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Difference between revisions of "1988 OIM Problems/Problem 1"

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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.
 
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.
  
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== Solution ==
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~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
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{{solution}}
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==Solution==
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Let the sides be <math>a,a+d_1,a+2d_1</math>, let the heights be <math>b,b-d_2,b-2d_2</math>, and assume for the sake of contradiction that <math>d_1,d_2</math> are both positive. Then, since longer heights must correspond to shorter bases (since multiplying yields equal areas) we must have <math>a</math> correspond to <math>b</math>, <math>a+d_1</math> correspond to <math>b-d_2</math>, and <math>a+2d_1</math> correspond to <math>b-2d_2</math>. Then, by triangle areas:
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<cmath>ab=(a+d_1)(b-d_2)=(a+2d_1)(b-2d_2)</cmath>
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From the first equality, we find that <math>bd_1=ad_2+d_1d_2</math>. From the equality <math>ab=(a+2d_1)(b-2d_2)</math>, we derive <math>bd_1=ad_2+2d_1d_2</math>. Substituting the first equation into the second results in
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<cmath>ad_2+d_1d_2=ad_2+2d_1d_2</cmath>
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so <math>d_1d_2=0</math>. But if only one of <math>d_1</math> and <math>d_2</math> is zero, then different heights (or bases) will correspond to the same base (or height), resulting in the same area, which is impossible. Thus <math>d_1=d_2=0</math>, and the conclusion follows.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
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== See also ==
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https://www.oma.org.ar/enunciados/ibe3.htm

Latest revision as of 23:28, 19 March 2025

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

See also

https://www.oma.org.ar/enunciados/ibe3.htm

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