Slovenija 2008

Let $C^{\prime }$ be the foot of the altitude from $C$. Since the triangle $ABC$ is isosceles with the apex at $C$, we have $|A{C}^{\prime }|=|C^{\prime }B|=\frac{1}{2}|AB|=|C{A}^{\prime }|$. Since $AB{A}^{\prime }$ and $CB{C}^{\prime }$ are right triangles and $\angle AB{A}^{\prime }=\angle CB{C}^{\prime }$, they are similar. This implies $$\frac{|AB|}{|B{A}^{\prime }|}=\frac{|CB|}{|B{C}^{\prime }|}.$$ Let $c$ be the length of the side $AB$ and $x=|B{A}^{\prime }|$. 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 $|C{A}^{\prime }|=\frac{c}{2}$ and $|A^{\prime }B|=a-\frac{c}{2}$. We can now calculate the length of the altitude $A{A}^{\prime }$ in two different ways, namely as the cathetus in triangles $AC{A}^{\prime }$ and $AB{A}^{\prime }$, respectively. Let $|A{A}^{\prime }|=v$. 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.