Difference between revisions of "1985 OIM Problems/Problem 2"

(Problem)
(Solution)
 
(2 intermediate revisions by one other user not shown)
Line 2: Line 2:
 
Let <math>P</math> be a point in the interior of equilateral triangle <math>ABC</math> such that:
 
Let <math>P</math> be a point in the interior of equilateral triangle <math>ABC</math> such that:
 
<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>
 +
 +
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
  
 
== Solution ==
 
== 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 23:36, 8 April 2024

Problem

Let $P$ be a point in the interior of equilateral triangle $ABC$ such that: \[PA=5,\;PB=7,\; and \; PC=8\] Find the length of one side of the triangle $ABC$

~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 $20$. It follows that the side length is $\boxed{\frac{40\sqrt3}{3}}$.

See also

https://www.oma.org.ar/enunciados/ibe1.htm