**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°.