USAMO 1973 #1 August 7, 2009Posted by lumixedia in Problem-solving.
Tags: contest math, geometry, olympiad math, USAMO, USAMO 1973
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.