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Belarus competition problems

Geometry Difficulty 5.1 AIME, harder Prove it Belarus

Two circles $\omega_1$ and $\omega_2$ intersect each other at points $X$ and $Y$ . Two lines pass through $Y$ : one intersects $\omega_1$ and $\omega_2$ for the second time at points $A$ and $B$ respectively and another intersects $\omega_1$ and $\omega_2$ for the second time at points $C$ and $D$ respectively. The line $AD$ intersects $\omega_1$ and $\omega_2$ for the second time at points $P$ and $Q$ respectively such that $YP = YQ$ .
Prove that circumcircles of $\triangle BCY$ and $\triangle PQY$ touch each other.
(Palina Chernikava)

Solution

The quadrilaterals $ACYP$ and $BDQY$ are cyclic, so
$\angle ACD = \angle ACY = \angle YPQ \text{ and } \angle ABD = \angle YBD = \angle YQP.$
Since the triangle $YPQ$ is isosceles, $\angle YPQ = \angle YQP$ whence $\angle ACD = \angle ABD$ . So the quadrilateral $ACBD$ is cyclic. Therefore
$\begin{aligned} \angle CYP &= \angle CYA + \angle AYP = \angle BYD + (\angle YPQ - \angle YAP) = \\ &= \angle YPQ + (\angle BYD - \angle BAD) = \angle YQP + \angle ADC = \\ &= \angle YQP + \angle ABC = \angle YQP + \angle YBC. \end{aligned}$
The last equality is equivalent to the tangency of the circumcircles of the triangles $BCY$ and $PQY$ .

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