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