Find the radius of the largest circle In the accompanying diagram, a circle of radius $r$ is tangent to both sides of the right-angled corner. What is the radius of the largest circle that will fit in the same corner between the larger circle and the corner?  
 

It's not difficult to show that if we drop a straight line from the center of the circle to the corner, we will thus form a right triangle with a hypotenuse of $r\sqrt{2}$ and two sides of length $r$.
Now, let $x$ be the radius of the largest circle that fits in the corner between the larger circle and the corner. It's also easy to see that the triangle with the hypotenuse of length $x + r$ and the two sides of length $r - x$ and the triangle we created above are similar. Therefore:
$$
\frac{x + r}{r\sqrt{2}}=\frac{r - x}{r}\implies\\
x + r=\sqrt{2}(r-x)\implies\\
x + r=r\sqrt{2}-x\sqrt{2}\implies\\
x + x\sqrt{2}=r\sqrt{2}-r\implies\\
x(1+\sqrt{2})=r(\sqrt{2}-1)\implies\\
x=r\frac{\sqrt{2}-1}{1+\sqrt{2}}
$$
Or, if you're so inclined, you can also rationalize the denominator:
$$
x=r\frac{\sqrt{2}-1}{\sqrt{2}+1}\implies\\
x=r\frac{\sqrt{2}-1}{\sqrt{2}+1}\cdot\frac{\sqrt{2}-1}{\sqrt{2}-1}\implies\\
x=r\frac{(\sqrt{2}-1)^2}{2-1}\implies\\
x=r(3-2\sqrt{2})
$$
Answer: $r(3-2\sqrt{2})$.
