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

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==Solution==
 
==Solution==
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{{solution}}
 
  
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By Vo Duc Dien
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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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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 &#8736;PAQ = &#8736;EAF
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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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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  &#8736;EAF < &#8736;IAJ
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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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But since E and F are interior of the tetrahedron, points I and J cannot be both at the vertices and IJ < <math>a</math>, &#8736;IAJ < &#8736;BAD = 60°. Therefore, &#8736;PAQ < 60°.
 
  
 
Solution with graphs posted at
 
Solution with graphs posted at

Revision as of 10:04, 8 July 2010

Problem

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

Solution

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

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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