1973 USAMO Problems/Problem 1

Problem

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

Solutions

Solution 1

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 $\angle PAQ = \angle EAF$

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 $\angle EAF < \angle IAJ$

But since $E$ and $F$ are interior of the tetrahedron, points $I$ and $J$ cannot be both at the vertices and $IJ < a$ , $\angle IAJ < \angle BAD = 60$ . Therefore, $\angle PAQ < 60$ .

Solution with graphs posted at

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

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1973 USAMO (Problems • Resources)
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1973 USAMO Problems/Problem 1

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Solution 1

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