1999 AIME Problems/Problem 15
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 1
As shown in the image above, let , , and be the midpoints of , , and , respectively. Suppose is the apex of the tetrahedron, and let be the foot of the altitude from to . The crux of this problem is the following lemma.
Lemma: The point is the orthocenter of .
Proof. Observe that the first equality follows by the Pythagorean Theorem, while the second follows from and . Thus, by the Perpendicularity Lemma, is perpendicular to and hence . Analogously, lies on the -altitude and -altitude of , and so is, indeed, the orthocenter of .
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 .
Solution 2
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 then use distance formula to find the distances from to each of the vertices of the medial triangle. We thus arrive at a fairly simple system of equations, yielding . The desired volume is thus .
Solution 3
The formed tetrahedron has pairwise parallel planar and oppositely equal length () edges and can be inscribed in a parallelepiped (rectangular box) with the six tetrahedral edges as non-intersecting diagonals of the box faces. Let the edge lengths of the parallelepiped be and solve (by Pythagoras)
to find that
Use the fact that the ratio of volumes between an inscribed tetrahedron and its circumscribing parallelepiped is and then the volume is
Solution by D. Adrian Tanner
Solution 4
Let Then define as the midpoints of . By Pythagorean theorem, Then let be the point in space which is the vertex of the tetrahedron with base .
Note that . Create point on the plane of such that (i.e by reflecting over the perpendicular bisector of ). Project onto as . Note by the definition of then is the dihedral angle between planes .
Now see that by Heron's, So , the hypotenuse has length . Similarly Further from Pythagoras Symmetrically Therefore
By Law of Cosines on ,
Therefore the altitude of the tetrahedron from vertex to base is So the area is
~ Aaryabhatta1
See also
1999 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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