Problem (Geometry):
Let $A B C$ be a triangle with $A B=A C$. The angle bisectors of $A$ and $B$ meet the sides $B C$ and $A C$ in $D$ and $E$, respectively. Let $K$ be the incenter of triangle $A D C$. Suppose that $\angle B E K=45^{\circ}$. Find all possible values of $\angle B A C$.

Solution:
Let $I$ be the incenter of triangle $A B C$, then $K$ lies on the line $C I$. Let $F$ be the point, where the incircle of triangle $A B C$ touches the side $A C$; then the segments $I F$ and $I D$ have the same length and are perpendicular to $A C$ and $B C$, respectively.



Figure 1



Figure 2

Let $P, Q$ and $R$ be the points where the incircle of triangle $A D C$ touches the sides $A D, D C$ and $C A$, respectively. Since $K$ and $I$ lie on the angle bisector of $\angle A C D$, the segments $I D$ and $I F$ are symmetric with respect to the line $I C$. Hence there is a point $S$ on $I F$ where the incircle of triangle $A D C$ touches the segment $I F$. Then segments $K P, K Q, K R$ and $K S$ all have the same length and are perpendicular to $A D, D C, C A$ and $I F$, respectively. So - regardless of the value of $\angle B E K$ - the quadrilateral $K R F S$ is a square and $\angle S F K=\angle K F C=45^{\circ}$.

Consider the case $\angle B A C=60^{\circ}$ (see Figure 1). Then triangle $A B C$ is equilateral. Furthermore we have $F=E$, hence $\angle B E K=\angle I F K=\angle S E K=45^{\circ}$. So $60^{\circ}$ is a possible value for $\angle B A C$.

Now consider the case $\angle B A C=90^{\circ}$ (see Figure 2). Then $\angle C B A=\angle A C B=45^{\circ}$. Furthermore, $\angle K I E=\frac{1}{2} \angle C B A+\frac{1}{2} \angle A C B=45^{\circ}, \angle A E B=180^{\circ}-90^{\circ}-22.5^{\circ}=67.5^{\circ}$ and $\angle E I A=\angle B I D=180^{\circ}-90^{\circ}-22.5^{\circ}=67.5^{\circ}$. Hence triangle $I E A$ is isosceles and a reflection of the bisector of $\angle I A E$ takes $I$ to $E$ and $K$ to itself. So triangle $I K E$ is symmetric with respect to this axis, i.e. $\angle K I E=\angle I E K=\angle B E K=45^{\circ}$. So $90^{\circ}$ is a possible value for $\angle B A C$, too.

If, on the other hand, $\angle B E K=45^{\circ}$ then $\angle B E K=\angle I E K=\angle I F K=45^{\circ}$. Then

- either $F=E$, which makes the angle bisector $B I$ be an altitude, i.e., which makes triangle $A B C$ isosceles with base $A C$ and hence equilateral and so $\angle B A C=60^{\circ}$,
- or $E$ lies between $F$ and $C$, which makes the points $K, E, F$ and $I$ concyclic, so $45^{\circ}=$ $\angle K F C=\angle K F E=\angle K I E=\angle C B I+\angle I C B=2 \cdot \angle I C B=90^{\circ}-\frac{1}{2} \angle B A C$, and so $\angle B A C=90^{\circ}$,



- or $F$ lies between $E$ and $C$, then again, $K, E, F$ and $I$ are concyclic, so $45^{\circ}=\angle K F C=$ $180^{\circ}-\angle K F E=\angle K I E$, which yields the same result $\angle B A C=90^{\circ}$. (However, for $\angle B A C=90^{\circ} E$ lies, in fact, between $F$ and $C$, see Figure 2. So this case does not occur.)

This proves $90^{\circ}$ and $60^{\circ}$ to be the only possible values for $\angle B A C$.

Denote angles at $A, B$ and $C$ as usual by $\alpha, \beta$ and $\gamma$. Since triangle $A B C$ is isosceles, we have $\beta=\gamma=90^{\circ}-\frac{\alpha}{2}<90^{\circ}$, so $\angle E C K=45^{\circ}-\frac{\alpha}{4}=\angle K C D$. Since $K$ is the incenter of triangle $A D C$, we have $\angle C D K=\angle K D A=45^{\circ}$; furthermore $\angle D I C=45^{\circ}+\frac{\alpha}{4}$. Now, if $\angle B E K=45^{\circ}$, easy calculations within triangles $B C E$ and $K C E$ yield

$$
\begin{aligned}
& \angle K E C=180^{\circ}-\frac{\beta}{2}-45^{\circ}-\beta=135^{\circ}-\frac{3}{2} \beta=\frac{3}{2}\left(90^{\circ}-\beta\right)=\frac{3}{4} \alpha, \\
& \angle I K E=\frac{3}{4} \alpha+45^{\circ}-\frac{\alpha}{4}=45^{\circ}+\frac{\alpha}{2} .
\end{aligned}
$$

So in triangles $I C E, I K E, I D K$ and $I D C$ we have (see Figure 3)

$$
\begin{array}{ll}
\frac{I C}{I E}=\frac{\sin \angle I E C}{\sin \angle E C I}=\frac{\sin \left(45^{\circ}+\frac{3}{4} \alpha\right)}{\sin \left(45^{\circ}-\frac{\alpha}{4}\right)}, & \frac{I E}{I K}=\frac{\sin \angle E K I}{\sin \angle I E K}=\frac{\sin \left(45^{\circ}+\frac{\alpha}{2}\right)}{\sin 45^{\circ}} \\
\frac{I K}{I D}=\frac{\sin \angle K D I}{\sin \angle I K D}=\frac{\sin 45^{\circ}}{\sin \left(90^{\circ}-\frac{\alpha}{4}\right)}, & \frac{I D}{I C}=\frac{\sin \angle I C D}{\sin \angle C D I}=\frac{\sin \left(45^{\circ}-\frac{\alpha}{4}\right)}{\sin 90^{\circ}} .
\end{array}
$$



Figure 3

Multiplication of these four equations yields

$$
1=\frac{\sin \left(45^{\circ}+\frac{3}{4} \alpha\right) \sin \left(45^{\circ}+\frac{\alpha}{2}\right)}{\sin \left(90^{\circ}-\frac{\alpha}{4}\right)} .
$$

But, since

$$
\begin{aligned}
\sin \left(90^{\circ}-\frac{\alpha}{4}\right) & =\cos \frac{\alpha}{4}=\cos \left(\left(45^{\circ}+\frac{3}{4} \alpha\right)-\left(45^{\circ}+\frac{\alpha}{2}\right)\right) \\
& =\cos \left(45^{\circ}+\frac{3}{4} \alpha\right) \cos \left(45^{\circ}+\frac{\alpha}{2}\right)+\sin \left(45^{\circ}+\frac{3}{4} \alpha\right) \sin \left(45^{\circ}+\frac{\alpha}{2}\right)
\end{aligned}
$$

this is equivalent to

$$
\sin \left(45^{\circ}+\frac{3}{4} \alpha\right) \sin \left(45^{\circ}+\frac{\alpha}{2}\right)=\cos \left(45^{\circ}+\frac{3}{4} \alpha\right) \cos \left(45^{\circ}+\frac{\alpha}{2}\right)+\sin \left(45^{\circ}+\frac{3}{4} \alpha\right) \sin \left(45^{\circ}+\frac{\alpha}{2}\right)
$$

and finally

$$
\cos \left(45^{\circ}+\frac{3}{4} \alpha\right) \cos \left(45^{\circ}+\frac{\alpha}{2}\right)=0 .
$$



But this means $\cos \left(45^{\circ}+\frac{3}{4} \alpha\right)=0$, hence $45^{\circ}+\frac{3}{4} \alpha=90^{\circ}$, i.e. $\alpha=60^{\circ}$ or $\cos \left(45^{\circ}+\frac{\alpha}{2}\right)=0$, hence $45^{\circ}+\frac{\alpha}{2}=90^{\circ}$, i.e. $\alpha=90^{\circ}$. So these values are the only two possible values for $\alpha$.

On the other hand, both $\alpha=90^{\circ}$ and $\alpha=60^{\circ}$ yield $\angle B E K=45^{\circ}$.

Final answer: $90^{\circ}$,$60^{\circ}$