1984 AIME Problems/Problem 9
Contents
[hide]Problem
In tetrahedron , edge has length 3 cm. The area of face is and the area of face is . These two faces meet each other at a angle. Find the volume of the tetrahedron in .
Solution 1
Position face on the bottom. Since , we find that . Because the problem does not specify, we may assume both and to be isosceles triangles. Thus, the height of forms a with the height of the tetrahedron. So, . The volume of the tetrahedron is thus .
Solution 2 (Rigorous)
It is clear that and where is the foot of the perpendicular from and to side . Thus where h is the height of the tetrahedron from . Hence, the volume of the tetrahedron is ~ Mathommill
(Note this actually isn't rigorous because they never proved that the height from to is the altitude of the tetrahedron.
Solution 3 (Sketchy)
Make faces and right triangles. This makes everything a lot easier. Then do everything in solution 1.
Solution 4 (coord/vector bash)
See also
1984 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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