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Difference between revisions of "2014 AMC 12A Problems/Problem 10"

(Created page with "==Problem== Three congruent isosceles triangles are constructed with their bases on the sides of an equilateral triangle of side length <math>1</math>. The sum of the areas of ...")
 
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Reflect each of the triangles over its respective side.  Then since the areas of the triangles total to the area of the equilateral triangle, it can be seen that the triangles fill up the equilateral one and the vertices of these triangles concur at the circumcenter of the equilateral triangle.  Hence the desired answer is just its circumradius, or <math>\boxed{\dfrac{\sqrt3}3\textbf{ (B)}}</math>.
 
Reflect each of the triangles over its respective side.  Then since the areas of the triangles total to the area of the equilateral triangle, it can be seen that the triangles fill up the equilateral one and the vertices of these triangles concur at the circumcenter of the equilateral triangle.  Hence the desired answer is just its circumradius, or <math>\boxed{\dfrac{\sqrt3}3\textbf{ (B)}}</math>.
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(Solution by djmathman)

Revision as of 20:21, 7 February 2014

Problem

Three congruent isosceles triangles are constructed with their bases on the sides of an equilateral triangle of side length $1$. The sum of the areas of the three isosceles triangles is the same as the area of the equilateral triangle. What is the length of one of the two congruent sides of one of the isosceles triangles?

$\textbf{(A) }\dfrac{\sqrt3}4\qquad \textbf{(B) }\dfrac{\sqrt3}3\qquad \textbf{(C) }\dfrac23\qquad \textbf{(D) }\dfrac{\sqrt2}2\qquad \textbf{(E) }\dfrac{\sqrt3}2$

Solution

Reflect each of the triangles over its respective side. Then since the areas of the triangles total to the area of the equilateral triangle, it can be seen that the triangles fill up the equilateral one and the vertices of these triangles concur at the circumcenter of the equilateral triangle. Hence the desired answer is just its circumradius, or $\boxed{\dfrac{\sqrt3}3\textbf{ (B)}}$.

(Solution by djmathman)

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