2013 AIME I Problems/Problem 7
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 , therefore:
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 (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 | ||
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