# 2007 AMC 10A Problems/Problem 15

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.

## Problem

Four circles of radius $1$ are each tangent to two sides of a square and externally tangent to a circle of radius $2$, as shown. What is the area of the square?

$[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); real h=3*sqrt(2)/2; pair O0=(0,0), O1=(h,h), O2=(-h,h), O3=(-h,-h), O4=(h,-h); pair X=O0+2*dir(30), Y=O2+dir(45); draw((-h-1,-h-1)--(-h-1,h+1)--(h+1,h+1)--(h+1,-h-1)--cycle); draw(Circle(O0,2)); draw(Circle(O1,1)); draw(Circle(O2,1)); draw(Circle(O3,1)); draw(Circle(O4,1)); draw(O0--X); draw(O2--Y); label("2",midpoint(O0--X),NW); label("1",midpoint(O2--Y),SE); [/asy]$

$\text{(A)}\ 32 \qquad \text{(B)}\ 22 + 12\sqrt {2}\qquad \text{(C)}\ 16 + 16\sqrt {3}\qquad \text{(D)}\ 48 \qquad \text{(E)}\ 36 + 16\sqrt {2}$

## Solution

Draw a square connecting the centers of the four small circles of radius $1$. This square has a diagonal of length $6$, as it includes the diameter of the big circle of radius $2$ and two radii of the small circles of radius $1$. Therefore, the side length of this square is $$\frac{6}{\sqrt{2}} = 3\sqrt{2}.$$ The radius of the large square has a side length $2$ units larger than the one found by connecting the midpoints, so its side length is $$2 + 3\sqrt{2}.$$ The area of this square is $(2+3\sqrt{2})^2 = 22 + 12\sqrt{2}$ $(B).$

## Solution 2

We draw the long diagonal of the square. This diagonal yields $2\sqrt{2}+1+1+2+2=2\sqrt{2}+6$

## See Also

 2007 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byProblem 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.

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