, 2023

Shortlist: Keep confidential!

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 $\triangle XAO$ and $\triangle YAO$ 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, $\triangle ABD$ and $\triangle ACD$ 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. $\square$