Subject (3). Consider the triangle $ABC$ with $\mathrm{m}(\angle BAC) = 90^{\circ}$. The bisector of $\angle ACB$ intersects the segment $(AB)$ at point $E$. If there exists $D \in (CE)$ such that $\mathrm{m}(\angle DAC) = \mathrm{m}(\angle BDE) = x^{\circ}$, then calculate $x$.

The statement in the problem is unfortunate. Such a point $D$ always exists, namely $D \equiv I$, the incenter of $\triangle ABC$, for which $\angle IAC = \angle BIE = 45^{\circ}$. ${ }^{2}$ Otherwise, by continuity considerations, as $D$ traverses $CE$, from $C$ to $E$, the angle $\angle DAC$ increases from $0^{\circ}$ to $90^{\circ}$, and the angle $\angle BDE$ increases from $\frac{1}{2} \angle C$ to $90^{\circ} - \frac{1}{2} \angle C$. This means that the continuous function $f(D) = \angle BDE - \angle DAC$ takes the values $f(C) = \frac{1}{2} \angle C > 0$ and $f(E) = -\frac{1}{2} \angle C < 0$, so there must exist (at least) one point $D$ for which $f(D) = 0$. If more than one exists, and since the problem asks for a (unique) $x$, the value must be 45. Perhaps this observation should lead to the elegant synthetic solution that follows, where it is demonstrated that $D \equiv I$ is the only such point. [^1]
## Solution. (Ioan Laurenţiu Ploscaru)

Let $T$ be the symmetric point of $A$ with respect to the line $C E$; clearly $T \in (B C)$, since $C E$ is the bisector of $\angle A C B$. Then $\angle B D E = \angle D A C = \angle D T C$, therefore $\angle B D C = \angle B T D$. This means that triangles $\triangle B D C$ and $\triangle B T D$ are similar, so $\angle T D B = \frac{1}{2} \angle C$. And $\angle T A B = \frac{1}{2} \angle C$, so $\angle T A B = \angle T D B$, and thus quadrilateral $A B T D$ is cyclic. From here it follows that $\angle B A D = \angle D T C = \angle D A C$, so $A D$ is the bisector of $\angle B A C$ (and obviously, then $B D$ is the bisector of $\angle A B C$, so $D$ is the incenter $I$ of $\triangle A B C$). Therefore $x=45$.