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USAMO 1973 #1 *August 7, 2009*

*Posted by lumixedia in Problem-solving.*

Tags: contest math, geometry, olympiad math, USAMO, USAMO 1973

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Tags: contest math, geometry, olympiad math, USAMO, USAMO 1973

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**USAMO 1973 #1.** Two points, and , lie in the interior of a regular tetrahedron . Prove that angle .

**Solution.** Let intersect the interior of triangle at and intersect the interior of triangle at . Suppose WLOG that and are on the same side of , and let intersect at and at . Clearly , so it would suffice to prove that . Note that the shortest side of a triangle is always opposite the smallest angle, and that the smallest angle in a triangle is at most . So we just need to show that is the shortest side of .

Consider the triangle . Let be on so that is parallel to . We can see that and , so . This implies that . But from SAS congruency on triangles and , we have . Thus . By symmetry, and is the shortest side of , as desired.

I didn’t entirely follow your solution- is this the heuristic principle?

We can at least show that the angle is for in the interior or on the boundary of the tetrahedron. Project into the triangle by extending , which doesn’t change the angle. Then will be as large as possible when are as far away as possible (it should be possible to make this rigorous), which occurs when are opposing vertices of the triangle . So just check by symmetry.

It’s just transitivity. We’re showing PAQ<60 by showing PAQ<P_2AQ_2 and P_2AQ_2<=60. The former is true because the angles being compared are in the same plane and one "contains" the other. The proof of the latter is basically what you just said, where much of my post is for the (it should be possible to make this rigorous) part.