1999 AIME Problems/Problem 15
Contents
[hide]Problem
Consider the paper triangle whose vertices are and The vertices of its midpoint triangle are the midpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid?
Solution
Let , , be the feet of the altitudes to sides , , , respectively, of . The base of the tetrahedron is the orthocenter of the large triangle, so we just need to find that, then it's easy from there.
To find the coordinates of , we need to find the intersection point of altitudes and . The equation of is simply . is perpendicular to line , so the slope of is equal to the negative reciprocal of the slope of . has slope , therefore . These two lines intersect at , so that's the base of the height of the tetrahedron.
Let be the foot of altitude in . From the Pythagorean Theorem, . However, since and are, by coincidence, the same point, and .
The area of the base is , so the volume is .
Alternate Solution
Consider the diagram provided in the previous solution. We first note that the medial triangle has coordinates , , and . We can compute the area of this triangle as . Suppose are the coordinates of the vertex of the resulting pyramid. Call this point . Clearly, the height of the pyramid is . The desired volume is thus .
We note that when folding the triangle to form the pyramid, some side lengths must stay the same. In particular, , , and . We thus arrive at a fairly simple system of equations, yielding . The desired volume is thus .
See also
1999 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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