Slovenian Mathematical Olympiad, 2008

Let $C'$ be the foot of the altitude from $C$. Since the triangle $ABC$ is isosceles with the apex at $C$, we have $|AC'| = |C'B| = \frac{1}{2}|AB| = |CA'|$. Since $ABA'$ and $CBC'$ are right triangles and $\angle ABA' = \angle CBC'$, they are similar. This implies
$$\frac{|AB|}{|BA'|} = \frac{|CB|}{|BC'|}.$$
Let $c$ be the length of the side $AB$ and $x = |BA'|$. Then
$$\frac{c}{x} = \frac{\frac{c}{2} + x}{\frac{c}{2}},$$

and we get the quadratic equation $2x^2 + cx - x^2 = 0$. The left-hand side can be factored as $(2x - c)(x + c) = 0$. Since $x$ and $c$ are positive, we have $x = \frac{c}{2}$. The length of the segment $BC$ is therefore equal to $c$, $|AC| = |BC| = |AB| = c$ and the triangle $ABC$ is equilateral.

Second solution

Denote the length of the side $AC$ by $a$ and the length of the side $AB$ by $c$. Then $|CA'| = \frac{c}{2}$ and $|A'B| = a - \frac{c}{2}$. We can now calculate the length of the altitude $AA'$ in two different ways, namely as the cathetus in triangles $ACA'$ and $ABA'$, respectively. Let $|AA'| = v$. Pythagoras's theorem for the first of the triangles implies $v^2 = a^2 - (c/2)^2$ and for the second $v^2 = c^2 - (a - (c/2))^2$, so
$$a^2 - \frac{c^2}{4} = c^2 - a^2 + ac - \frac{c^2}{4}$$
or $2a^2 - ac - c^2 = 0$. This quadratic equation can be factored as $(2a + c)(a - c) = 0$, which implies $a = c$ (since $a$ and $c$ are positive the condition $2a + c = 0$ is never satisfied). From $|BC| = |AC| = a = c$ we conclude that the triangle $ABC$ is equilateral.