Difference between revisions of "2013 AMC 10A Problems/Problem 20"

Latest revision as of 16:55, 23 July 2021

Problem

A unit square is rotated $45^\circ$ about its center. What is the area of the region swept out by the interior of the square?

$\textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8}$

Solution 1

First, we need to see what this looks like. Below is a diagram.

$[asy] size(200); defaultpen(linewidth(0.8)); path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square; fill(square^^square2,grey); for(int i=0;i<=3;i=i+1) { path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1)); draw(arcrot); fill(arcrot--(0,0)--cycle,grey); draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow); } draw(square^^square2);[/asy]$

For this square with side length 1, the distance from center to vertex is $r = \frac{\sqrt{2}}{2}$ , hence the area is composed of a semicircle of radius $r$ , plus $4$ times a parallelogram (or a kite with diagonals of $(\sqrt{2}-1)$ and $r \text{ or} \frac{\sqrt{2}}{2}$ ) with height $\frac{1}{2}$ and base $\frac{\sqrt{2}}{2(1+\sqrt{2})}$ . That is to say, the total area is $\frac{1}{2} \pi \left(\frac{\sqrt{2}}{2}\right)^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } 2 - \sqrt{2} + \frac{\pi}{4}}$ .

$[asy] size(150);defaultpen(linewidth(0.8)); path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;//fill(square^^square2,grey); for(int i=0;i<=3;i=i+1){path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));draw(arcrot); fill(arcrot--(0,0)--cycle,grey);} //draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);} draw(square^^square2); //draw((-.5,.5)--(.5,-.5)^^(0,sqrt(.5))--(0,-sqrt(.5)),dotted);draw((.5,.5)--(-.5,-.5),dotted); [/asy]$ (To turn each dart-shaped piece into a parallelogram, cut along the dashed line and flip over one half.) $[asy] size(150,Aspect);real r=sqrt(2);real b=2-2/r; draw((0,0)--(-1,1)--(b-1,1)--(0,r)--cycle);draw((0,1)--(b-1,1)--(b/2-1,1-b/2));draw((0,0)--(b-1,1),dashed); fill((2,0)--(b+1,1)--(b+2,0)--cycle,lightgray);draw((.5,.5)--(1,.5),EndArrow); draw((2,0)--(1,1)--(b+1,1)--(b+2,0)--(2,0)^^(b+1,1)--(b/2+1,1-b/2)^^(2,0)--(2+b/2,b/2)); draw((2,0)--(b+1,1),dashed); [/asy]$ Alternatively, you can move the dart-shaped piece to the other side and make a kite. $[asy] size(75,Aspect);real r=sqrt(2);real b=2-2/r; draw((r-1,1)--(b-1,1)); draw((0,0)--(b-1,1)--(0,r)--(r-1,1)--cycle); draw((0,r)--(0,0),dashed); [/asy]$

Solution 2

$[asy] size(200); defaultpen(linewidth(0.8)); path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square; for(int i=0;i<=6;i=i+1) { path arcrot=arc(origin,sqrt(2)/2,45+270*i,270*(i+1)); draw(arcrot); } draw(square^^square2);[/asy]$

$\textbf{(high res image; no labels)}$

Let $O$ be the center of the square and $C$ be the intersection of $OB$ and $AD$ . The desired area consists of the unit square, plus $4$ regions congruent to the region bounded by arc $AB$ , $\overline{AC}$ , and $\overline{BC}$ , plus $4$ triangular regions congruent to right triangle $BCD$ . The area of the region bounded by arc $AB$ , $\overline{AC}$ , and $\overline{BC}$ is $\frac{\text{Area of Circle}-\text{Area of Square}}{8}$ . Since the circle has radius $\dfrac{1}{\sqrt {2}}$ , the area of the region is $\dfrac{\dfrac{\pi}{2}-1}{8}$ , so 4 times the area of that region is $\dfrac{\pi}{4}-\dfrac{1}{2}$ . Now we find the area of $\triangle BCD$ . $BC=BO-OC=\dfrac{\sqrt {2}}{2}-\dfrac{1}{2}$ . Since $\triangle BCD$ is a $45-45-90$ right triangle, the area of $\triangle BCD$ is $\dfrac{BC^2}{2}=\dfrac {\left (\dfrac {\sqrt {2}}{2}-\dfrac{1}{2} \right)^2}{2}$ , so $4$ times the area of $\triangle BCD$ is $\dfrac{3}{2}-\sqrt {2}$ . Finally, the area of the whole region is $1+ \left(\dfrac {3}{2}-\sqrt {2} \right) + \left(\dfrac{\pi}{4}-\dfrac{1}{2} \right)=\dfrac{\pi}{4}+2-\sqrt {2}$ , which we can rewrite as $\boxed{\textbf{(C) } 2 - \sqrt{2} + \frac{\pi}{4}}$ .

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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
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Difference between revisions of "2013 AMC 10A Problems/Problem 20"

Latest revision as of 16:55, 23 July 2021

Contents

Problem

Solution 1

Solution 2

See Also

@@ Line 20: / Line 20: @@
 path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));
 draw(arcrot);
-fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);
+fill(arcrot--(0,0)--cycle,grey);
 draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);
 }
 draw(square^^square2);</asy>
-For this square with side length 1, the distance from center to vertex is <math>r = \frac{1}{\sqrt{2}}</math>, hence the area is composed of a semicircle of radius <math>r</math>, plus <math>4</math> times a parallelogram with height <math>\frac{1}{2}</math> and base <math>\frac{\sqrt{2}}{2(1+\sqrt{2})}</math>. That is to say, the total area is <math>\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}</math>.
+For this square with side length 1, the distance from center to vertex is <math>r = \frac{\sqrt{2}}{2}</math>, hence the area is composed of a semicircle of radius <math>r</math>, plus <math>4</math> times a parallelogram (or a kite with diagonals of <math>(\sqrt{2}-1)</math> and <math>r \text{ or} \frac{\sqrt{2}}{2}</math>) with height <math>\frac{1}{2}</math> and base <math>\frac{\sqrt{2}}{2(1+\sqrt{2})}</math>. That is to say, the total area is <math>\frac{1}{2} \pi \left(\frac{\sqrt{2}}{2}\right)^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } 2 - \sqrt{2} + \frac{\pi}{4}}</math>.
+<asy>
+size(150);defaultpen(linewidth(0.8));
+path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;//fill(square^^square2,grey);
+for(int i=0;i<=3;i=i+1){path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));draw(arcrot);
+fill(arcrot--(0,0)--cycle,grey);}
+//draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);}
+draw(square^^square2);
+//draw((-.5,.5)--(.5,-.5)^^(0,sqrt(.5))--(0,-sqrt(.5)),dotted);draw((.5,.5)--(-.5,-.5),dotted);
+</asy>
+(To turn each dart-shaped piece into a parallelogram, cut along the dashed line and flip over one half.)
+<asy>
+size(150,Aspect);real r=sqrt(2);real b=2-2/r;
+draw((0,0)--(-1,1)--(b-1,1)--(0,r)--cycle);draw((0,1)--(b-1,1)--(b/2-1,1-b/2));draw((0,0)--(b-1,1),dashed);
+fill((2,0)--(b+1,1)--(b+2,0)--cycle,lightgray);draw((.5,.5)--(1,.5),EndArrow);
+draw((2,0)--(1,1)--(b+1,1)--(b+2,0)--(2,0)^^(b+1,1)--(b/2+1,1-b/2)^^(2,0)--(2+b/2,b/2));
+draw((2,0)--(b+1,1),dashed);
+</asy>
+Alternatively, you can move the dart-shaped piece to the other side and make a kite.
+<asy>
+size(75,Aspect);real r=sqrt(2);real b=2-2/r;
+draw((r-1,1)--(b-1,1));
+draw((0,0)--(b-1,1)--(0,r)--(r-1,1)--cycle);
+draw((0,r)--(0,0),dashed);
+</asy>
 ==Solution 2==
+<asy>
+size(200);
+defaultpen(linewidth(0.8));
+path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;
+for(int i=0;i<=6;i=i+1)
+{
+path arcrot=arc(origin,sqrt(2)/2,45+270*i,270*(i+1));
+draw(arcrot);
+}
+draw(square^^square2);</asy>
+<center><math>\textbf{(high res image; no labels)}</math></center>
 [[Image:AMC 10A 2013 20.jpg]]
-Let <math>O</math> be the center of the square and <math>C</math> be the intersection of <math>OB</math> and <math>AD</math>. The desired area consists of the unit square, plus 4 regions congruent to the region bounded by arc <math>AB</math>, <math>\overline{AC}</math>, and <math>\overline{BC}</math>, plus 4 triangular regions congruent to right triangle <math>BCD</math>. The area of the region bounded by arc <math>AB</math>, <math>\overline{AC}</math>, and <math>\overline{BC}</math> is (Area of Circle-Area of Square)/8. Since the circle has radius <math>\dfrac{1}{\sqrt {2}}</math>, the area of the region is <math>\dfrac{\dfrac{\pi}{2}-1}{8}</math>, so 4 times the area of that region is <math>\dfrac{\pi}{4}-\dfrac{1}{2}</math>. Now we find the area of <math>\triangle BCD</math>. <math>BC=BO-OC=\dfrac{\sqrt {2}}{2}-\dfrac{1}{2}</math>. Since <math>\triangle BCD</math> is a 45-45-90 right triangle, the area of <math>\triangle BCD</math> is <math>\dfrac{BC^2}{2}=\dfrac{({\dfrac{\sqrt {2}}{2}-\dfrac{1}{2})^2}{2}</math>, so 4 times the area of <math>\triangle BCD</math> is <math>\dfrac{3}{2}-\sqrt {2}</math>. Finally, the area of the whole region is <math>1+ \left(\dfrac {3}{2}-\sqrt {2} \right) + \left(\dfrac{\pi}{4}-\dfrac{1}{2} \right)=</math>\dfrac{\pi}{4}+2-\sqrt {2}$.
+Let <math>O</math> be the center of the square and <math>C</math> be the intersection of <math>OB</math> and <math>AD</math>. The desired area consists of the unit square, plus <math>4</math> regions congruent to the region bounded by arc <math>AB</math>, <math>\overline{AC}</math>, and <math>\overline{BC}</math>, plus <math>4</math> triangular regions congruent to right triangle <math>BCD</math>. The area of the region bounded by arc <math>AB</math>, <math>\overline{AC}</math>, and <math>\overline{BC}</math> is <math>\frac{\text{Area of Circle}-\text{Area of Square}}{8}</math>. Since the circle has radius <math>\dfrac{1}{\sqrt {2}}</math>, the area of the region is <math>\dfrac{\dfrac{\pi}{2}-1}{8}</math>, so 4 times the area of that region is <math>\dfrac{\pi}{4}-\dfrac{1}{2}</math>. Now we find the area of <math>\triangle BCD</math>. <math>BC=BO-OC=\dfrac{\sqrt {2}}{2}-\dfrac{1}{2}</math>. Since <math>\triangle BCD</math> is a <math>45-45-90</math> right triangle, the area of <math>\triangle BCD</math> is <math>\dfrac{BC^2}{2}=\dfrac {\left (\dfrac {\sqrt {2}}{2}-\dfrac{1}{2} \right)^2}{2}</math>, so <math>4</math> times the area of <math>\triangle BCD</math> is <math>\dfrac{3}{2}-\sqrt {2}</math>. Finally, the area of the whole region is <math>1+ \left(\dfrac {3}{2}-\sqrt {2} \right) + \left(\dfrac{\pi}{4}-\dfrac{1}{2} \right)=\dfrac{\pi}{4}+2-\sqrt {2}</math>, which we can rewrite as <math>\boxed{\textbf{(C) }    2 - \sqrt{2} + \frac{\pi}{4}}</math>.
 ==See Also==