BMO 2024 | Problem 1 | Solution - 30 April 2024

Problem 1.
Let ABC be an acute-angled triangle with AC>AB and let D be the foot of the A -angle bisector on BC. The reflections of lines AB and AC in line BC meet AC and AB at points E and F respectively. A line through D meets AC and AB at G and H respectively such that G lies strictly between A and C while H lies strictly between B and F. Prove that the circumcircles of △EDG and △FDH are tangent to each other.

Solution 1

Note that $\angle BEC = \angle BFC =\beta-\gamma$ . Moreover, $CD$ is the internal angle bisector of $\angle ACF$ while $BD$ is the external angle bisector of $\angle ABE.$
Hence, $D$ is the incenter in $\triangle ACF$ and the $E-$ excenter in $\triangle ABE$ . Therefore, $\angle DEG = \angle DFH =\frac{\beta-\gamma}{2}.$

We will be done if we show that $\angle HDE =\angle DGE+\angle HFD.$

However, we know that $\angle HDE = \angle DEG+\angle DGE=\angle HFD+\angle DGE$ , so we are done.

Solution 2

Let $A'$ be the reflection of $A$ to $BC$ . $A',B,E$ and $A',C,F$ are collinear.
$\frac{EC}{EB}=\frac{sin 180-B}{sin C}=\frac{sin B}{sin C}=\frac{AC}{AB}=\frac{DC}{DB}$ Thus $ED$ is the angle bisector of $\angle CEB$ . Similarily $FD$ is the angle bisector of $\angle BFC$ .
$\angle FDE=\angle FCE+\angle DFC+\angle CED=\angle B+\angle C$ $\angle HFD+\angle DGE=\angle B+\angle C-\angle FDH=\angle HDE$ As desired.

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