Points $K,L,M$ on the sides $AB,BC,CA$ respectively of a triangle $ABC$ satisfy $\frac{AK}{KB} = \frac{BL}{LC} = \frac{CM}{MA}$. Show that the triangles $ABC$ and $KLM$ have a common orthocenter if and only if $\triangle ABC$ is equilateral.
Source: Czech-Polish-Slovak 2004 Q5
Tags: geometry, vector, circumcircle, geometry unsolved
Points $K,L,M$ on the sides $AB,BC,CA$ respectively of a triangle $ABC$ satisfy $\frac{AK}{KB} = \frac{BL}{LC} = \frac{CM}{MA}$. Show that the triangles $ABC$ and $KLM$ have a common orthocenter if and only if $\triangle ABC$ is equilateral.