HMMT February

Geometry Difficulty 5.4 AIME, harder Prove it United States

Problem:

Let $ABCD$ be a square, and let $M$ be the midpoint of side $BC$ . Points $P$ and $Q$ lie on segment $AM$ such that $\angle BPD = \angle BQD = 135^{\circ}$ . Given that $AP < AQ$ , compute $\frac{AQ}{AP}$ .

Solution

Solution:

Notice that $\angle BPD = 135^{\circ} = 180^{\circ} - \frac{\angle BAD}{2}$ and $P$ lying on the opposite side of $BD$ as $C$ means that $P$ lies on the circle with center $C$ through $B$ and $D$ . Similarly, $Q$ lies on the circle with center $A$ through $B$ and $D$ .

Let the side length of the square be $1$ . We have $AB = AQ = AD$ , so $AQ = 1$ . To compute $AP$ , let $E$ be the reflection of $D$ across $C$ . We have that $E$ lies both on $AM$ and the circle centered at $C$ through $B$ and $D$ . Since $AB$ is tangent to this circle,
$AB^2 = AP \cdot AE$
by power of a point. Thus, $1^2 = AP \cdot \sqrt{5} \Longrightarrow AP = \frac{1}{\sqrt{5}}$ . Hence, the answer is $\sqrt{5}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.