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2013 AIME I Problems/Problem 7

Revision as of 16:52, 13 March 2015 by Mathgeek2006 (talk | contribs)

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 1

Let the height of the box be $x$.

After using the Pythagorean Theorem three times, we can quickly see that the sides of the triangle are 10, $\sqrt{\left(\frac{x}{2}\right)^2 + 64}$, and $\sqrt{\left(\frac{x}{2}\right)^2 + 36}$. Since the area of the triangle is $30$, the altitude of the triangle from the base with length $10$ is $6$.

Considering the two triangles created by the altitude, we use the Pythagorean theorem twice to find the lengths of the two line segments that make up the base of $10$.

We find: \[10 = \sqrt{\left(28+x^2/4\right)}+x/2\]

Solving for $x$ gives us $x=\frac{36}{5}$. Since this fraction is simplified: \[m+n=\boxed{041}\]

Solution 2

We may use vectors. Let the height of the box be $2h$. Without loss of generality, let the front bottom left corner of the box be $(0,0,0)$. Let the center point of the bottom face be $P_1$, the center of the left face be $P_2$ and the center of the front face be $P_3$.

We are given that the area of the triangle $\triangle P_1 P_2 P_3$ is $30$. Thus, by a well known formula, we note that $\frac{1}{2}|\vec{P_1P_2} \text{x} \vec{P_1P_3}|=30$ We quickly attain that $\vec{P_1P_2}=<-6,0,h>$ and $\vec{P_1P_3}=<0,-8,h>$ (We can arbitrarily assign the long and short ends due to symmetry)

Computing the cross product, we find: \[\vec{P_1P_2} x \vec{P_1P_3}=-<6h,8h,48>\]

Thus: \[\sqrt{(6h)^2+(8h)^2+48^2}=2*30=60\] \[h=3.6\] \[2h=7.2\]

\[2h=36/5\]

\[m+n=\boxed{041}\]

Solution 3

Let the height of the box be $x$.

After using the Pythagorean Theorem 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 semiperimeter is $\left(10 + \sqrt{(x/2)^2 + 64} + \sqrt{(x/2)^2 + 36}\right)/2$. Therefore, when we square Heron's formula, we find

\begin{align*}900 &= \frac{1}{2}\left(\left(10 + \sqrt{(x/2)^2 + 64} + \sqrt{(x/2)^2 + 36}\right)/2\right)\times\left(\left(10 + \sqrt{(x/2)^2 + 64} + \sqrt{(x/2)^2 + 36}\right)/2 - 10\right)\\&\qquad\times\left(\left(10 + \sqrt{(x/2)^2 + 64} + \sqrt{(x/2)^2 + 36}\right)/2 - \sqrt{(x/2)^2 + 64}\right)\\&\qquad\times\left(\left(10 + \sqrt{(x/2)^2 + 64} + \sqrt{(x/2)^2 + 36}\right)/2 - \sqrt{(x/2)^2 + 36}\right).\end{align*}

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

See also

2013 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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