1988 OIM Problems/Problem 1

Problem

The measurements of the sides of a triangle are in arithmetic progression and the lengths of the heights of the same triangle are also in arithmetic progression. Prove that the triangle is equilateral.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

Let the sides be $a,a+d_1,a+2d_1$ , let the heights be $b,b-d_2,b-2d_2$ , and assume for the sake of contradiction that $d_1,d_2$ are both positive. Then, since longer heights must correspond to shorter bases (since multiplying yields equal areas) we must have $a$ correspond to $b$ , $a+d_1$ correspond to $b-d_2$ , and $a+2d_1$ correspond to $b-2d_2$ . Then, by triangle areas: $ab=(a+d_1)(b-d_2)=(a+2d_1)(b-2d_2)$ From the first equality, we find that $bd_1=ad_2+d_1d_2$ . From the equality $ab=(a+2d_1)(b-2d_2)$ , we derive $bd_1=ad_2+2d_1d_2$ . Substituting the first equation into the second results in $ad_2+d_1d_2=ad_2+2d_1d_2$ so $d_1d_2=0$ . But if only one of $d_1$ and $d_2$ is zero, then different heights (or bases) will correspond to the same base (or height), resulting in the same area, which is impossible. Thus $d_1=d_2=0$ , and the conclusion follows.

~ eevee9406

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1988 OIM Problems/Problem 1

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