1973 USAMO Problems/Problem 1

Revision as of 20:57, 29 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

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See also

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

Solution by Vo Duc Dien

Link and extend AP to meet the plane containing triangle BCD at P”; link AQ and extend it to meet the same plane at Q”. We know that P” and Q” are inside triangle BCD and that /_PAQ = /_P”AQ”

Now let’s look at the triangle BCD with interior points P” and Q”. Since they are interior, one can always be able to draw two circles C1 and C2 sharing the same center being at one of the vertices of triangle BCD with C1 to pass through point Q” and intercept one side of triangle BCD at Q’ and C2 to pass through point P” and intercept the same side at P’. And we have /_P’AQ’ = /_P”AQ” = /_PAQ. But /_P’AQ’ < /_BAC = 60°

Therefore /_PAQ < 60°.