Difference between revisions of "2013 AIME I Problems/Problem 7"
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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. | 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 semiperimeter is (10 + | + | The semiperimeter is <math>\left(10 + \sqrt{(x/2)^2 + 64} + \sqrt{(x/2)^2 + 36}\right)/2</math>. Therefore, when we square Heron's formula, we find |
− | 900 = | + | <cmath>\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*}</cmath> |
Solving, we get <math>\boxed{041}</math>. | Solving, we get <math>\boxed{041}</math>. |
Revision as of 16:52, 13 March 2015
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
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 . Since the area of the triangle is , the altitude of the triangle from the base with length is .
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 .
We find:
Solving for gives us . Since this fraction is simplified:
Solution 2
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 3
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 semiperimeter is . Therefore, when we square Heron's formula, we find
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.