Difference between revisions of "2007 AMC 10A Problems/Problem 15"
Mariekitty (talk | contribs) m (→Solution 1) |
|||
(25 intermediate revisions by 8 users not shown) | |||
Line 1: | Line 1: | ||
− | ==Problem== | + | == Problem == |
Four circles of radius <math>1</math> are each tangent to two sides of a square and externally tangent to a circle of radius <math>2</math>, as shown. What is the area of the square? | Four circles of radius <math>1</math> are each tangent to two sides of a square and externally tangent to a circle of radius <math>2</math>, as shown. What is the area of the square? | ||
− | + | <center> | |
− | + | <asy> | |
− | + | unitsize(5mm); | |
+ | defaultpen(linewidth(.8pt)+fontsize(8pt)); | ||
+ | real h=3*sqrt(2)/2; | ||
+ | pair O0=(0,0), O1=(h,h), O2=(-h,h), O3=(-h,-h), O4=(h,-h); | ||
+ | pair X=O0+2*dir(30), Y=O2+dir(45); | ||
+ | draw((-h-1,-h-1)--(-h-1,h+1)--(h+1,h+1)--(h+1,-h-1)--cycle); | ||
+ | draw(Circle(O0,2)); | ||
+ | draw(Circle(O1,1)); | ||
+ | draw(Circle(O2,1)); | ||
+ | draw(Circle(O3,1)); | ||
+ | draw(Circle(O4,1)); | ||
+ | draw(O0--X); | ||
+ | draw(O2--Y); | ||
+ | label("$2$",midpoint(O0--X),NW); | ||
+ | label("$1$",midpoint(O2--Y),SE); | ||
+ | </asy> | ||
+ | </center> | ||
<math>\text{(A)}\ 32 \qquad \text{(B)}\ 22 + 12\sqrt {2}\qquad \text{(C)}\ 16 + 16\sqrt {3}\qquad \text{(D)}\ 48 \qquad \text{(E)}\ 36 + 16\sqrt {2}</math> | <math>\text{(A)}\ 32 \qquad \text{(B)}\ 22 + 12\sqrt {2}\qquad \text{(C)}\ 16 + 16\sqrt {3}\qquad \text{(D)}\ 48 \qquad \text{(E)}\ 36 + 16\sqrt {2}</math> | ||
− | + | ==Solution 1== | |
− | |||
− | |||
− | |||
− | |||
− | <cmath> | + | Draw a square connecting the centers of the four small circles of radius <math>1</math>. This square has a diagonal of length <math>6</math>, as it includes the diameter of the big circle of radius <math>2</math> and two radii of the small circles of radius <math>1</math>. Therefore, the side length of this square is <cmath>\frac{6}{\sqrt{2}} = 3\sqrt{2}.</cmath> The radius of the large square has a side length <math>2</math> units larger than the one found by connecting the midpoints, so its side length is <cmath>2 + 3\sqrt{2}.</cmath> The area of this square is <math>(2+3\sqrt{2})^2 = 22 + 12\sqrt{2}</math> <math>(B).</math> |
− | == | + | == Solution 2 == |
− | + | We draw the long diagonal of the square. This diagonal yields <math>2\sqrt{2}+1+1+2+2=2\sqrt{2}+6</math>. We know that the side length <math>s</math> in terms of the diagonal <math>d</math> is <math>s=\frac{d}{\sqrt{2}}</math>, so our side length is <math>\frac{2\sqrt{2}+6}{\sqrt{2}}</math>. However, we are trying to look for the area, so squaring <math>\frac{2\sqrt{2}+6}{\sqrt{2}}</math> yields <math>\frac{44+24\sqrt{2}}{2}=\boxed{\text{(B)}22+12\sqrt{2}}</math> | |
==See Also== | ==See Also== | ||
Line 22: | Line 34: | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:07, 1 May 2021
Contents
Problem
Four circles of radius are each tangent to two sides of a square and externally tangent to a circle of radius , as shown. What is the area of the square?
Solution 1
Draw a square connecting the centers of the four small circles of radius . This square has a diagonal of length , as it includes the diameter of the big circle of radius and two radii of the small circles of radius . Therefore, the side length of this square is The radius of the large square has a side length units larger than the one found by connecting the midpoints, so its side length is The area of this square is
Solution 2
We draw the long diagonal of the square. This diagonal yields . We know that the side length in terms of the diagonal is , so our side length is . However, we are trying to look for the area, so squaring yields
See Also
2007 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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.