2012 AIME II Problems/Problem 5
Problem 5
In the accompanying figure, the outer square has side length . A second square of side length is constructed inside with the same center as and with sides parallel to those of . From each midpoint of a side of , segments are drawn to the two closest vertices of . The result is a four-pointed starlike figure inscribed in . The star figure is cut out and then folded to form a pyramid with base . Find the volume of this pyramid.
Solution
The volume of this pyramid can be found by the equation , where is the base and is the height. The base is easy, since it is a square and has area .
To find the height of the pyramid, the height of the four triangles is needed, which will be called . By drawing a line through the middle of the larger square, we see that its length is equal to the length of the smaller rectangle and two of the triangle's heights. Then , which means that .
When the pyramid is made, you see that the height is the one of the legs of a right triangle, with the hypotenuse equal to and the other leg having length equal to half of the side length of the smaller square, or . So, the Pythagorean Theorem can be used to find the height.
Finally, .
See Also
2012 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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