Balti Tee 1994 · Ülesanne 15

Solution:

Consider a triangle $ABC$ with all its sides and heights having integer lengths. From the cosine theorem we conclude that $\cos \angle A$, $\cos \angle B$ and $\cos \angle C$ are rational numbers. Let $AH$ be one of the heights of the triangle $ABC$, with the point $H$ lying on the straight line determined by the side $BC$. Then $|BH|$ and $|CH|$ must be rational and hence integer (consider the Pythagorean theorem for the triangles $ABH$ and $ACH$). Now, if $|BH|$ and $|CH|$ have different parity then $|AB|$ and $|AC|$ also have different parity and $|BC|$ is odd. If $|BH|$ and $|CH|$ have the same parity then $|AB|$ and $|AC|$ also have the same parity and $|BC|$ is even. In both cases the perimeter of triangle $ABC$ is an even number and hence cannot be equal to $1995$.