Difference between revisions of "1988 OIM Problems/Problem 1"
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~translated into English by Tomas Diaz. ~orders@tomasdiaz.com | ~translated into English by Tomas Diaz. ~orders@tomasdiaz.com | ||
| − | == Solution == | + | ==Solution== |
| − | + | 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: | |
| + | <cmath>ab=(a+d_1)(b-d_2)=(a+2d_1)(b-2d_2)</cmath> | ||
| + | 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 | ||
| + | <cmath>ad_2+d_1d_2=ad_2+2d_1d_2</cmath> | ||
| + | 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. | ||
| + | |||
| + | ~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406] | ||
== See also == | == See also == | ||
https://www.oma.org.ar/enunciados/ibe3.htm | 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 
, let the heights be 
, and assume for the sake of contradiction that 
are both positive. Then, since longer heights must correspond to shorter bases (since multiplying yields equal areas) we must have 
correspond to 
, 
correspond to 
, and 
correspond to 
. Then, by triangle areas:
![\[ab=(a+d_1)(b-d_2)=(a+2d_1)(b-2d_2)\]](http://latex.artofproblemsolving.com/5/a/2/5a2eda5689628543cfa7e32176842aa94914cab6.png)
From the first equality, we find that 
. From the equality 
, we derive 
. Substituting the first equation into the second results in
![\[ad_2+d_1d_2=ad_2+2d_1d_2\]](http://latex.artofproblemsolving.com/4/2/7/4276473ff99a78463fd942f1a7dfa12706a4a368.png)
so 
. But if only one of 
and 
is zero, then different heights (or bases) will correspond to the same base (or height), resulting in the same area, which is impossible. Thus 
, and the conclusion follows.
