## USAMO 1973 #1 August 7, 2009

Posted by lumixedia in Problem-solving.
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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.

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.