2021 AMC Fall 10B Problem 15

Rotate the square $90^\circ$ about its center so that $A \to B$, $B \to C$, $D \to A$. Side $AD$ goes to side $BA$, and segment $BP$ goes to a segment from $C$ to a point of $AB$, rotated $90^\circ$ from $BP$. Since $CQ$ is also perpendicular to $BP$ and passes through $C$, that image is exactly $CQ$. Hence $CQ = BP = 6 + 7 = 13$.

Now look at right triangle $QBC$ (right angle at $B$). Its hypotenuse is $QC = 13$, and $BR \perp QC$ means $R$ is the foot of the altitude from $B$. The altitude-to-hypotenuse relations give
$$BR^2 = QR \cdot RC \quad\Longrightarrow\quad 36 = QR \cdot RC, \qquad QR + RC = 13.$$
So $\{QR, RC\} = \{4, 9\}$, and $BC^2 = RC \cdot QC$ is either $9 \cdot 13 = 117$ or $4 \cdot 13 = 52$.

If $RC = 4$, then $BQ^2 = QR \cdot QC = 117 > 52 = BC^2 = AB^2$, so $Q$ would lie outside side $AB$. Hence $RC = 9$ and the area is $BC^2 = 117$.

The answer is $\boxed{\textbf{(D)}\ 117}$.

Two perpendicular segments joining sides of a square are images of each other under a quarter turn, so they are equal in length; this converts the strange piece of data ($BR$ and $PR$) into the hypotenuse of a right triangle whose altitude is known. The mean-proportional relations $h^2 = mn$ and $\text{leg}^2 = (\text{adjacent segment})(\text{hypotenuse})$ then finish without any variable.

Let the side be $s$ and $\theta = \angle ABP$. In right triangle $BRC$, $\angle BCR = \theta$ (from the congruent triangles), so $\sin\theta = 6/s$; from triangle $ABP$, $\cos\theta = s/13$. Then $36/s^2 + s^2/169 = 1$ gives $s^4 - 169s^2 + 6084 = 0$, so $s^2 = 117$ or $52$; only $117$ is a choice.

• Trying to solve with the side as unknown in two similar triangles and getting a quartic, or picking the root 52 without checking that Q must lie on side AB.
• Assuming $BR$ is half of $BP$ or that $R$ is the center of the square; nothing in the problem forces that.