Difference between revisions of "2007 AMC 10A Problems/Problem 15"

(Solution 2)
m (Problem)
Line 1: Line 1:
==Problem==
+
Three circles are inscribed in a rectangle of width w and height h as shown. Two
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?
+
of the circles are congruent and are each tangent to two adjacent sides of the
 
+
rectangle and to each other. The other circle is larger and is tangent to three sides
[[Image:2007 AMC 10A -15 for wiki.png]]
+
of the rectangle and to the two smaller circles. What the ratio of h to w? Express
 
+
your answer as a decimal to the nearest hundredth.
<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===
 
 
 
 
 
The diagonal has length <math>\sqrt{2}+1+2+2+1+\sqrt{2}=6+2\sqrt{2}</math>. Therefore the sides have length <math>2+3\sqrt{2}</math>, and the area is
 
 
 
<cmath>A=(2+3\sqrt{2})^2=4+6\sqrt{2}+6\sqrt{2}+18=22+12\sqrt{2}  \Rightarrow \text{(B)}</cmath>
 
 
 
=== Solution 2 ===
 
 
 
 
 
 
 
Extend two radii from the larger circle to the centers of the two smaller circles above. This forms a right triangle of sides <math>3, 3, 3\sqrt{2}</math>. The length of the hypotenuse of the right triangle plus twice the radius of the smaller circle is equal to the side of the square. It follows, then <cmath> A = (2+3\sqrt{2})^2 = 22 + 12\sqrt{2} \Rightarrow \text{(B)}</cmath>
 
  
 
==See Also==
 
==See Also==

Revision as of 12:20, 21 October 2018

Three circles are inscribed in a rectangle of width w and height h as shown. Two of the circles are congruent and are each tangent to two adjacent sides of the rectangle and to each other. The other circle is larger and is tangent to three sides of the rectangle and to the two smaller circles. What the ratio of h to w? Express your answer as a decimal to the nearest hundredth.

See Also

2007 AMC 10A (ProblemsAnswer KeyResources)
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. AMC logo.png