2013 AMC 12A Problems/Problem 23

Problem

$ABCD$ is a square of side length $\sqrt{3} + 1$ . Point $P$ is on $\overline{AC}$ such that $AP = \sqrt{2}$ . The square region bounded by $ABCD$ is rotated $90^{\circ}$ counterclockwise with center $P$ , sweeping out a region whose area is $\frac{1}{c} (a \pi + b)$ , where $a$ , $b$ , and $c$ are positive integers and $\text{gcd}(a,b,c) = 1$ . What is $a + b + c$ ?

$\textbf{(A)} \ 15 \qquad \textbf{(B)} \ 17 \qquad \textbf{(C)} \ 19 \qquad \textbf{(D)} \ 21 \qquad \textbf{(E)} \ 23$

Solution

We first note that diagonal $\overline{AC}$ is of length $\sqrt{6} + \sqrt{2}$ . It must be that $\overline{AP}$ divides the diagonal into two segments in the ratio $\sqrt{3}$ to $1$ . It is not difficult to visualize that when the square is rotated, the initial and final squares overlap in a rectangular region of dimensions $2\sqrt{3}$ by $\sqrt{3} + 1$ . The area of the overall region (of the initial and final squares) is therefore twice the area of the original square minus the overlap, or $2 (\sqrt{3} + 1)^2 - 2 (\sqrt{3} + 1) = 2 (4 + 2 \sqrt{3}) - 2 \sqrt{3} - 2 = 6 + 2 \sqrt{3}$ .

The area also includes $4$ circular segments. Two are quarter-circles centered at $P$ of radii $\sqrt{2}$ (the segment bounded by $\overline{PA}$ and $\overline{PA'}$ ) and $\sqrt{6}$ (that bounded by $\overline{PC}$ and $\overline{PC'}$ ). Assuming $A$ is the bottom-left vertex and $B$ is the bottom-right one, it is clear that the third segment is formed as $B$ swings out to the right of the original square [recall that the square is rotated counterclockwise], while the fourth is formed when $D$ overshoots the final square's left edge. To find these areas, consider the perpendicular from $P$ to $\overline{BC}$ . Call the point of intersection $E$ . From the previous paragraph, it is clear that $PE = \sqrt{3}$ and $BE = 1$ . This means $PB = 2$ , and $B$ swings back inside edge $\overline{BC}$ at a point $1$ unit above $E$ (since it left the edge $1$ unit below). The triangle of the circular sector is therefore an equilateral triangle of side length $2$ , and so the angle of the segment is $60^{\circ}$ . Imagining the process in reverse, it is clear that the situation is the same with point $D$ .

The area of the segments can be found by subtracting the area of the triangle from that of the sector; it follows that the two quarter-segments have areas $\frac{1}{4} \pi (\sqrt{2})^2 - \frac{1}{2} \sqrt{2} \sqrt{2} = \frac{\pi}{2} - 1$ and $\frac{1}{4} \pi (\sqrt{6})^2 - \frac{1}{2} \sqrt{6} \sqrt{6} = \frac{3 \pi}{2} - 3$ . The other two segments both have area $\frac{1}{6} \pi (2)^2 - \frac{(2)^2 \sqrt{3}}{4} = \frac{2 \pi}{3} - \sqrt{3}$ .

The total area is therefore $(6 + 2 \sqrt{3}) + (\frac{\pi}{2} - 1) + (\frac{3 \pi}{2} - 3) + 2 (\frac{2 \pi}{3} - \sqrt{3})$ $= 2 + 2 \sqrt{3} + 2 \pi + \frac{4 \pi}{3} - 2 \sqrt{3}$ $= \frac{10 \pi}{3} + 2$ $= \frac{1}{3} (10 \pi + 6)$

Since $a = 10$ , $b = 6$ , and $c = 3$ , the answer is $a + b + c = 10 + 6 + 3 = \boxed{\textbf{(C)} \ 19}$ .

Video Solution by Richard Rusczyk

https://www.youtube.com/watch?v=-3d31c-7xm8

~dolphin7

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2013 AMC 12A Problems/Problem 23

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Problem

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Video Solution by Richard Rusczyk

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