2011 USAMO Problems/Problem 5

Problem

Let $P$ be a given point inside quadrilateral $ABCD$ . Points $Q_1$ and $Q_2$ are located within $ABCD$ such that $\angle Q_1 BC = \angle ABP$ , $\angle Q_1 CB = \angle DCP$ , $\angle Q_2 AD = \angle BAP$ , $\angle Q_2 DA = \angle CDP$ . Prove that $\overline{Q_1 Q_2} \parallel \overline{AB}$ if and only if $\overline{Q_1 Q_2} \parallel \overline{CD}$ .

Solution

First note that $\overline{Q_1 Q_2} \parallel \overline{AB}$ if and only if the altitudes from $Q_1$ and $Q_2$ to $\overline{AB}$ are the same, or $|Q_1B|\sin \angle ABQ_1 =|Q_2A|\sin \angle BAQ_2$ . Similarly $\overline{Q_1 Q_2} \parallel \overline{CD}$ iff $|Q_1C|\sin \angle DCQ_1 =|Q_2D|\sin \angle CDQ_2$ .

If we define $S =\frac{|Q_1B|\sin \angle ABQ_1}{|Q_2A|\sin \angle BAQ_2} \times \frac{|Q_2D|\sin \angle CDQ_2}{|Q_1C|\sin \angle DCQ_1}$ , then we are done if we can show that S=1.

By the law of sines, $\frac{|Q_1B|}{|Q_1C|}=\frac{\sin\angle Q_1CB}{\sin\angle Q_1BC}$ and $\frac{|Q_2D|}{|Q_2A|}=\frac{\sin\angle Q_2AD}{\sin\angle Q_2DA}$ .

So, $S=\frac{\sin \angle ABQ_1}{\sin \angle BAQ_2}\cdot\frac{\sin \angle CDQ_2}{\sin \angle DCQ_1}\cdot\frac{\sin \angle BCQ_1}{\sin \angle CBQ_1}\cdot\frac{\sin \angle DAQ_2}{\sin \angle ADQ_2}$

By the terms of the problem, $S=\frac{\sin \angle PBC}{\sin \angle PAD}\cdot\frac{\sin \angle PDA}{\sin \angle PCB}\cdot\frac{\sin \angle PCD}{\sin \angle PBA}\cdot\frac{\sin \angle PAB}{\sin \angle PDC}$ . (If two subangles of an angle of the quadrilateral are equal, then their complements at that quadrilateral angle are equal as well.)

Rearranging yields $S= \frac{\sin \angle PBC}{\sin \angle PCB}\cdot\frac{\sin \angle PDA}{\sin \angle PAD}\cdot\frac{\sin \angle PCD}{\sin \angle PDC}\cdot\frac{\sin \angle PAB}{\sin \angle PBA}$ .

Applying the law of sines to the triangles with vertices at P yields $S=\frac{|PC|}{|PB|}\frac{|PA|}{|PD|}\frac{|PD|}{|PC|}\frac{|PB|}{|PA|}=1$ .

Solution 2

Lemma. If $AB$ and $CD$ are not parallel, then $AB, CD, Q_1 Q_2$ are concurrent.

Proof. Let $AB$ and $CD$ meet at $R$ . Notice that with respect to triangle $ADR$ , $P$ and $Q_2$ are isogonal conjugates. With respect to triangle $BCR$ , $P$ and $Q_1$ are isogonal conjugates. Therefore, $Q_1$ and $Q_2$ lie on the reflection of $RP$ in the angle bisector of $\angle{DRA}$ , so $R, Q_1, Q_2$ are collinear. Hence, $AB, CD, Q_1 Q_2$ are concurrent at $R$ .

Now suppose $Q_1 Q_2 \parellel AB$ (Error compiling LaTeX. Unknown error_msg) but $Q_1 Q_2$ is not parallel to $CD$ . Then $AB$ and $CD$ are not parallel and thus intersect at a point $R$ . But then $Q_1 Q_2$ also passes through $R$ , contradicting $Q_1 Q_2 \parallel AB$ . A similar contradiction occurs if $Q_1 Q_2 \parallel CD$ but $Q_1 Q_2$ is not parallel to $CD$ , so we can conclude that $Q_1 Q_2 \parallel AB$ if and only if $Q_1 Q_2 \parallel CD$ .

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

2011 USAMO (Problems • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6
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2011 USAMO Problems/Problem 5

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Problem

Solution

Solution 2

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