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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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By Vo Duc Dien
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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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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
  
 
http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1973Problem1
 
http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1973Problem1
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 +
=== Solution 2 (similar to solution 1, but full proof) ===
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Let the side length of the regular tetrahedron <math>ABCD</math> be <math>s</math>.
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Extend <math>AP</math> and <math>AQ</math> to intersect the plane <math>BCD</math> at points <math>P'</math> and <math>Q'</math> inside <math>\Delta BCD</math>, respectively. We can observe that <math>\angle PAQ=\angle P'AQ'</math>.
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Extend <math>P'Q'</math> to intersect the sides of <math>\Delta BCD</math> at <math>P''</math> and <math>Q''</math>. We can observe that <math>\angle P'AQ'<\angle P''AQ''</math>. We assume that, WLOG, <math>P''</math> is on <math>BD</math> and <math>Q''</math> is on <math>BC</math>.
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 +
Let the ratios <math>\frac{BP''}{BD}=\mu</math> and <math>\frac{BQ''}{BC}=\lambda</math>. We can see that <math>\mu,\lambda\in (0,1)</math>. It follows from Stewart's theorem that
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<math>
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AP''^2=AD^2\mu+AB^2(1-\mu)-\mu(1-\mu)BD^2=(\mu^2-\mu+1)s\\
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AQ''^2=AC^2\lambda+AB^2(1-\lambda)-\lambda(1-\lambda)BC^2=(\lambda^2-\lambda^2+1)s
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</math>
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Furthermore, we have
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<math>
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P''C^2=CD^2\mu+CB^2(1-\mu)-\mu(1-\mu)BD^2=(\mu^2-\mu+1)s
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</math>
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Using this, we can conclude that
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<math>
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P''Q''^2=P''C^2\lambda+(BD\mu)^2(1-\lambda)-\lambda(1-\lambda)BC^2\\
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=(\mu\lambda-\mu-\lambda+2)s^2
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</math>
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By cosine law, we have
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<math>
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\cos \angle P''AQ''=\frac{P''A^2+Q''A^2-P''Q''^2}{2P''A\cdot Q''A}\\
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=\frac{\mu\lambda-\mu-\lambda+2}{2\sqrt{(\mu^2-\mu+1)(\lambda^2-\lambda+1)}}
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</math>
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for <math>\mu,\lambda\in(0,1)</math>.
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 +
We first find a lower bound of the numerator. We write <math>\mu\lambda-\mu-\lambda+2=(\lambda-1)\mu-\lambda+1</math> as a linear function of <math>\mu</math> by letting <math>\lambda</math> be constant. Since <math>\lambda\in (0,1)</math>, the coefficient <math>\lambda-1</math> of <math>\mu</math> in this linear function is negative, meaning this expression is least when <math>\mu</math> is least "at" <math>1</math>. Hence, <math>\mu\lambda-\mu-\lambda+2<1</math>.
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Now we find an upper bound of the denominator. We complete the square in the expression <math>2\sqrt{(\mu^2-\mu+1)(\lambda^2-\lambda+1)}=2\sqrt{((\mu-\frac{1}{2})^2+\frac{3}{4})((\lambda-\frac{1}{2})^2+\frac{3}{4})}</math> which is least when <math>\lambda</math> and <math>\mu</math> "are" both 1. Hence, <math>2\sqrt{(\mu^2-\mu+1)(\lambda^2-\lambda+1)}>2</math>.
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Hence, <math>\cos \angle P''AQ''<\frac{1}{2}</math>, implying that <math>\angle P''AQ''<60^\circ</math>.
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Therefore, <math>\angle PAQ=\angle P'AQ'<\angle P''AQ''<60^\circ</math>.
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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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{{MAA Notice}}

Latest revision as of 00:16, 19 April 2025

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

Solution 2 (similar to solution 1, but full proof)

Let the side length of the regular tetrahedron $ABCD$ be $s$.

Extend $AP$ and $AQ$ to intersect the plane $BCD$ at points $P'$ and $Q'$ inside $\Delta BCD$, respectively. We can observe that $\angle PAQ=\angle P'AQ'$.

Extend $P'Q'$ to intersect the sides of $\Delta BCD$ at $P''$ and $Q''$. We can observe that $\angle P'AQ'<\angle P''AQ''$. We assume that, WLOG, $P''$ is on $BD$ and $Q''$ is on $BC$.

Let the ratios $\frac{BP''}{BD}=\mu$ and $\frac{BQ''}{BC}=\lambda$. We can see that $\mu,\lambda\in (0,1)$. It follows from Stewart's theorem that $AP''^2=AD^2\mu+AB^2(1-\mu)-\mu(1-\mu)BD^2=(\mu^2-\mu+1)s\\ AQ''^2=AC^2\lambda+AB^2(1-\lambda)-\lambda(1-\lambda)BC^2=(\lambda^2-\lambda^2+1)s$

Furthermore, we have $P''C^2=CD^2\mu+CB^2(1-\mu)-\mu(1-\mu)BD^2=(\mu^2-\mu+1)s$

Using this, we can conclude that $P''Q''^2=P''C^2\lambda+(BD\mu)^2(1-\lambda)-\lambda(1-\lambda)BC^2\\ =(\mu\lambda-\mu-\lambda+2)s^2$

By cosine law, we have $\cos \angle P''AQ''=\frac{P''A^2+Q''A^2-P''Q''^2}{2P''A\cdot Q''A}\\ =\frac{\mu\lambda-\mu-\lambda+2}{2\sqrt{(\mu^2-\mu+1)(\lambda^2-\lambda+1)}}$ for $\mu,\lambda\in(0,1)$.

We first find a lower bound of the numerator. We write $\mu\lambda-\mu-\lambda+2=(\lambda-1)\mu-\lambda+1$ as a linear function of $\mu$ by letting $\lambda$ be constant. Since $\lambda\in (0,1)$, the coefficient $\lambda-1$ of $\mu$ in this linear function is negative, meaning this expression is least when $\mu$ is least "at" $1$. Hence, $\mu\lambda-\mu-\lambda+2<1$.

Now we find an upper bound of the denominator. We complete the square in the expression $2\sqrt{(\mu^2-\mu+1)(\lambda^2-\lambda+1)}=2\sqrt{((\mu-\frac{1}{2})^2+\frac{3}{4})((\lambda-\frac{1}{2})^2+\frac{3}{4})}$ which is least when $\lambda$ and $\mu$ "are" both 1. Hence, $2\sqrt{(\mu^2-\mu+1)(\lambda^2-\lambda+1)}>2$.

Hence, $\cos \angle P''AQ''<\frac{1}{2}$, implying that $\angle P''AQ''<60^\circ$.

Therefore, $\angle PAQ=\angle P'AQ'<\angle P''AQ''<60^\circ$.

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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