AMC10 2007 10A, Problem 15

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]

Solution: We draw the diagonal of the square. This diagonal yields $2\sqrt{2}+1+1+2+2=2\sqrt{2}+6$. We know that the side length $s$ in terms of the diagonal $d$ is $s=\frac{d}{\sqrt{2}}$, so our side length is $\frac{2\sqrt{2}+6}{\sqrt{2}}$. However, we are trying to look for the area, so squaring $\frac{2\sqrt{2}+6}{\sqrt{2}}$ yields $\frac{44+24\sqrt{2}}{2}=\fbox{22 + 12\sqrt {2}}$

Answer: 22 + 12\sqrt {2}