Difference between revisions of "1973 USAMO Problems/Problem 1"
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+ | [[Solution by Vo Duc Dien]] | ||
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+ | Let M and N be the midpoints of AB and AC and let P” and Q” be points where AP and | ||
+ | AQ intercept the plane DMN. We have /_PAQ = /_P”AQ”. | ||
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+ | Now let’s look at the plane DMN. Sine the two points P and Q are in the interior of the tetrahedron or even inside the airspace Ax (AB extension), Ay (AC extension) and Az (AD extension), one can always be able to draw two circles C1 and C2 with the same center at one of the vertices of triangle DMN with C1 to pass through point Q” and intercept one side of DMN 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’ < /_MAN = 60° Therefore, /_PAQ < 60°. |
Revision as of 20:30, 29 January 2010
Problem
Two points and
lie in the interior of a regular tetrahedron
. Prove that angle
.
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See also
1973 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
Let M and N be the midpoints of AB and AC and let P” and Q” be points where AP and AQ intercept the plane DMN. We have /_PAQ = /_P”AQ”.
Now let’s look at the plane DMN. Sine the two points P and Q are in the interior of the tetrahedron or even inside the airspace Ax (AB extension), Ay (AC extension) and Az (AD extension), one can always be able to draw two circles C1 and C2 with the same center at one of the vertices of triangle DMN with C1 to pass through point Q” and intercept one side of DMN 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’ < /_MAN = 60° Therefore, /_PAQ < 60°.