Geometry

In triangle ABC, X lies on AB and Y lies on AC such that BY bisects ∠ABC and CX bisects ∠ACB. BY and CX intersect at a point P. Suppose that P lies on the circumcircle of triangle AXY . If AX = 15 and AY = 24, find AP^2.

So far I've worked out that angle BAC = 60 degrees, and after doing some trig computations, I was also able to find that angle XPA = 60 - arctan(3*sqrt(3)/13) and YPA = 60 + arctan(3*sqrt(3)/13), but AP still seems out of reach. Any assistance would be most appreciated.

OP

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Original post by Prasiortle

In triangle ABC, X lies on AB and Y lies on AC such that BY bisects ∠ABC and CX bisects ∠ACB. BY and CX intersect at a point P. Suppose that P lies on the circumcircle of triangle AXY . If AX = 15 and AY = 24, find AP^2.

So far I've worked out that angle BAC = 60 degrees, and after doing some trig computations, I was also able to find that angle XPA = 60 - arctan(3*sqrt(3)/13) and YPA = 60 + arctan(3*sqrt(3)/13), but AP still seems out of reach. Any assistance would be most appreciated.

How did u work out BAC?

Reply 3

Prasiortle

OP

13

Original post by RickHendricks

How did u work out BAC?

By simple angle chasing. Anyway, if you're not good enough at geometry to even see how to find angle BAC, I doubt you will be able to help me.

Reply 4

ghostwalker

17

Original post by Prasiortle

...

Someone posted the idea of joining X and Y by a straight line, and working from there, but unfortunately they've now deleted it. Although their subsequent working was flawed, the initial move was sound.

And XY=21 by cosine rule on AXY

Spoiler

(edited 5 years ago)

Reply 5

RickHendricks

19

Original post by Prasiortle

By simple angle chasing. Anyway, if you're not good enough at geometry to even see how to find angle BAC, I doubt you will be able to help me.