Difference between revisions of "2013 AMC 10A Problems/Problem 20"
Csasomachar (talk | contribs) (→Solution 2) |
Made in 2016 (talk | contribs) (Rewrote the final equation and boxed) |
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[[Image:AMC 10A 2013 20.jpg]] | [[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 <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}}{28}</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 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)=\dfrac{\pi}{4}+2-\sqrt {2}</math> | + | 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}}{28}</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 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)=\dfrac{\pi}{4}+2-\sqrt {2}</math>, which we can rewrite as $\boxed{2 - \sqrt{2} + \frac{\pi}{4}} \text{(C)} |
==See Also== | ==See Also== |
Revision as of 17:34, 31 December 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
, which we can rewrite as $\boxed{2 - \sqrt{2} + \frac{\pi}{4}} \text{(C)}
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.