1973 USAMO Problems/Problem 1
Problem
Two points 
and 
lie in the interior of a regular tetrahedron 
. Prove that angle 
.
Solution
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See also
| 1973 USAMO (Problems • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
Let M and N be the midpoints of AB and AC and let P” and Q” be points where AP and AQ intercept the plane DMN. We have /_PAQ = /_P”AQ”.
Now let’s look at the plane DMN. Sine the two points P and Q are in the interior of the tetrahedron or even inside the airspace Ax (AB extension), Ay (AC extension) and Az (AD extension), one can always be able to draw two circles C1 and C2 with the same center at one of the vertices of triangle DMN with C1 to pass through point Q” and intercept one side of DMN at Q’ and C2 to pass through point P” and intercept the same side at P’. And we have /_P’AQ’ = /_P”AQ” = /_PAQ. But /_P’AQ’ < /_MAN = 60° Therefore, /_PAQ < 60°.
