Problem 2. Let $ABC$ be an acute triangle. Let $D$ be the foot of the altitude corresponding to side $BC$; $M$ the midpoint of side $BC$ and $F$ the intersection point of the internal angle bisector of $\angle BAC$ with side $BC$. Determine all triangles for which $F$ is the midpoint of segment $DM$.

Note. It is understood that the determination of a triangle consists in giving the length of its sides, in terms of some parameter if necessary.
Solution. The condition $D F=F M$ can be rewritten as

$$
2 B F=B M+B D
$$

If we call $a, b$ and $c$ the lengths of the sides $B C, C A$ and $A B$, respectively, and apply the Angle Bisector Theorem, we will have that

$$
\frac{B F}{c}=\frac{F C}{b}=\frac{a}{b+c}
$$

Consequently,

$$
B F=\frac{a c}{b+c}
$$

On the other hand, applying trigonometry first in the triangle $A B D$ and then the Law of Cosines in the triangle $A B C$,

$$
B D=c \cos \beta=\frac{c^{2}+a^{2}-b^{2}}{2 a}
$$

Substituting the values of (4) and (5) into (3), we get

$$
\frac{2 a c}{b+c}=\frac{a}{2}+\frac{c^{2}+a^{2}-b^{2}}{2 a} .
$$

By simplifying, we arrive at

$$
\begin{aligned}
& 2 a^{2} b-2 a^{2} c+c^{3}-b^{2} c+c^{2} b-b^{3}=(b-c)\left(2 a^{2}-c^{2}-b c-b c-b^{2}\right) \\
& =(b-c)\left(2 a^{2}-(b+c)^{2}\right)=(b-c)(\sqrt{2} a-b-c)(\sqrt{2} a+b+c)=0
\end{aligned}
$$

Of the three factors, the last one clearly cannot be zero, leaving the possible options as

$$
b=c \quad \text { or } \quad b+c=\sqrt{2} a
$$

That is, the triangle is either isosceles at $A$ or the side $a$ is given by the arithmetic mean of $b$ and $c$ multiplied by the factor $\sqrt{2}$.

Finally, observe that a triangle of type $(a, b, b)$ is acute if and only if $a<\sqrt{2} b$. On the other hand, one of type $((b+c) / \sqrt{2}, b, c)$ is acute if and only if $c / 3<b<c$.