Baltic Way 2023 Shortlist

Let $O$ be the circumcentre of $\odot (ABC)$. We proceed in several steps.

Step 1: Point $O$ lies on $\odot (ABD)$ and $\odot (ACE)$. Proof. Note that since $|AD|=|DC|$, we have $\angle CDA=180^{\circ }-2\angle ACB$, therefore $\angle ADB=2\angle ACB=\angle AOB$, which means that quadrilateral $AODB$ is cyclic. Similarly, we can prove that quadrilateral $AOEC$ is cyclic. $\square$

Step 2: Points $Y$, $A$, $Z$ are collinear. Moreover, $YZ$ is tangent to $\odot (ABC)$ at point $A$. Proof. Note that since $YB$ is tangent to $\odot (ABC)$, we have that: $$\angle AOY=\angle YBA=\angle ACB.$$ Since $\angle AOB=2\angle ACB$, we get that: $$\angle YAB=\angle YOB=\angle AOB-\angle AOY=\angle ACB.$$ This means that $YA$ is tangent to $\odot (ABC)$. Consequently, $|YA|=|YB|$ implies that $Y$ lies on the perpendicular bisector of $AB$. Similarly, we can prove that $AZ$ is tangent to $\odot (ABC)$ and $Z$ lies on the perpendicular bisector of $AC$. Since the tangent line at a fixed point is unique, we conclude that points $Y$, $A$, $Z$ all lie on the tangent to $\odot (ABC)$ at the point $A$. $\square$

Step 3: Points $Y$, $O$, $E$ and $Z$, $O$, $D$ are collinear. Moreover, $O$ is the orthocentre of $△YZX$. Proof. The first part follows from the previous result that $Y$ lies on the perpendicular bisector of $AB$, which is $OE$. A similar argument applies to $Z$, $O$, $D$. Since the radius of a circle is perpendicular to the corresponding tangent, note that: $$\angle OBY=\angle ODY=\angle YAO=90^{\circ }\text{and}\angle OCZ=\angle OEZ=90^{\circ }.$$ This gives us that $ZD\perp YX$ and $YE\perp ZX$, implying that $O$ is the orthocentre of $△YZX$. Also note that this immediately gives us that points $A$, $O$, $X$ are collinear, too. $\square$

To finish the problem, note that quadrilateral $YDEZ$ is cyclic from $\angle YDZ=\angle YEZ=90^{\circ }$. Therefore: $$\angle DAO=\angle DYO=\angle OZE=\angle OAE.$$

Since $OX$ is the diameter of $\odot (DEX)$ (note that $\angle ODX=\angle OEX=90^{\circ }$) and $OX$ is the bisector of $\angle DAE$, by symmetry we have that $DV$ and $UE$ are parallel lines. It is well-known that the diagonals of an isosceles trapezoid intersect on the perpendicular bisector of their parallel sides. Therefore, lines $UV$, $DE$, $AX$ are concurrent, as desired.