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

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==Problem==
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== Problem ==
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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^o</math>.
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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>.
  
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==Solution==
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== Solutions ==
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{{solution}}
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=== Solution 1 ===
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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
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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]]
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[[Category:3D Geometry Problems]]
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[[Solution by Vo Duc Dien]]
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{{MAA Notice}}
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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
 
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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°.
 

Latest revision as of 08:10, 4 January 2022

Problem

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

Solutions

Solution 1

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 $\angle PAQ = \angle EAF$

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 $\angle EAF < \angle IAJ$

But since $E$ and $F$ are interior of the tetrahedron, points $I$ and $J$ cannot be both at the vertices and $IJ < a$, $\angle IAJ < \angle BAD = 60$. Therefore, $\angle PAQ < 60$.

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

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