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

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A rectangular box has width <math>12</math> inches, length <math>16</math> inches, and height <math>\frac{m}{n}</math> inches, where <math>m</math> and <math>n</math> 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 <math>30</math> square inches. Find <math>m+n</math>.
 
A rectangular box has width <math>12</math> inches, length <math>16</math> inches, and height <math>\frac{m}{n}</math> inches, where <math>m</math> and <math>n</math> 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 <math>30</math> square inches. Find <math>m+n</math>.
  
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==Solution==
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==Solution 1==
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After using the pythagorean formula three times, we can quickly see that the sides of the triangle are 10, <math>\sqrt{(x/2)^2 + 64}</math>, and <math>\sqrt{(x/2)^2 + 36}</math>. Therefore, we can use Heron's formula to set up an equation for the area of the triangle.
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We may use vectors. Let the height of the box be <math>2h</math>. Without loss of generality, let the front bottom left corner of the box be <math>(0,0,0)</math>. Let the center point of the bottom face be <math>P_1</math>, the center of the left face be <math>P_2</math> and the center of the front face be <math>P_3</math>.
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We are given that the area of the triangle <math>\triangle P_1 P_2 P_3</math> is <math>30</math>. Thus, by a well known formula, we note that <math>\frac{1}{2}|\vec{P_1P_2} \text{x} \vec{P_1P_3}|=30</math>
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We quickly attain that <math>\vec{P_1P_2}=<-6,0,h></math> and <math>\vec{P_1P_3}=<0,-8,h></math> (We can arbitrarily assign the long and short ends due to symmetry)
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Computing the cross product, we find:
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<cmath>\vec{P_1P_2} x \vec{P_1P_3}=-<6h,8h,48></cmath>
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Thus:
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<cmath>\sqrt{(6h)^2+(8h)^2+48^2}=2*30=60</cmath>
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<cmath>h=3.6</cmath>
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<cmath>2h=7.2</cmath>
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<cmath>2h=36/5</cmath>
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<cmath>m+n=\boxed{041}</cmath>
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==Solution 2==
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Let the height of the box be <math>x</math>.
 +
 
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After using the Pythagorean Theorem three times, we can quickly see that the sides of the triangle are 10, <math>\sqrt{(x/2)^2 + 64}</math>, and <math>\sqrt{(x/2)^2 + 36}</math>. Therefore, we can use Heron's formula to set up an equation for the area of the triangle.
  
 
The semi perimeter is (10 + <math>\sqrt{(x/2)^2 + 64}</math> + <math>\sqrt{(x/2)^2 + 36}</math>)/2
 
The semi perimeter is (10 + <math>\sqrt{(x/2)^2 + 64}</math> + <math>\sqrt{(x/2)^2 + 36}</math>)/2

Revision as of 13:44, 30 March 2013

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

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 2

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 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 $\boxed{041}$.

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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