1973 USAMO Problems/Problem 1

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.

1973 USAMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

Page

Toolbox

Search

1973 USAMO Problems/Problem 1

Contents

Problem

Solutions

Solution 1

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

See also