Serbian Mathematical Olympiad

Solution:

Let $I_{c}$ be the center of the excircle of $\triangle ABC$ opposite vertex $C$. Then $AI_{c}$ is the bisector of angle $CAD$, from which $\triangle I_{c}AD \cong \triangle I_{c}AC$. It follows that $\text{I c DB = I c DA = I c CA = I c CB}$, so $I_{c}$ lies on the circle $BCD$. Similarly, $I_{c}$ lies on the circle $ACE$, which means that $X \equiv I_{c}$. Also, since $XD = XC$ and analogously $XE = XC$, the point $X$ is the center of the circle $CDEY$.

Further, $\text{DYA = CYA - CYD = CBA - CED = CED = CXA = DXA}$ and similarly $\text{EYB = EXB}$. So let us consider the point $X'$ symmetric to point $X$ with respect to $AB$. Because of $\text{DX'A = DYA}$ this point lies on the circle $ADY$, and analogously also on the circle $BEY$.

If $\text{C < 60}$, then $\text{AX'B = AXB = 90 - 1 2 C > C = AYB}$, so the point $X'$ lies inside $\triangle ABY$, and hence also inside $\triangle DEY$. It follows that $DY + EY > DX' + EX' = DX + EX = 2XY$. Similarly, if $\text{C > 60}$, the point $Y$ lies inside $\triangle DEX$, so then $DY + EY < 2XY$. Therefore, if $DY + EY = 2XY$, we must have $\text{C = 60}$, and then $X' \equiv Y$ so the equality indeed holds.

Second solution. We denote the angles of triangle $ABC$ in the usual way by $\alpha, \beta, \gamma$. As in the first solution, the points $C, D, E, Y$ lie on a circle with center $X$. Also, because of $\text{CYD = CED = 2}$, the line $DY$ contains the midpoint $N$ of the arc $AC$ of the circle $ABC$.

Let us denote $\text{YDE = x}$ and $\text{YED = y}$. The condition $DY + EY = 2XY$ gives $\sin x + \sin y = \sin x + \sin \left(90^{\circ} - \frac{\gamma}{2} - x\right) = 1$.

$$\left(1 - \sin \frac{\gamma}{2}\right) \sin x + \cos \frac{\gamma}{2} \cos x = 1$$

On the other hand, the sine rule in $\triangle ADN$ gives $2 \sin x \sin \frac{\alpha + \gamma}{2} = \cos \left(\frac{\alpha - \gamma}{2} - x\right)$, from which it follows that $\operatorname{tg} x = \frac{\cos \frac{\alpha - \gamma}{2}}{2 \sin \frac{\alpha + \gamma}{2} - \sin \frac{\alpha - \gamma}{2}}$. From this we find
$$\sin x = \frac{\cos \frac{\alpha - \gamma}{2}}{\sqrt{3 - 2 \cos \gamma + 4 \sin \frac{\gamma}{2} \cos \varphi}}, \quad \cos x = \frac{2 \sin \frac{\alpha + \gamma}{2} - \sin \frac{\alpha - \gamma}{2}}{\sqrt{3 - 2 \cos \gamma + 4 \sin \frac{\gamma}{2} \cos \varphi}}$$
where $\varphi = \alpha + \frac{\gamma}{2} - 90^{\circ} = \frac{\alpha - \beta}{2}$. Substituting into equation (*) we obtain
$$2 \cos \frac{3\gamma - 180^{\circ}}{4} \cos \frac{\varphi}{2} = \sqrt{3 - 2 \cos \gamma + 4 \sin \frac{\gamma}{2} \cos \varphi}$$
which after squaring reduces to
$$\cos \varphi = \frac{2 - 2 \cos \gamma - \sin \frac{3\gamma}{2}}{1 - 4 \sin \frac{\gamma}{2} + \sin \frac{3\gamma}{2}} = \frac{-3t + 4t^{2} + 4t^{3}}{1 - t - 4t^{3}}, \quad \text{where} \quad t = \sin \frac{\gamma}{2}$$
If $\gamma \neq 60^{\circ}$, cancelling $2t - 1 \neq 0$ gives $\cos \varphi = -\frac{2t^{2} + 3t}{2t^{2} + t + 1} < 0$, which is impossible. Therefore, we must have $\gamma = 60^{\circ}$.