Difference between revisions of "2013 AMC 10A Problems/Problem 20"
m (→Solution 1) |
(added more diagrams to Solution 1) |
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path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1)); | path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1)); | ||
draw(arcrot); | draw(arcrot); | ||
− | fill(arcrot--( | + | fill(arcrot--(0,0)--cycle,grey); |
draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow); | draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow); | ||
} | } | ||
Line 27: | Line 27: | ||
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) } 2 - \sqrt{2} + \frac{\pi}{4}}</math>. | 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) } 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> | ||
==Solution 2== | ==Solution 2== | ||
[[Image:AMC 10A 2013 20.jpg]] | [[Image:AMC 10A 2013 20.jpg]] |
Revision as of 13:28, 27 February 2016
Contents
Problem
A unit square is rotated about its center. What is the area of the region swept out by the interior of the square?
Solution 1
First, we need to see what this looks like. Below is a diagram.
For this square with side length 1, the distance from center to vertex is , hence the area is composed of a semicircle of radius , plus times a parallelogram with height and base . That is to say, the total area is .
(To turn each dart-shaped piece into a parallelogram, cut along the dashed line and flip over one half.)
Solution 2
Let be the center of the square and be the intersection of and . The desired area consists of the unit square, plus regions congruent to the region bounded by arc , , and , plus triangular regions congruent to right triangle . The area of the region bounded by arc , , and is . Since the circle has radius , the area of the region is , so 4 times the area of that region is . Now we find the area of . . Since is a right triangle, the area of is , so 4 times the area of is . Finally, the area of the whole region is .
See Also
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.