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Difference between revisions of "2013 AMC 12A Problems/Problem 23"

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Revision as of 11:27, 7 April 2013

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 $2$ 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 $2 (\sqrt{3} + 1)^2 - 2 (\sqrt{3} + 1) = 2 (4 + 2 \sqrt{3}) - 2 \sqrt{3} - 2 = 6 + 2 \sqrt{3}$.


Apart from the rectangular region of the initial and final squares, the swept out area includes $4$ circular segments. Two of them are quarter-segments of 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 the areas of these segments, 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 $D$.


To see that no other area is included, note that the four vertices represent local maxima of distance from point $P$, so the path of any other point in or on the square lies in one of the four circular segments if it does not lie entirely within the initial or final squares.


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 \[(6 + 2 \sqrt{3}) + (\frac{\pi}{2} - 1) + (\frac{3 \pi}{2} - 3) + 2 (\frac{2 \pi}{3} - \sqrt{3})\] \[= 2 + 2 \sqrt{3} + 2 \pi + \frac{4 \pi}{3} - 2 \sqrt{3}\] \[= \frac{10 \pi}{3} + 2\] \[= \frac{1}{3} (10 \pi + 6)\]


This means $a = 10$, $b = 6$, and $c = 3$. So $a + b + c = 10 + 6 + 3 = 19$; answer choice $\boxed{\textbf{(C)} \ 19}$ is correct.

See also

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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