1984 AIME Problems/Problem 9
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)
We can use 3D coordinates.
Let and WLOG, let , because the area of and the tetrahedron area won't change if we put it somewhere else with
To find , we can again let the -coordinate be for simplicity. Note that is units away from because the area of is . Since the angle between and is , we can form a 30-60-90 triangle between , , and an altitude dropped from onto face . Since is the hypotenuse, we get and as legs. Then and , so
(I highly advise you to draw both the tetrahedron and 30-60-90 triangle to get a better perspective.)
Now, we can move onto vectors. To find the volume of the tetrahedron, we use the formula Letting be the base we have (from the problem statement). We need to find the distance between and , and to do this, we should find the projection of onto face .
Note that we can simplify this to projecting onto This is because we know the projection will have the same -coordinate as and , as both are Now we find , or plugging in our coordinates, .
Let the -coordinates for both be for simplicity, because we can always add a at the end. Using the projection formula, we get
Finally, we calculate the distance between and to be . So the height is , and plugging into our tetrahedron formula we get
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