. # Area.
Per. # Perimeter.
L. # Line from V to B, bi$ecting V.
λ. # Line from V to B, cutting B in a given ratio.
l. # Any other line $pecified how drawn.
R. # Radius of in$cribed circle.
<036>. # Circle.
<017>. # Square.
: # Ratio: thus, S: s $ignifies the ratio of the $ides.
[0126]
_N. B._ Between each of the Data a full $top is placed. Moreover, m and n are
$ometimes u$ed for different $egments than tho$e of B by P, but then it is $ignified in
words: - the $ame al$o is to be ob$eved of P.
Ob$erve likewi$e, for the more ready finding any propo$ed Problem in the Synop-
$is, that the data are ranged in the $ame order as the Symbols here recited; _viz_.
all data whereof V is one are placed fir$t; next tho$e where B is given, either $im-
ply by it$elf, or combined with any other datum; next P, &c.
Moreover, when in the references you find this mark*, it $ignifies that $uch Authors
have only con$tructed the Problem _partially_, and not _generally_ as propo$ed in the Synop-
$is, e. g. for a right-angled triangle, when it is propo$ed for a triangle in general;
and again, for a line bi$ecting another, when it is propo$ed to cut it $o that the $eg-
ments may be in any given ratio.
[0127]
INDEX
OF THE
Authors refered to in the SYNOPSIS.
ANDERSONI Var. Prob. Practice, cum Supplemento Apollonii
# Redivivi, 4to. # _Pari$iis_ 1612
Ander$oni Exercitationum Math. Deas ima. 4to. # _ibid_. 1619
A$hby’s Algebra, 2d Edit. 12mo. # _Lond_. 1741
Briti$h Oracle (Vol. I. being all that was publi$hed) 12mo. # _ibid_. 1769
Ca$tillioneus inNewtoni Arith. Univer$alem, 4to. # _Am$telod_. 1761
Clavius in Euclidem, var. Ed.
Court Magazine.
D’Omerique (Hugonis) Analy$is Geometrica, 4to. # _Gadibus_ 1699
Diarian Repo$itory, Periodical Work, printed for Robin$on, 4to. # _Lond_. 1770, &c.
Fo$ter’s Mi$cellanies, or Math. Lugubrations, fo. # _ibid_. 1659
General Magazine.
Gentleman’s Magazine.
Gentleman’s Diary.
Ghetaldi (Marini) Var. Prob. Collectio, 4to. # _Venetiis_ 1607
Ghetaldus de Re$olutione & Compo$itione Math. fo. # _Romæ_ 1640
Gregorius a Sancto Vincentio, fo. # _Antverp_. 1647
Herigoni Cur$us Math. Lat. & Gallicè, 8vo. 5 Tom. # _Paris_ 1644
Hutton’s Ladies Diaries.
# Mathematical Mi$cellany.
Imperial Magazine.
Ladies Diaries.
Martin’s Math. Corre$pondence, in his Magazine.
Mathematical Magazine.
Mathematician, Periodical Work. 8vo. # _Lond_. 1751
[0128]
Mi$cellanea Curio$æ, Periodical Work. 8vo. 6 Nos. # _rork_ 1734-5
Mi$cellanea Cur. Math. Period Work, by Holliday, 4to. 2 vols. # _Lond_. 1745
Mi$cellanea Scienti$ica Curio$a, 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. # _Ba$iliæ_ 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
Saunder$on’s Algebra, 2 vols. 4to. # _Camb_. 1740
Schooten’s (Franci$cus à) Exercitationes Math. 4to. # _Lug. Bat_. 1657
Simp$on’s Algebra, 8vo. 3d Edit. \\ 1$t Edit. # _Lond_. 1767 \\ _ibid_. 1745
# Select Exerci$es, 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 Exerci$es, Periodical Work, 8vo. # _ibid_. 1750
Univer$al Magazine.
Vietæ Opera, fo. # _Lug. Bat_. 1646
We$t’s Mathematics, 2d Edit. 8vo. # _Lond_. 1763
Wolfius’s Algebra, tran$lated by Hanna, 8vo. # _ibid_. 1739
Lately was publi$hed by the $ame AUTHOR;
[Price SIX SHILLINGS in Boards.]
APOLLONIUS concerning TANGENCIES, as re$tored by Vieta & Ghetaldus,
with a Supplement; the 2d Edit. To which is now added a Second Supplement,
being Fermat’s Treati$e on SPHERICAL TANGENCIES. Likewi$e Apollonius con-
cerning DETERMINATE SECTION, as re$tored by Willebrordus Snellius; to which
is added an entire new Work, being the $ame re$tored by Mr. W. Wales.
[0129]
SYNOPSIS.
1. V. B. P. # SIMPSON’s Alg. pr. 5. - Mis. Cur. Math. Vol. I. \\ page 31. - Vieta I$t. Ap. to Apollonius Gallus, pr. 5. \\ - *A$hby’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. - Univer$al Mag. Mar. 1749.
4. V. B. S: s. # Simp$on’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. # Simp$on’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. - \\ Saunder$on’s Alg. art. 332. - *Rudd’s Prac. Geom. part \\ 2d, qu. 38. - *A$hby’s Alg. pag. 102.
6. V. B. S - s. # Simp$on’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. - *Simp$on’s Sel. Ex. pr. \\ 2. - *Wolfius’s Alg. pr. 12
8. - *A$hby’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. Ca$tillionei, pr. 5.
9. V. B. Ar. # *Simp$on’s Alg. pr. 33. - Simp. Geom. pr. 5. - *Saunder - \\ $on’s Alg. art. 329. - *D’Omerique L. III. pr. 36. - *Oughtred’s
[0130]
# 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. # Simp$on’s Alg. pr. 72. - Simp. Geom. pr. 21.
12. V. B. λ. # *Simp$on’s Alg. pr. 58. - *Ghetaldus var. prob. 3. - Regio - \\ montanus de triangulis, Lib. II. pr. 29.
13. V. B. Direction \\ of l thro’ V. # }Simp$on’s Sel. Ex. pr. 48.
14. V. B. R. # *Simp$on’s Sel. Ex. pr. 29. - Br. Oracle, qu. 67, Cor. - \\ *We$t’s Mathematics, 2d. Ed. pag. 45.
15. V. B. Side of \\ ins. <017>. # }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. Ca$tillionei, pr. 7. - *Simp$on’s Sel. Ex. pr. 31. - \\ *Simp$on’s Alg. I$t. 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. - *Ander$on var. prob. 2; \\ - *Simp. Sel. Ex. pr. 1. - Twy$den in Fo$ter’s Mi$cell. pr. I. \\ - Renaldinus pag. 526, 529, *437. - *Mathematician, pr. \\ 42. - *A$hby’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 - \\ der$on, art. 327. - *Ander$on var. prob. 15, 16. - *L. Diary, \\ 1770, p. 35.
22. V. B<_>2: m<_>2. P. # *Mis. Scient. Cur. qu. 54.
[0131]
23. V. Point in B. \\ : \\ {S±s.\x} # }Simp$on’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. # *Ander$on var. prob. 1, 3. - See V. B±S. s.
28. V. B ‖ to a given \\ line. Ar. # }Simp$on’s Sel. Ex. pr. 41. - Simp. Geom. pr. 4.
29. V. Pointin B. Ar. # Simp$on’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 Repo$itory, 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. - Fo$ter’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. Ca$tillionei, 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. <036>. # }Gent. Diary, 1767, qu. 300.
41. V. P + s. S - s: \\ m - n. # }Gent. Diary, 1751, qu. 108.
[0132]
42. V. S or s. S x s: \\ m x n, $egts. 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. # Simp$on’s Alg. pr. 3.
# <035> When V and S: s are given, the triangle is given in \\ $pecies, and therefore may be con$tructed 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. # Simp$on’s Alg. pr. 7, 9. - *York Mis. Cur. qu. 46.
47. V. S±s. m - n. # Simp$on’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. # Ander$on var. prob. 22. - *Simp. Alg. pr. 34.
49. V. S±s. L. # Deducible from Simp$on’s Geo. pr. 19, 20.
50. V. S - s. R. # *Br. Oracle, qu. 20.
51. V. S<_>2 + s<_>2. λ. # Hutton’s Mis. qu. 24.
52. V. m. n. # Simp$on’s Alg. pr. 4. - York Mis. Cur. qu. 17, Ca$e 3d. - \\ *Rudd’s Pract. Geo. part 2d, qu. 2. - Court Mag. Feb. 1763, \\ m and n being $egts. by L.
53. V. Ar. Per. # *D’Omerique L. III. pr. 34. - *Simp$on’s Alg. pr. 35. - \\ *Simp$on’s Sel. Ex. pr. 30. - Arith. Uni. Ca$tillionei, 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. <017>. # }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 I$t, qu. 19. - Reducible to \\ V. B. S + s.
57. V. L. m: n, \\ $egts. by L. # }Mis. Cur. Math. V. I. qu. 26.
58. V. L. m - n, \\ $egts. by L. # }L. Diary, 1773, qu. 662.
[0133]
59. V. L. Side of \\ ins. <017>. # }T. and Country Mag. Nov. and Dec. 1772.
60. V. λ. Ra. of \\ circ. <036>. # }*Gent. Diary, 1766, qu. 282. - Reducible to V. B. λ.
61. B. P. S: s. # D’Omerique L. 1. 49. - Vieta 1$t. 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 \\ Mi$cel. qu. 58.
62. B. P. S + s. # D’Omerique L. III. 25. - Vieta ib. pr. 3. - Gregorius a S. \\ Vinc. pr. 82, pag. 48. - Ander$on var. prob. 20, Cor. - \\ Simp. Alg. pr. 77. - Simp. Geom. pr. 15. - Arith. Un. \\ Ca$tillionei, 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. - Simp$. Sel. Ex. pr. 21. - Simp. Geom. \\ pr. 16.
65. B. P. A - a. # Simp$. Alg. pr. 15.
66. B. P. L. # Ander$on 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 Mi$cel. 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.
[0134]
78. B. S±s. Side \\ of ins. <017>. # }Reducible to B. P. S±s.
79. B. A. Ar. # Gent. Diary, 1741, qu. 5.
80. B. A - a. λ. # *Simp$. Algebra, pr. 59.
81. B. A or a. R. # Imperial Mag. Nov. 1760.
82. B. R. Ra. of \\ circums. <036>. # }Mathematician, pr. 66.
83. B±P. A - a. m - n. # L. Diary, qu. 646.
84. B±S. All the \\ angles. # }*Briti$h 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, \\ $egts. by L. # }Ghetaldus var. prob. 4. - Regiomontanus de triang. L. II. \\ 33. - Mi$. 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. # Simp$. Alg. pr. 24.
98. P. S + s. m: n. # D’Omerique L. III. 31.
[0135]
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. <036>. # } 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. <036>. # }Gent. Diary, qu. 363.
108. S. s. Ar. # Mis. Cur. Math. qu. 121. - Saunder$on, 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, \\ $egts. 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.
[0136]
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. <017>. # } *Simp$. 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.
# <035> When all the angles are given, the triangle is given in \\ $pecies, and therefore may be con$tructed, by $imilar triangles, \\ with any other datum.
[0137]
Continuation of the SYNOPSIS,
Containing $uch Data as cannot readily be expre$$ed by the
Symbols before u$ed without more words at length.
124. IT is required to con$truct an i$oceles triangle $uch, as to have its equal legs
to a given line, and moreover $uppo$ing a circle in$cribed therein, and a di-
ameter thereof drawn parallel to the ba$e, and continued to meet the equal legs;
$uch line $hall divide the area in a given ratio.
Rudd’s Prac. Geom. part 2d, qu. 4<_>6. - Supp. to Gent. Diary, 1741, 1742, qu. II.
125. In a plane triangle, ABC, there is given the angle
at C, and the parts or $egments of the ba$e AD, AE;
to con$truct the triangle $o, that if BD be drawn, the
angle ABD may be a _maximum_, and BC to EC in
a given ratio.
L. Diary, 1773, qu. 659.
126. V. B. Line from A or a to divide S or s in a given ratio.
Town and Coun. Mag. Dec. 1772.
127. V. B. Difference of two lines from the angles at the ba$e to the centre of the
in$cribed <036>.
*Simp$on’s Sel. Ex. pr. 18.
128. V. B. Two lines from A and a meeting in a point O given in po$ition, as al$o
the angle of a line from V to O with either of the $ides about the vertical angle.
Mis. Cur. Math. qu. 107.
129. V. A point in B. Rect. of the $egments of B made by that point.
Simp. Sel. Ex. pr. 44.
130. Po$ition of a line through V. B. S - s.
Simp. Sel. Ex. pr. 49.
131. Po$ition of a line through V. B. S, line bi$ect. B, & s in geometrical progre$$ion.
Simp. Sel. Ex. pr. 51.
132. V. S. The angle of a line from extreme of S with the ba$e, and $egment of
s cut off thereby adjacent to the ba$e.
*Mis. Cur. Math. Vol. II. qu. 30.
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133. V. The $quare of the ba$e equal to the rectangle of one $ide 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 $quare of a line, drawn in a given direction from V to B,
to the rectangle of the $egments of B made thereby.
Ander$oni Exercitationes Math. No. 8.
137. V. S or s. m: n, the$e being $egts. 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 bi$ecting S. Line from a bi$ecting s.
*Simp$on’s Sel. Ex. pr. 15.
140. V. Per. Line parallel to B bi$ecting the area.
*Gent. Di. 1750, qu. 98. - Reducible to V. B. S + s.
141. S. s. Line from V to centre of ins. <036>.
L. Diary, 1771, qu. 635.
142. Ba$e. One $ide. Ratio of a line from V, making a given angle with $aid $ide,
to alternate part of B.
Hutton’s Mis. pag. 63, Cor.
143. B. Point of contact therein of ins. <036>. mxn.
Hutton’s L. Diaries, 1722, qu. 94.
144. L. A perpendicular thereto from one of the angles at the ba$e. The other
angle at the ba$e.
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 ba$e. Perpendicular therefrom to oppo$ite $ide a _max_-
_imum_. A line from the other angle at the ba$e bi$ecting its oppo$ite $ide.
*L. Diary, 1769, 1770, qu. 607.
147. S. s. m: n, the$e being $egts. 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.
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148. S. s. Line from V making an angle with S=A.
Diarian Repo$itory, qu. 197.
149. S. m. n, the$e being $egments of B by l from V making a given angle with S.
T. and Country Mag. 1769, pag 662.
150. One angle at the ba$e. The $i le adjacent. The ratio of the other $ide to a line
drawn from V to an unknown point in B, and the length of a line drawn from
the $aid point parallel to the given $ide to terminate in the unknown $ide.
Hutton’s Mi$c. 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 bi$ecting the oppo$ite $ides.
Mathematician, pr. 48. - Simp. Sel. Ex. pr. 22. \\ - Palladium, 1752. qu. 43.
153. Sxs: mxn, $egts by L. m - n. Angle made by L & I bi$ecting the ba$e.
Brit. Oracle, qu. 93.
154. The ba$e of an i$oceles triangle, and the di$tance of the vertical angle from
the foot of a perpendicular from one of the equal angles upon the oppo$ite $ide.
Brit. Oracle, qu. 8.
155. P - n. m of an i$oceles triangle, m and n being $egts. by a perpendicular from
one of the angles at the ba$e on one of the equal $ides.
Court Mag. Sep. 1761.
156. V. Line bi$ecting A or a. Neare$t di$tance from V to periphery of ins. <036>.
*Gent. Diary, qu. 129.
157. V. The $egments 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 in$cribed 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 $egments of B made by a line dividing V into given angles.
Mis. Cur. Math. Vol. II. qu. 48.
161. V. One $ide. Ratio of the $egments of the ba$e made by a line dividing V
into given angles.
Palladium, 1756, pa: 43.
[0140]
SYNOPSIS
Of Data for Right-angled Triangles which have not yet
been con$tructed in general, the vertical angle being
$uppo$ed acute or obtu$e.
<035> The much greater part of the$e problems are purpo$ely le$t without any reference.
The Compiler has $een an Author from whom Con$tructions to them all may
be derived, but he forbears to name him, in order to leave them as Exerci$es
for young GEOMETRICIANS.
1. H. H x P + P<_>2.
2. H. P<_>2 - n<_>2.
3. H. m<_>2 - P<_>2.
4. H. S±m, or H. s±n.
5. H. S<_>2 + m<_>2, or H. s<_>2 + n<_>2.
6. H. S<_>2 + n<_>2.
7. H. m<_>2 - n<_>2.
8. H. l bi$ecting A or a. # Univer. Mu$eum, July, 1767. - Ladies \\ Diary, 1772, qu. 633, Cor. - Mis. \\ Scient. Curio$a, pag. 196, Cor. II.
[0141]
9. H<_>2: S x s. Any other datum.
10. H<_>2: S<_>2 + m<_>2, or H<_>2: s<_>2 + n<_>2. any other.
11. H + P. H x P.
12. H + P. H x P + P<_>2.
13. H + P. S<_>2 + n<_>2.
14. H + P. m<_>2 + n<_>2.
15. H x P + P<_>2. P.
16. H x P: S<_>2 + n<_>2. any other.
17. H x P + P<_>2: S<_>2 + n<_>2. 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. H<_>2 + S<_>2: P<_>2, or H<_>2 + s<_>2: P<_>2. any other.
22. H<_>2 + S<_>2: S<_>2 + m<_>2, or H<_>2 + s<_>2: s<_>2 + n<_>2. any other.
23. H<_>2 + S<_>2. m, or H<_>2 + s<_>2. n.
24. H<_>2 + S<_>2. n, or H<_>2 + s<_>2. m.
25. H<_>2 + S<_>2: m<_>2, or H<_>2 + s<_>2: n<_>2. 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: s<_>2, or H x s: S<_>2. any other
28. H + S + m. H<_>2 + S<_>2 + m<_>2.
29. H + S + m. H x S + S x m + S<_>2.
30. H + S + m. H x S + S x m.
31. H + S + m. H<_>2 + m<_>2.
32. H<_>2 + S<_>2 + m<_>2: H x S + S x m. any other.
33. H<_>2 + S<_>2 + m<_>2: H x S + S x m + S<_>2. any other.
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34. H<_>2 + S<_>2 + m<_>2. H + m.
35. H<_>2 + S<_>2 + m<_>2: H<_>2 - m<_>2. any other.
36. H<_>2 + S<_>2 + m<_>2: H<_>2 + 2m<_>2. any other.
37. H<_>2 + S<_>2 + m<_>2: P<_>2. any other.
38. H<_>2 + S<_>2 + m<_>2. S.
39. H<_>2 + S<_>2 + m<_>2: s<_>2. any other.
40. H<_>2 + S<_>2 + m<_>2: 2S<_>2 + m<_>2. any other.
41. H<_>2 + S<_>2 + m<_>2.: S<_>2 + 2m<_>2. any other.
42. H<_>2 + S<_>2 + m<_>2: Sxn. any other.
43. H<_>2 + S<_>2 + m<_>2. n.
44. H x S + S x m + S<_>2. H + m.
45. H x S + S x m + S<_>2. S.
46. H + m. H<_>2 - m<_>2.
47. H + m. P.
48. H + m. s.
49. H<_>2 + 2m<_>2: P<_>2. any other.
50. H<_>2±m<_>2: S<_>2. any other.
51. H<_>2 - m<_>2: S<_>2 + 2m<_>2. any other.
52. H<_>2 - m<_>2. n.
53. P. m<_>2 + n<_>2.
54. P<_>2: m<_>2±n<_>2. 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. P<_>2 + s<_>2: m<_>2 - P<_>2. any other.
61. P<_>2 + s<_>2: S<_>2 + n<_>2. any other.
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62. P<_>2 - n<_>2. m<_>2 - P<_>2.
63. P<_>2 - n<_>2: S<_>2. any other.
64. m<_>2 - P<_>2: s<_>2. any other.
65. m<_>2 - P<_>2: S<_>2 + n<_>2. any other.
66. P<_>2 - n<_>2: S<_>2 + n<_>2. any other.
67. P<_>2 - n<_>2. m.
68. m<_>2 - P<_>2. n.
69. P<_>2 - n<_>2: m<_>2. any other.
70. m<_>2 - P<_>2: n<_>2. any other.
71. P<_>2 - n<_>2. m - n.
72. m<_>2 - P<_>2. m - n.
73. P<_>2 - n<_>2: m<_>2 + 2n<_>2. any other.
74. Pxm - P x n: S<_>2 + n<_>2. 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. - Fo$ter’s Math. Lug. pr. 18.
77. S<_>2: n<_>2, or s<_>2: m<_>2. any other.
78. S<_>2: s<_>2 + n<_>2, or s<_>2: S<_>2 + m<_>2. any other.
79. S<_>2: 2s<_>2 + n<_>2, or s<_>2: 2S<_>2 + m<_>2. any other.
80. S + m. n, or s + n. m.
81. S<_>2 + m<_>2. n, or s<_>2 + n<_>2. m.
82. S<_>2 + n<_>2. m, or S<_>2 + n<_>2. n.
83. S<_>2 + n<_>2: s<_>2 + n<_>2. any other.
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84. S<_>2 + n<_>2. m - n.
85. S<_>2 + n<_>2: m<_>2 - n<_>2. any other.
86. S<_>2 + n<_>2: m<_>2 + 2n<_>2. any other.
87. s<_>2 + n<_>2: m<_>2 - n<_>2. any other.
88. m - n. m<_>2 - n<_>2.
89. Ar. Sides in arithmetical progre$$ion.
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 I$t. App. to Apoll. Gallus, pr. 7.
91. Per. Sides in geom. progre$$ion.
Simps. Alg. pr. 39.
92. I bi$ecting an acute angle. I from right angle bi$ecting the foregoing given line.
Gent. Diary, qu. 266.
93. H. Part of S adjacent to the right angle intercepted by a perpendicular to H
from middle of H.
L. Diary, qu. 633.
94. One leg and a line parallel thereto intercepted by the hypothenu$e and
the other leg being given; to determine the triangle $uch, that the rectangle un-
der the hypothenu$e and a line from the acute angle, adjacent to the given leg,
to the point of inter$ection of the parallel and the other leg may be of a given
magnitude.
L. Diary. qu. 648.
N. B. In all the Nos. from 28 to 52 inclu$ive, S may be changed for s, and m
for n, though this be not expre$$ed as in others.
FINIS.
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[0146]
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