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Iranian Mathematical Olympiad

Geometry Difficulty 6.3 National Olympiad Prove it Iran

Given an inscribed pentagon $ABCDE$ with circumcircle $\Gamma$ . Line $\ell$ passes through vertex $A$ and is tangent to $\Gamma$ . Points $X, Y$ lie on $\ell$ so that $A$ lies between $X$ and $Y$ . Circumcircle of triangle $\triangle XED$ intersects segment $AD$ at $Q$ and circumcircle of triangle $\triangle YBC$ intersects segment $AC$ at $P$ . Lines $XE, YB$ intersect at $S$ , and lines $XQ, YP$ at $Z$ . Prove that circumcircle of triangles $\triangle XYZ$ and $\triangle BES$ are tangent.

Solution

Assume the circumcircles of $\triangle ABY$ and $\triangle AEX$ meet for the second time at $K$ . Since
$\angle KEX = \angle KAX = 180^\circ - \angle KAY = 180^\circ - \angle KBY = \angle KBS,$
we find out that $KBES$ is concyclic. We have
$\angle YKB = \angle YAB = \angle BCA = \angle PYB,$

$\angle EKX = \angle EAX = \angle ADE = \angle QXE.$

$\begin{align*} \angle YZX &= \angle YSX - \angle SYZ - \angle SXZ \\ &= 180^\circ - (\angle BKE + \angle YKB + \angle EKX) \\ &= 180^\circ - \angle YKX. \end{align*}$
Therefore, $YKX$ is inscribed in a circle called $\omega$ . Let's say the tangent line to $\omega$ at $K$ meets $XY$ at $T$ . We have
$\angle TKB = \angle TKY + \angle YKB = \angle KXT + \angle YAB = \angle KEA + \angle AEB = \angle KEB.$
As a result, $TK$ is tangent to the circumcircle of $\triangle KEB$ . So, the two circles are tangent at $K$ .

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