41st Balkan Mathematical Olympiad

Let $M$ be the midpoint of $CD$ and let $Q$ and $R$ be the points of intersection of $XT$ and $YZ$ with $CD$ respectively. Since $$\angle RZX=\angle YZX=90^{\circ }=\angle RMX\text{and}\angle QTY=\angle XTY=90^{\circ }=\angle QMY$$ the quadrilaterals $XZRM$ and $YTQM$ are cyclic.



By the power of the point $A$ with respect to the circumcircle of $XZRM$ and with respect to $\omega$ we have $$AR\cdot AM=AZ\cdot AX=AC\cdot AD$$ It follows that $$AR=\frac{AC\cdot AD}{AM}$$ which is independent of the circle $\omega$. So $Z$ is a point on the fixed circle of diameter $AR$. Similarly, $Q$ is independent of the circle $\omega$ and $T$ is a point on the fixed circle of diameter $BQ$. Let $P$ be the point of intersection of $ZT$ with $CD$. We will show that $P$ is a fixed point independent of $\omega$.

Since $$\angle QTZ=\angle XTZ=\angle XYZ=90^{\circ }-\angle YXZ=\angle ZAQ$$ then $ATQZ$ is cyclic, thus $$PT\cdot PZ=PA\cdot PQ.$$ Letting $U$ be the point of intersection of $ZT$ with the circumcircle of $△AZR$ we also have $$PU\cdot PZ=PA\cdot PR.$$ We deduce that $$\frac{PT}{PU}=\frac{PQ}{PR}$$ from which it follows that $P$ is the centre of homothety of the two fixed circles with diameters $AR$ and $BQ$. Thus $P$ is indeed a fixed point. $\square$

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Alternative solution.

Applying the properties of cross (double) ratio we get: $$(A,C;P,D)=Z(X,C;T,D)\stackrel{\omega }{\cong }Y(X,C;T,D)=(M,C;B,D)$$ It follows that $$\frac{AP}{CP}:\frac{AD}{CD}=\frac{MB}{BC}:\frac{MD}{CD}\Rightarrow \frac{AP}{CP}=\frac{MB}{BC}\cdot \frac{AD}{MD}=\text{const}$$ Therefore $P$ is a fixed point. $\square$