66th Czech and Slovak Mathematical Olympiad

Points $K$, $L$, $M$ are defined as feet of altitudes and we need to express perpendicularity. Since the statement involves lengths and not angles, we characterise the perpendicularity by lengths of segments. Comparing Pythagorean theorems in triangles $AMC$, $BMC$ we learn $A{C}^2-B{C}^2=A{M}^2-B{M}^2$. Denoting the lengths of $BC$, $CA$, $AB$ by $a$, $b$, $c$, respectively, this implies $$AM-BM=\frac{b^2-a^2}{AM+BM}=\frac{b^2-a^2}{c}$$ and likewise $$BK-CK=\frac{c^2-b^2}{a}\text{and}CL-AL=\frac{a^2-c^2}{b}.$$

The equality from the problem statement rewrites as $$\begin{array}{c}\frac{a^2-b^2}{c}+\frac{b^2-c^2}{a}+\frac{c^2-a^2}{b}=0,\\ ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)=0.\end{array}$$ Let's try to factor the left-hand side by viewing it as a cubic polynomial $P$ in variable $a$. It is easy to check that for isosceles triangle the left-hand side is zero, hence $P(b)=0$ and $P(c)=0$ and we know two roots of $P$. After dividing $P(a)$ by $(a-b)(a-c)$ we are left with linear polynomial $a(b-c)+b^2-c^2$ which can be factored easily. To sum up, the equality from problem statement is equivalent with $$(a-b)(b-c)(c-a)(a+b+c)=0.$$ It's obvious that this equality holds if and only if triangle $ABC$ is isosceles.