Difference between revisions of "2013 AIME I Problems/Problem 7"

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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 $\boxed{041}$ .

Solution 4

It isn't hard to see that the triangle connecting the centers of the faces of the rectangular prism is congruent to the triangle connecting the midpoints of three edges that concur. So we can now apply de Guas theorem to see that:

$30^2=24^2+(3x)^2+(4x)^2$

Where $x$ is half the desired length of the height.

Solving yields $2x=\frac{36}{5}$

And thus $36+5=\boxed{041}$

---Solution 4 contributed by Siddharth Namachivayam

@@ Line 54: / Line 54: @@
 And thus <math>36+5=\boxed{041}</math>
+---Solution 4 contributed by Siddharth Namachivayam
 == See also ==
 {{AIME box|year=2013|n=I|num-b=6|num-a=8}}
 {{MAA Notice}}

2013 AIME I (Problems • Answer Key • Resources)
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