This math problem was brought to my attention by my math Ph.D. friends. It was the first problem of the United States Mathematical Olympiad in year 1975, but until today there has been no solution posted. It’s a three dimenional geometrical problem, and is really difficult if trying to solve by using the three dimensional methods with sine, cosine, reflective angles, etc… It’s just a little trick that solved it. Hội Toán Học Hoa Kỳ wants the solution posted at a well-known website www.cut-the-knot.org so they can go get it. So the solution was also posted there.

Problem 1 of USA Mathematical Olympiad 1973

Two points, P and Q, lie in the interior of a regular tetrahedron ABCD. Prove that angle PAQ < 60°.

Solution by Vo Duc Dien

Let the side length of the regular tetrahedron be a. Link and extend AP to meet the plane containing triangle BCD at E; link AQ and extend it to meet the same plane at F. We know that E and F are inside triangle BCD and that /_PAQ = /_EAF

Now let’s look at the plane containing triangle BCD with points E and F inside the triangle. Link and extend EF on both sides to meet the sides of the triangle BCD at I and J, I on BC and J on DC. We have /_EAF < /_IAJ

But since E and F are interior of the tetrahedron, points I and J cannot be both at the vertices and IJ < a, /_IAJ < /_BAD = 60°.

Therefore /_PAQ < 60°.