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== | ||
+ | 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:
From the first equality, we find that
. From the equality
, we derive
. Substituting the first equation into the second results in
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.