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?  
 
 A: 
I think this image makes this problem much easier. By the Pythagorean Theorem we have $$r^2 + r^2 = (x\sqrt 2 + x + r)^2$$
Solving for $x$ we get:
$$r\sqrt 2 = x(\sqrt 2 + 1) + r$$
$$r(\sqrt 2 - 1) = x(\sqrt 2 + 1)$$
$$x = \frac{\sqrt 2 - 1}{\sqrt 2 + 1}r=\frac{(\sqrt 2 - 1)(\sqrt 2 - 1)}{(\sqrt 2 + 1)(\sqrt 2 - 1)}r=\frac{2-2\sqrt 2 + 1}{1}r=(3-2\sqrt 2)r$$
A: Let's look at this geometrically first. The radius of the small square plus the radius$\cdot \sqrt2$ plus the radius of the large circle is equal to the radius of the large circle$\cdot \sqrt2$
If $x$ is the radius we want, then
$$x + x\sqrt2 + r = r\sqrt2 \quad\text{and now we solve for x}$$
$$x(1+\sqrt2)  = r(\sqrt2-1)$$
$$x = \frac{r(\sqrt2-1)}{1+\sqrt2} = \frac{r(\sqrt2-1)}{1+\sqrt2} \frac{\sqrt2-1}{\sqrt2-1} = r(3-2\sqrt2)$$
A: 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: $r(3-2\sqrt{2})$.
