Zaslavsky A.A.

A line passing through the center of the circumscribed circle and the intersection point of the altitudes of a non-equilateral triangle $ABC$ divides its perimeter and area in the same ratio. Find this ratio.
According to problem $\underline{5462}$, the specified line passes through the center $I$ of the inscribed circle. This line contains no more than one vertex of the triangle. Let it not pass through vertices $A$ and $B$. Since $A I, B I$ are the angle bisectors of angles $H A O, H B O$, we get that $A H: A O=H I: I O=B H: B O$. Since $A O=B O$, then $A H=B H$, which means triangle $A B C$ is isosceles and the desired ratio is $1: 1$.

## Answer

$1: 1$.