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USAMO 1973 #1 August 7, 2009

Posted by lumixedia in Problem-solving.
Tags: , , , ,

USAMO 1973 #1. Two points, {P} and {Q}, lie in the interior of a regular tetrahedron {ABCD}. Prove that angle {PAQ<60^{\circ}}.

Solution. Let {AP} intersect the interior of triangle {BCD} at {P_1} and {AQ} intersect the interior of triangle {BCD} at {Q_1}. Suppose WLOG that {C} and {D} are on the same side of {P_1Q_1}, and let {P_1Q_1} intersect {BC} at {P_2} and {BD} at {Q_2}. Clearly {\angle PAQ<\angle P_2AQ_2}, so it would suffice to prove that {\angle P_2AQ_2\le60^{\circ}}. Note that the shortest side of a triangle is always opposite the smallest angle, and that the smallest angle in a triangle is at most {60^{\circ}}. So we just need to show that {P_2Q_2} is the shortest side of {\triangle P_2AQ_2}.

Consider the triangle {P_2Q_2D}. Let {E} be on {CD} so that {Q_2E} is parallel to {BC}. We can see that {\angle P_2Q_2D>\angle EQ_2D=60^{\circ}} and {\angle P_2DQ_2<\angle EDQ_2=60^{\circ}}, so {\angle P_2Q_2D>\angle P_2DQ_2}. This implies that {P_2D>P_2Q_2}. But from SAS congruency on triangles {P_2CA} and {P_2CD}, we have {P_2A=P_2D}. Thus {P_2A>P_2Q_2}. By symmetry, {Q_2A>P_2Q_2} and {P_2Q_2} is the shortest side of {\triangle P_2AQ_2}, as desired.


1. Akhil Mathew - August 7, 2009

I didn’t entirely follow your solution- is this the heuristic principle?

We can at least show that the angle is \leq 60 for P in the interior or on the boundary of the tetrahedron. Project P, Q into the triangle BCD by extending AP, AQ, which doesn’t change the angle. Then PAQ will be as large as possible when P, Q are as far away as possible (it should be possible to make this rigorous), which occurs when P,Q are opposing vertices of the triangle BCD. So just check P=B, Q=C by symmetry.

2. lumixedia - August 8, 2009

It’s just transitivity. We’re showing PAQ<60 by showing PAQ<P_2AQ_2 and P_2AQ_2<=60. The former is true because the angles being compared are in the same plane and one "contains" the other. The proof of the latter is basically what you just said, where much of my post is for the (it should be possible to make this rigorous) part.

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