2014 AMC 12A Problems/Problem 10
Contents
Problem
Three congruent isosceles triangles are constructed with their bases on the sides of an equilateral triangle of side length 
. 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?
Solution 1
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 
.
(Solution by djmathman)
Solution 2
Since the total area of each congruent isosceles triangle is the same, the area of each is ${\dfrac{1}3$ (Error compiling LaTeX. Unknown error_msg) the total area of the equilateral triangle of side length 1, or ${\dfrac{1}3$ (Error compiling LaTeX. Unknown error_msg) x ${\dfrac{\sqrt3}4$ (Error compiling LaTeX. Unknown error_msg). Likewise, the area of each can be defined as ${\dfrac{bh}2$ (Error compiling LaTeX. Unknown error_msg) with base b equaling 1, meaning that h = ${\dfrac{3\sqrt3}2$ (Error compiling LaTeX. Unknown error_msg). A side length of the isosceles triangle is the hypotenuse with legs ${\dfrac{b}2$ (Error compiling LaTeX. Unknown error_msg) and h. Using the Pythagorean Theorem, the side length is .
(Solution by johnstucky)
See Also
| 2014 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
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