Baltic Way 2023 Shortlist

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Step 1: Points $X,P,O,Q,Y$ lie on a circle. Proof. Since point $X$ lies on the perpendicular bisector of $AB$, we have that $XA,XB$ are tangents to $\odot (ABC)$. This means that $\angle XAO=\angle OBX=90^{\circ }$, therefore quadrilateral $OAXB$ is cyclic. From Power of the Point we have $$|XM|\cdot |MO|=|AM|\cdot |MB|=|PM|\cdot |MQ|$$ since quadrilateral $APBQ$ is cyclic too. This implies that points $X,P,O,Q$ lie on the same circle. Similarly, we can prove that points $Y,P,O,Q$ lie on the same circle, so points $X,P,O,Q,Y$ lie on a circle as desired. $\square$

Step 2: $\angle APX=\angle YQA$ Proof. Observe that $\angle XAP=\angle AQP$. Moreover, $\angle AYP=\angle XQP$, since quadrilateral $XPQY$ is cyclic. We deduce that $$\angle AQX=\angle AQP-\angle XQP=\angle XAP-\angle AYP=\angle APY.$$ Remember that $\angle XPY=\angle XQY$ as again quadrilateral $XPQY$ is cyclic. Consequently, we have $$\angle XPA=\angle XPY-\angle APY=\angle XQY-\angle AQX=\angle AQY$$ as desired. $\square$

Notice that $△XAO$ and $△YAO$ are right angle triangles, therefore $|XA|=|AO|\cdot \tan \angle ACB$ and $|AY|=|AO|\cdot \tan \angle CBA$. This means that $\frac{|AX|}{|AY|}=\frac{\tan \angle ACB}{\tan \angle CBA}$. Similarly, $△ABD$ and $△ACD$ are right angle triangles, therefore $|AD|=|BD|\cdot \tan \angle CBA$ and $|AD|=|CD|\cdot \tan \angle ACB$. This means that $\frac{|BD|}{|DC|}=\frac{\tan \angle ACB}{\tan \angle CBA}$. We deduce that $$\frac{|TX|}{|TY|}=\frac{|AX|}{|AY|}=\frac{\tan \angle ACB}{\tan \angle CBA}=\frac{|BD|}{|DC|}$$ as desired. $\square$