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>. | ||
− | ==Solution== | + | ==Solution 1== |
− | After using the | + | 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>. |
+ | |||
+ | 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> | ||
+ | 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) | ||
+ | |||
+ | Computing the cross product, we find: | ||
+ | <cmath>\vec{P_1P_2} x \vec{P_1P_3}=-<6h,8h,48></cmath> | ||
+ | |||
+ | Thus: | ||
+ | <cmath>\sqrt{(6h)^2+(8h)^2+48^2}=2*30=60</cmath> | ||
+ | <cmath>h=3.6</cmath> | ||
+ | <cmath>2h=7.2</cmath> | ||
+ | |||
+ | <cmath>2h=36/5</cmath> | ||
+ | |||
+ | <cmath>m+n=\boxed{041}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let the height of the box be <math>x</math>. | ||
+ | |||
+ | 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 12:44, 30 March 2013
Contents
Problem 7
A rectangular box has width inches, length inches, and height inches, where and 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 square inches. Find .
Solution 1
We may use vectors. Let the height of the box be . Without loss of generality, let the front bottom left corner of the box be . Let the center point of the bottom face be , the center of the left face be and the center of the front face be .
We are given that the area of the triangle is . Thus, by a well known formula, we note that We quickly attain that and (We can arbitrarily assign the long and short ends due to symmetry)
Computing the cross product, we find:
Thus:
Solution 2
Let the height of the box be .
After using the Pythagorean Theorem three times, we can quickly see that the sides of the triangle are 10, , and . Therefore, we can use Heron's formula to set up an equation for the area of the triangle.
The semi perimeter is (10 + + )/2
900 = ((10 + + )/2)((10 + + )/2 - 10)((10 + + )/2 - )((10 + + )/2 - ).
Solving, we get .
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 |