3. Given is a triangle $ABC$. Points $D$ and $E$ on line $AB$ are such that $AD = AC$ and $BE = BC$, with the arrangement $D-A-B-E$. The circumcircles of triangles $DBC$ and $EAC$ intersect at point $X \neq C$, and the circumcircles of triangles $DEC$ and $ABC$ intersect at point $Y \neq C$. If $DY + EY = 2XY$, determine $\angle ACB$.

(Miloš Milosavljević)

## Ministry of Education, Science and Technological Development Mathematical Society of Serbia  14th SERBIAN MATHEMATICAL OLYMPIAD FOR HIGH SCHOOL STUDENTS

## Second day
3. Let $I_{c}$ be the center of the excircle of $\triangle ABC$ opposite vertex $C$. Then $A I_{c}$ is the angle bisector of $\angle CAD$, from which $\triangle I_{c} A D \cong \triangle I_{c} A C$. It follows that $\angle I_{c} D B = \angle I_{c} D A = \angle I_{c} C A = \angle I_{c} C B$, so $I_{c}$ lies on the circle $BCD$. Similarly, $I_{c}$ lies on the circle $ACE$, which means $X \equiv I_{c}$. Also, since $X D = X C$ and similarly $X E = X C$, point $X$ is the center of the circle $CDEY$.

Further, $\angle D Y A = \angle C Y A - \angle C Y D = \angle C B A - \angle C E D = \angle C E D = \angle C X A = \angle D X A$ and similarly $\angle E Y B = \angle E X B$. Therefore, consider the point $X^{\prime}$ symmetric to point $X$ with respect to $AB$. Because $\angle D X^{\prime} A = \angle D Y A$, this point lies on the circle $ADY$, and similarly on the circle $BEY$.

![](https://cdn.mathpix.com/cropped/2024_06_04_fe4fa2bd25c3abd0cb36g-4.jpg?height=368&width=583&top_left_y=427&top_left_x=1222)

If $\angle C = \angle A Y B$, then point $X^{\prime}$ lies inside $\triangle A B Y$, and thus inside $\triangle D E Y$. It follows that $D Y + E Y > D X^{\prime} + E X^{\prime} = D X + E X = 2 X Y$. Similarly, if $\angle C > 60^{\circ}$, point $Y$ lies inside $\triangle D E X$, and then $D Y + E Y < 2 X Y$. Therefore, if $D Y + E Y = 2 X Y$, it must be that $\angle C = 60^{\circ}$, and then $X^{\prime} \equiv Y$, so the equality indeed holds.

Another solution. Let the angles of $\triangle ABC$ be denoted by $\alpha, \beta, \gamma$ as usual. As in the first solution, points $C, D, E, Y$ lie on a circle with center $X$. Also, since $\angle C Y D = \angle C E D = \frac{\beta}{2}$, line $DY$ contains the midpoint $N$ of arc $AC$ of the circle $ABC$.

Let $\angle Y D E = x$ and $\angle Y E D = y$. The condition $D Y + E Y = 2 X Y$ 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 theorem in $\triangle ADN$ gives $2 \sin x \sin \frac{\alpha + \gamma}{2} = \cos \left(\frac{\alpha - \gamma}{2} - x\right)$, from which $\operatorname{tg} x = \frac{\cos \frac{\alpha - \gamma}{2}}{2 \sin \frac{\alpha + \gamma}{2} - \sin \frac{\alpha - \gamma}{2}}$. 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 (*) gives

$$
2 \cos \frac{3 \gamma - 180^{\circ}}{4} \cos \frac{\varphi}{2} = \sqrt{3 - 2 \cos \gamma + 4 \sin \frac{\gamma}{2} \cos \varphi}
$$

which, after squaring, reduces to

$$
\cos \varphi = \frac{2 - 2 \cos \gamma - \sin \frac{3 \gamma}{2}}{1 - 4 \sin \frac{\gamma}{2} + \sin \frac{3 \gamma}{2}} = \frac{-3 t + 4 t^{2} + 4 t^{3}}{1 - t - 4 t^{3}}, \quad \text{where} \quad t = \sin \frac{\gamma}{2}
$$

If $\gamma \neq 60^{\circ}$, canceling $2 t - 1 \neq 0$ gives $\cos \varphi = -\frac{2 t^{2} + 3 t}{2 t^{2} + t + 1} < 0$, which is impossible. Therefore, it must be that $\gamma = 60^{\circ}$.