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

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.

@@ Line 1: / Line 1: @@
-==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^o</math>.
+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}}
+=== 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>
+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>
+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>.
+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 <math>ABCD</math> be <math>s</math>.
+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>.
+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>.
+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
+<math>
+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
+</math>
+Furthermore, we have
+<math>
+P''C^2=CD^2\mu+CB^2(1-\mu)-\mu(1-\mu)BD^2=(\mu^2-\mu+1)s
+</math>
+Using this, we can conclude that
+<math>
+P''Q''^2=P''C^2\lambda+(BD\mu)^2(1-\lambda)-\lambda(1-\lambda)BC^2\\
+=(\mu\lambda-\mu-\lambda+2)s^2
+</math>
+By cosine law, we have
+<math>
+\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)}}
+</math>
+for <math>\mu,\lambda\in(0,1)</math>.
+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>.
+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>.
+Hence, <math>\cos \angle P''AQ''<\frac{1}{2}</math>, implying that <math>\angle P''AQ''<60^\circ</math>.
+Therefore, <math>\angle PAQ=\angle P'AQ'<\angle P''AQ''<60^\circ</math>.
+{{alternate solutions}}
+hurdler: Remark on solution 1: This proof is not rigorous, in the very last step. The last step needs more justification.
 ==See also==
@@ Line 10: / Line 63: @@
 [[Category:Olympiad Geometry Problems]]
+[[Category:3D Geometry Problems]]
-[[Solution by Vo Duc Dien]]
+{{MAA Notice}}
-Let M and N be the midpoints of AB and AC and let P” and Q” be points where AP and
-AQ intercept the plane DMN. We have /_PAQ = /_P”AQ”.
-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°.

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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