2013 AIME I Problems/Problem 7

Problem 7

A rectangular box has width $12$ inches, length $16$ inches, and height $\frac{m}{n}$ inches, where $m$ and $n$ are relatively prime positive integers. Three faces of the box meet at a corner of the box. The center points of those three faces are the vertices of a triangle with an area of $30$ square inches. Find $m+n$ .

Solution

After using the pythagorean formula three times, we can quickly see that the sides of the triangle are 10, $\sqrt{(x/2)^2 + 64}$ , and $\sqrt{(x/2)^2 + 36}$ . Therefore, we can use Heron's formula to set up an equation for the area of the triangle.

The semi perimeter is (10 + $\sqrt{(x/2)^2 + 64}$ + $\sqrt{(x/2)^2 + 36}$ )/2

900 = $\frac{1}{2}$ ((10 + $\sqrt{(x/2)^2 + 64}$ + $\sqrt{(x/2)^2 + 36}$ )/2)((10 + $\sqrt{(x/2)^2 + 64}$ + $\sqrt{(x/2)^2 + 36}$ )/2 - 10)((10 + $\sqrt{(x/2)^2 + 64}$ + $\sqrt{(x/2)^2 + 36}$ )/2 - $\sqrt{(x/2)^2 + 64}$ )((10 + $\sqrt{(x/2)^2 + 64}$ + $\sqrt{(x/2)^2 + 36}$ )/2 - $\sqrt{(x/2)^2 + 36}$ ).

Solving, we get $\boxed{041}$ .

P.S. No expressions cancel out. Good luck.

2013 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2013 AIME I Problems/Problem 7

Problem 7

Solution

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