Difference between revisions of "1973 USAMO Problems/Problem 1"

Revision as of 21:53, 29 January 2010

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

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 [[Category:Olympiad Geometry Problems]]
-[[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°.

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