2021 AMC Fall 10B Problem 15

1. Show $CQ = BP$. Label the square $A$ top-left, $B$ bottom-left, $C$ bottom-right, $D$ top-right, so $P$ is on the top side $AD$ and $Q$ is on the left side $AB$. Look at triangles $ABP$ and $BCQ$.
- Both have a right angle (at $A$ and at $B$).
- $AB = BC$ (sides of the square).
- Because $BP$ and $CQ$ cross at a right angle, angle $ABP$ equals angle $BCQ$. (In triangle $BRC$, angle $RBC$ + angle $BCR = 90^\circ$; also angle $ABP$ + angle $RBC = 90^\circ$.)

So the triangles are congruent, and $CQ = BP = 6 + 7 = 13$.
2. Focus on triangle $QBC$. It has a right angle at $B$, and its long side is $QC = 13$. The segment $BR$ drops from the right angle and meets $QC$ at a right angle, so $BR = 6$ is the height to the long side.
3. Use the right-triangle height facts. The height splits the long side into $QR$ and $RC$. The two small triangles $QRB$ and $BRC$ have the same shape as each other, which gives
$$BR \times BR = QR \times RC, \quad\text{so}\quad QR \times RC = 36.$$
Also $QR + RC = 13$. Two numbers that add to $13$ and multiply to $36$: they are $4$ and $9$.
4. Find the side. Triangle $BRC$ has the same shape as triangle $QBC$, so $\frac{BC}{RC} = \frac{QC}{BC}$, which gives $BC^2 = RC \times QC$.
- If $RC = 9$: $BC^2 = 9 \times 13 = 117$.
- If $RC = 4$: $BC^2 = 4 \times 13 = 52$.
5. Which one is real? In the same way, $BQ^2 = QR \times QC$. If $RC = 4$, then $QR = 9$ and $BQ^2 = 117$, which is bigger than the side squared, $52$. Then $Q$ would be past the end of side $AB$, which is not allowed. So $RC = 9$ and the area is $BC^2 = 117$.

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

Two segments inside a square that cross at a right angle and each run between sides like this are always the same length: a quarter turn of the square carries one onto the other. That turns the odd-looking lengths $6$ and $7$ into the long side of a right triangle with a known height. The "height to the long side" facts, height squared equals the product of the two pieces and a leg squared equals its piece times the whole, then finish the job.

Call the side $s$. From step 1, triangle $BRC$ (right angle at $R$) has the same shape as triangle $PAB$ (right angle at $A$). Matching sides: $\frac{BR}{BC} = \frac{AP}{BP}$ and $\frac{RC}{BC} = \frac{AB}{BP}$. So $AP = \frac{6 \times 13}{s} = \frac{78}{s}$ and $AB = s$. Pythagoras in triangle $ABP$: $s^2 + \frac{78^2}{s^2} = 169$. Trying the choices, $s^2 = 117$ gives $117 + \frac{6084}{117} = 117 + 52 = 169$. It works, and $117$ is choice (D). (The equation has one other solution, $s^2 = 52$, but it is not a choice, and step 5 shows it would put $Q$ off the side.)