Difference between revisions of "1973 USAMO Problems/Problem 1"
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| − | ==Problem== | + | == Problem == |
| − | Two points <math>P</math> and <math>Q</math> lie in the interior of a regular tetrahedron <math>ABCD</math>. Prove that angle <math>PAQ<60^ | + | Two points <math>P</math> and <math>Q</math> lie in the interior of a regular tetrahedron <math>ABCD</math>. Prove that angle <math>PAQ < 60^\circ</math>. |
| − | ==Solution== | + | == Solutions == |
| − | {{ | + | |
| + | === Solution 1 === | ||
| + | Let the side length of the regular tetrahedron be <math>a</math>. Link and extend <math>AP</math> to meet the plane containing triangle <math>BCD</math> at <math>E</math>; link <math>AQ</math> and extend it to meet the same plane at <math>F</math>. We know that <math>E</math> and <math>F</math> are inside triangle <math>BCD</math> and that <math>\angle PAQ = \angle EAF</math> | ||
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| + | Now let’s look at the plane containing triangle <math>BCD</math> with points <math>E</math> and <math>F</math> inside the triangle. Link and extend <math>EF</math> on both sides to meet the sides of the triangle <math>BCD</math> at <math>I</math> and <math>J</math>, <math>I</math> on <math>BC</math> and <math>J</math> on <math>DC</math>. We have <math>\angle EAF < \angle IAJ</math> | ||
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| + | But since <math>E</math> and <math>F</math> are interior of the tetrahedron, points <math>I</math> and <math>J</math> cannot be both at the vertices and <math>IJ < a</math>, <math>\angle IAJ < \angle BAD = 60</math>. Therefore, <math>\angle PAQ < 60</math>. | ||
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| + | Solution with graphs posted at | ||
| + | |||
| + | http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1973Problem1 | ||
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| + | {{alternate solutions}} | ||
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| + | hurdler: Remark on solution 1: This proof is not rigorous, in the very last step. The last step needs more justification. | ||
==See also== | ==See also== | ||
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
| + | [[Category:3D Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 07:10, 4 January 2022
Contents
Problem
Two points 
and 
lie in the interior of a regular tetrahedron 
. Prove that angle 
.
Solutions
Solution 1
Let the side length of the regular tetrahedron be 
. Link and extend 
to meet the plane containing triangle 
at 
; link 
and extend it to meet the same plane at 
. We know that and are inside triangle and that 
Now let’s look at the plane containing triangle with points and inside the triangle. Link and extend 
on both sides to meet the sides of the triangle at 
and 
, on 
and on 
. We have 
But since and are interior of the tetrahedron, points and cannot be both at the vertices and 
, 
. Therefore, 
.
Solution with graphs posted at
http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1973Problem1
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
hurdler: Remark on solution 1: This proof is not rigorous, in the very last step. The last step needs more justification.
See also
| 1973 USAMO (Problems • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
