Difference between revisions of "1985 OIM Problems/Problem 2"
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<cmath>PA=5,\;PB=7,\; and \; PC=8</cmath> | <cmath>PA=5,\;PB=7,\; and \; PC=8</cmath> | ||
Find the length of one side of the triangle <math>ABC</math> | Find the length of one side of the triangle <math>ABC</math> | ||
| + | |||
| + | ~translated into English by Tomas Diaz. ~orders@tomasdiaz.com | ||
== Solution == | == Solution == | ||
| − | {{ | + | By Viviani's Theorem, the altitude of the triangle is the sum of the given lengths, or <math>20</math>. It follows that the side length is <math>\boxed{\frac{40\sqrt3}{3}}</math>. |
| + | |||
| + | == See also == | ||
| + | https://www.oma.org.ar/enunciados/ibe1.htm | ||
Latest revision as of 22:36, 8 April 2024
Problem
Let 
be a point in the interior of equilateral triangle 
such that:
![\[PA=5,\;PB=7,\; and \; PC=8\]](http://latex.artofproblemsolving.com/d/a/6/da6667e0d97ec0389974191bbe91ab057c9df9db.png)
Find the length of one side of the triangle
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
By Viviani's Theorem, the altitude of the triangle is the sum of the given lengths, or 
. It follows that the side length is 
.
