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

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==Solution==
 
==Solution==
  
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We first note that diagonal <math> \overline{AC} </math> is of length <math> \sqrt{6} + \sqrt{2} </math>. It must be that <math> \overline{AP} </math> divides the diagonal into two segments in the ratio <math>\sqrt{3}</math> to <math>1</math>. It is not difficult to visualize that when the square is rotated, the initial and final squares overlap in a rectangular region of dimensions <math>2</math> by <math>\sqrt{3} + 1</math>. 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 <math> 2 (\sqrt{3} + 1)^2 - 2 (\sqrt{3} + 1) = 2 (4 + 2 \sqrt{3}) - 2 \sqrt{3} - 2 = 6 + 2 \sqrt{3} </math>.
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We first note that diagonal <math> \overline{AC} </math> is of length <math> \sqrt{6} + \sqrt{2} </math>. It must be that <math> \overline{AP} </math> divides the diagonal into two segments in the ratio <math>\sqrt{3}</math> to <math>1</math>. It is not difficult to visualize that when the square is rotated, the initial and final squares overlap in a rectangular region of dimensions <math>2\sqrt{3}</math> by <math>\sqrt{3} + 1</math>. 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 <math> 2 (\sqrt{3} + 1)^2 - 2 (\sqrt{3} + 1) = 2 (4 + 2 \sqrt{3}) - 2 \sqrt{3} - 2 = 6 + 2 \sqrt{3} </math>.
  
  
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The area also includes <math>4</math> circular segments. Two of them are quarter-segments of circles centered at <math>P</math> of radii <math>\sqrt{2}</math> (the segment bounded by <math> \overline{PA} </math> and <math> \overline{PA'}</math>) and <math>\sqrt{6}</math> (that bounded by <math>\overline{PC}</math> and <math>\overline{PC'}</math>). Assuming <math>A</math> is the bottom-left vertex and <math>B</math> is the bottom-right one, it is clear that the third segment is formed as <math>B</math> swings out to the right of the original square [recall that the square is rotated counterclockwise], while the fourth is formed when <math>D</math> overshoots the final square's left edge. To find the areas of these segments, consider the perpendicular from <math>P</math> to <math>\overline{BC}</math>. Call the point of intersection <math>E</math>. From the previous paragraph, it is clear that <math>PE = \sqrt{3}</math> and <math>BE = 1</math>. This means <math>PB = 2</math>, and <math>B</math> swings back inside edge <math>\overline{BC}</math> at a point <math>1</math> unit above <math>E</math> (since it left the edge <math>1</math> unit below). The triangle of the circular sector is therefore an equilateral triangle of side length <math>2</math>, and so the angle of the segment is <math>60^{\circ}</math>. Imagining the process in reverse, it is clear that the situation is the same with point <math>D</math>.
+
The area also includes <math>4</math> circular segments. Two are quarter-circles centered at <math>P</math> of radii <math>\sqrt{2}</math> (the segment bounded by <math> \overline{PA} </math> and <math> \overline{PA'}</math>) and <math>\sqrt{6}</math> (that bounded by <math>\overline{PC}</math> and <math>\overline{PC'}</math>). Assuming <math>A</math> is the bottom-left vertex and <math>B</math> is the bottom-right one, it is clear that the third segment is formed as <math>B</math> swings out to the right of the original square [recall that the square is rotated counterclockwise], while the fourth is formed when <math>D</math> overshoots the final square's left edge. To find these areas, consider the perpendicular from <math>P</math> to <math>\overline{BC}</math>. Call the point of intersection <math>E</math>. From the previous paragraph, it is clear that <math>PE = \sqrt{3}</math> and <math>BE = 1</math>. This means <math>PB = 2</math>, and <math>B</math> swings back inside edge <math>\overline{BC}</math> at a point <math>1</math> unit above <math>E</math> (since it left the edge <math>1</math> unit below). The triangle of the circular sector is therefore an equilateral triangle of side length <math>2</math>, and so the angle of the segment is <math>60^{\circ}</math>. Imagining the process in reverse, it is clear that the situation is the same with point <math>D</math>.
  
  
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This means <math>a = 10</math>, <math>b = 6</math>, and <math>c = 3</math>. So <math>a + b + c = 10 + 6 + 3 = 19</math>; answer choice <math>\boxed{\textbf{(C)} \ 19}</math> is correct.
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Since <math>a = 10</math>, <math>b = 6</math>, and <math>c = 3</math>, the answer is <math>a + b + c = 10 + 6 + 3 = \boxed{\textbf{(C)} \ 19}</math>.
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==Video Solution by Richard Rusczyk==
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https://artofproblemsolving.com/videos/amc/2013amc12a/362
 +
 
 +
~dolphin7
  
 
== See also ==
 
== See also ==

Latest revision as of 15:51, 3 April 2020

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://artofproblemsolving.com/videos/amc/2013amc12a/362

~dolphin7

See also

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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All AMC 12 Problems and Solutions

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