Czech-Slovak-Polish Match

A point $V$ of the plane containing a triangle $ABC$ is its orthocenter if and only if, at the same time, $AV\perp BC$ and $BV\perp AC$; that is, $AV\cdot BC=0$ and $BV\cdot AC=0$. Substituting $BC=BV-CV$ and $AC=AV-CV$, an easy manipulation leads to the equivalent condition in the form of the equality of the scalar products $$AV\cdot BV=AV\cdot CV=BV\cdot CV.(1)$$ Our goal is thus to find out when the system (1) is satisfied together with the analogous system $$KV\cdot LV=KV\cdot MV=LV\cdot MV,(2)$$ expressing the fact that the point $V$ is the orthocenter of the triangle $KLM$. We now express the vectors from (2) as linear combinations of the vectors from (1). By hypothesis, there exists a number $p$, $0<p<1$, for which $$AK=pAB,BL=pBC,CM=pCA.$$

Substituting into the first equality $AK=AV-KV$ and $AB=AV-BV$, we get after a small manipulation the first of the following three equalities $$KV=(1-p)AV+pBV,LV=(1-p)BV+pCV,MV=(1-p)CV+pAV;$$ the other two can be derived similarly. Taking products, we get $$\begin{gathered} KV\cdot LV=(1-p{)}^{2}AV\cdot BV+p(1-p)AV\cdot CV+p(1-p)B{V}^2= \\ =(1-p)s+p(1-p)B{V}^2, \end{gathered}$$ where $s$ denotes the common value of the products from (1). Similarly, $$KV\cdot MV=(1-p)s+p(1-p)A{V}^2\text{and}LV\cdot MV=(1-p)s+p(1-p)B{V}^2.$$ We see that the system (2) is equivalent to the system of equalities $$p(1-p)A{V}^2=p(1-p)B{V}^2=p(1-p)C{V}^2,$$ which, in view of the condition $p(1-p)\ne 0$, is fulfilled if and only if $|AV|=|BV|=|CV|$. The last condition means that the orthocenter $V$ of the triangle $ABC$ coincides with its circumcenter. This happens if and only if the triangle $ABC$ is equilateral.