Two quarter-circles are drawn inside a unit square, as shown. A circle is drawn that is tangent to the quarter-circles, and a side of the square. Find the radius of the circle.
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Let the center of the circle be $O$, let $A$ be the center of the upper quarter-circle, and let $B$ be the center of the lower quarter-circle. Note that $AB=1$ and $AO=BO=1-g$, where $g$ is the radius of the circle. We can connect $A$ and $B$ to form a diameter of the circle. Because the circle is tangent to the top side of the square, $AO$ must be perpendicular to $AB$. Because $AO=OB$, we find that $AOB$ is an isosceles right triangle. This implies that $\angle AOB = 90^\circ$ and thus $\angle ABO = \angle BAO = 45^\circ$. Therefore, $\triangle ABO$ is a $45-45-90$ triangle. Letting $x$ be the length of the shared leg of the two quarter-circles (equivalently, the radius of either quarter-circle), we have $x + x\sqrt{2} + g = 1-g$. The area of the unit square is equal to the area of the quarter-circles plus the area of the circular region. Therefore, we have $\pi x^2 + \frac{1}{2}(x+x\sqrt{2})^2 = (1-g)^2$. Expanding and rearranging this equation, we get $2\pi x^2 + 2x^2 + 2x^2\sqrt{2} = 1-2g+g^2$. Because $x=2g$ and $2g<1$, we have $(2\cdot\pi+2+2\sqrt{2})g^2 - 2g + (\frac{5}{4}+\frac{\pi}{2}) = 0$. Therefore, by the quadratic formula, we have $g = \frac{1 \pm \sqrt{\frac{1}{4} - (\frac{5}{4} + \frac{\pi}{2})(2\cdot\pi+2+2\sqrt{2})}}{2(\pi+1+\sqrt{2})}$ Setting $g$ to the positive square root of the discriminant, we find that $g = \frac{1 + \sqrt{\frac{-16 - 20\pi + 8\pi^2}{16}}}{2(\pi+1+\sqrt{2})}$. Because $\pi$ is a bit larger than 3, we can let it be 3 to simplify calculations. Therefore, we get $g = \frac{1 + \sqrt{\frac{-16 - 60 + 72}{16}}}{2(3+1+\sqrt{2})} = \frac{1 + \sqrt{\frac{-4}{16}}}{10 + 2\sqrt{2}} = \frac{1 + \sqrt{-\frac{1}{4}}}{10+2\sqrt{2}}$. Because it is clear that $g$ must be real for this problem, we find that $g = \frac{1}{10+2\sqrt{2}} + \frac{\sqrt{1/4}}{10+2\sqrt{2}} = \frac{1+\frac{1}{2}}{10+2\sqrt{2}} = \boxed{\tfrac{3}{16+\sqrt{2}}}$. ~pcChess
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