SOLUTIONS BY MEANS OF CONICS
243
Then, by the above proposition,
ZPBA = 2 ZPAB.
Therefore their doubles are equal,
or ¿POA = 2 ¿POB,
and OP accordingly trisects the arc APB and the angle AOB.
2. ‘ Some says Pappus, set out another solution not in
volving recourse to a vevans, as follows.
Let BPS be an arc of a circle which it is required to
trisect.
Suppose it done, and let the arc SP be one-third of the
arc SPR.
Join BP, SP.
Then the angle BSP is equal
to twice the angle SBP.
Let SE bisect the angle BSP, R x n
meeting BP in E, and draw EX, PX perpendicular to RS.
Then Z ERS = Z ESR, so that RE = ES.
Therefore RX = XS, and X is given.
Again BS: SP = RE: EP = RX : XX]
therefore RS: RX = SP: XX.
But
therefore
RS= 2 RX;
SP = 2 XX.
It follows that P lies on a hyperbola with S as focus and XE
as directrix, and with eccentricity 2.
Hence, in order to trisect the arc, we have only to bisect RS
at X, draw XE at right angles to RS, and then draw a hyper
bola with S as focus, XE as directrix, and 2 as the eccentricity.
The hyperbola is the same as that used in the first solution.
The passage of Pappus from which this solution is taken is
remarkable as being one of three passages in Greek mathe
matical works still extant (two being in Pappus and one in
a fragment of Anthemius on burning mirrors) which refer to
the focus-and-directrix property of conics. The second passage
in Pappus comes under the heading of Lemmas to the Surface-
Loci of Euclid. 1 Pappus there gives a complete proof of the
1 Pappus, vii, pp. 1004-1114.
R 2