Question on circles geometry

Question

The circle ω touches the circle Ω internally at P. The centre O of Ω is outside ω. Let XY be a diameter of Ω which is also tangent to ω. Assume PY > PX. Let PY intersect ω at Z. If Y Z = 2PZ, what is the magnitude of angle PYX in degrees?

Answer 1

Consider $h$ the dilation of center $p$ and ratio $3$. Then $h(\omega)$ is a circle that passes through $P$ and $h(Z)=Y$ and has its center on the line $PO'$, ($O'$ being the center of $\omega$.) Thus $h(\omega)=\Omega$. In particular, $h(O')=O$, and this implies that $OO'=2O'P$. Let $T$ be the common point between $XY$ and $\omega$, we conclude by considering the triangle $OO'T$ that $\sin(O' OT)=\dfrac{TO'}{OO'}=\dfrac12$. Thus $\angle O' OT=30^\circ$. Finally, $\angle PYX=\dfrac12 \angle O' OT=15^\circ$.