Difference between revisions of "1973 USAMO Problems/Problem 1"
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
| − | Let the side length of the regular tetrahedron be a. Link and extend AP to meet the plane containing triangle BCD at E; link AQ and extend it to meet the same plane at F. We know that E and F are inside triangle BCD and that | + | Let the side length of the regular tetrahedron be {<math>a</math>}. Link and extend AP to meet the plane containing triangle BCD at E; link AQ and extend it to meet the same plane at F. We know that E and F are inside triangle BCD and that ∠PAQ = ∠EAF |
| − | Now let’s look at the plane containing triangle BCD with points E and F inside the triangle. Link and extend EF on both sides to meet the sides of the triangle BCD at I and J, I on BC and J on DC. | + | Now let’s look at the plane containing triangle BCD with points E and F inside the triangle. Link and extend EF on both sides to meet the sides of the triangle BCD at I and J, I on BC and J on DC. We have ∠EAF < ∠IAJ |
| − | + | But since I and J are on the sides and not on the vertices, IJ {<math>a</math>}, ∠IAJ < ∠BAD = 60°. Therefore, ∠PAQ < 60°. | |
| − | |||
| − | But since I and J are on the sides and not on the vertices, IJ < a, | ||
| − | |||
Revision as of 00:27, 30 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 the side length of the regular tetrahedron be {
}. Link and extend AP to meet the plane containing triangle BCD at E; link AQ and extend it to meet the same plane at F. We know that E and F are inside triangle BCD and that ∠PAQ = ∠EAF
Now let’s look at the plane containing triangle BCD with points E and F inside the triangle. Link and extend EF on both sides to meet the sides of the triangle BCD at I and J, I on BC and J on DC. We have ∠EAF < ∠IAJ
But since I and J are on the sides and not on the vertices, IJ {}, ∠IAJ < ∠BAD = 60°. Therefore, ∠PAQ < 60°.
