Let $\triangle ABC$ be a triangle with circumcenter P and $\angle BAC = 60^{\circ}$. Suppose line $ BP$ intersects $AC$ at point $X$, and line $CP$ intersects $AB$ at point $Y$. If $CY = 6$ and $\angle PXY = 15^{\circ}$, then find the length of $XY.$
What I tried so far,
Using angle chasing methods, I found below angles:
$\angle$BPC = 120$^\circ$
$\angle$PBC = 30$^\circ$
$\angle$PCB = 30$^\circ$
$\angle$PYX = 45$^\circ$
Other angles can also be found easily.
Also, AP = BP = CP = circumradius
I tried using trigonometry, sine/cosine rule, but not getting clue how to get XY's length. If I get the circumradius from the CY's value, the we can find PY. then if we get the value of PX, then can apply law of cosine, to get the value of XY.
How to proceed from here?
Observe, $AXPY$ is a cyclic quadrilateral $\implies \angle XAP=\angle XYP=45^{\circ}$.
Since $P$ is the circumcenter, $PA=PC\implies \angle XCP=\angle XAP=45^{\circ}.$
Hence, $\triangle XYC$ is isosceles with $\angle YXC=90^{\circ}$. $$\therefore ~~ XY=3\sqrt{2} $$