1983 AIME Problems/Problem 11
First, we find the height of the figure by drawing a perpendicular from the midpoint of to . The hypotenuse of the triangle is the median of equilateral triangle , and one of the legs is . We apply the Pythagorean Theorem to find that the height is equal to .
Next, we complete the figure into a triangular prism, and find the volume, which is .
Now, we subtract off the two extra pyramids that we included, whose combined volume is .
Thus, our answer is .
Extend and to meet at , and and to meet at . Now, we have a regular tetrahedron , which has twice the volume of our original solid. This tetrahedron has side length . Using the formula for the volume of a regular tetrahedron, which is , where S is the side length of the tetrahedron, the volume of our original solid is:
We can also find the volume by integrating horizontal cross-sections of the solid. As in solution 1, we can find the height of the solid is ; thus, we will integrate with respect to height from to , noting that each cross section of height is a rectangle. The problem becomes , where is the width of the rectangle and is the length. We can express in terms of as since it decreases linearly with respect to , and since the length increases linearly with respect to . Now we solve:
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