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 (dedicated to my former teacher Lê Chí Đệ)

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°.