1973 USAMO Problems/Problem 1

Revision as of 02:19, 30 January 2010 by Vo Duc Dien (talk | contribs) (See also)

Problem

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

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See also

1973 USAMO (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5
All USAMO Problems and Solutions

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 I and J are on the sides and not on the vertices, IJ < $a$, ∠IAJ < ∠BAD = 60°. Therefore, ∠PAQ < 60°.

Solution with graphs posted at

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1973Problem1