Three circles are inscribed inside a unit square. Find the radius of the 3rd circle. One circle is inscribed inside a unit square such that all sides of square are tangent to it. 
A second circle is at the upper-left corner of the square such that it is tangent to the two sides of the square and touches the first circle externally. 
A third circle touches both circles externally and the upper side of the square is tangent to the third circle. (it is on the right side of the second circle.)
Find the radius of the 3rd circle. 

 A: 
Let the three radii be $r_1$, $r_2$ and $r_3$. From matching various lengths, the following relationships among the radii hold,
$$\sqrt{2}r_1=r_1+(1+\sqrt{2})r_2\tag{1}$$
$$(r_1+r_3)^2-(r_1-r_3)^2=a^2\tag{2}$$
$$(r_2+r_3)^2-(r_2-r_3)^2=b^2\tag{3}$$
$$r_1-r_2=a+b\tag{4}$$
Solve (1) for the radius of the second circle,
$$r_2=(3-2\sqrt{2})r_1$$
Simplify (2) and (3) to get,
$$r_3=\frac{a^2}{4r_1}\tag{5}$$
$$\frac{a^2}{b^2}=\frac{r_1}{r_2}\tag{6}$$
Combine (6) and (4) to get
$$a=\sqrt{r_1}(\sqrt{r_1}-\sqrt{r_2})\tag{7}$$
Then, plug (7) into (5) to obtained the radius of the third circle,
$$r_3=\frac 14  (\sqrt{r_1}-\sqrt{r_2})^2 $$
Use the result of the second radius $r_2$ derived above to express $r_3$ in terms of $r_1 = \frac 12$,
$$r_3=\frac 12 (3-2\sqrt{2})r_1=\frac 14 (3-2\sqrt{2})$$
A: The radius of the first circle is $\frac{1}{2}$, so its curvature is $k_1=2$. The radius of the second circle is $\frac{(\sqrt{2}-1)/2}{2\sqrt{2}}$, so its curvature is $k_2=\frac{4\sqrt{2}}{\sqrt{2}-1}$. The upper side of the square can be considered a circle with infinite radius and zero curvature. Thus,  the Descartes theorem in this case reduces to $$k_4=k_1+k_2+2 \sqrt{k_1 k_2}$$ This gives
$$k_4=2+\frac{4\sqrt{2}}{\sqrt{2}-1}+2 \sqrt{ \frac{8\sqrt{2}}{\sqrt{2}-1}  }\approx 26.109...$$
which corresponds to a radius of $$\frac{1}{26.109}\approx 0.038$$
A: As @Jaap Scherphuis commented,
this is a special case of 
Descartes' Theorem
\begin{align} 
k_3&=k_1+k_2+k_4+2\sqrt{k_1k_2+k_2k_4+k_4k_1}
=k_1+k_2+2\sqrt{k_1k_2}
,
\end{align}
where 
\begin{align} 
k_1&=1/r_1
,\\
k_2&=1/r_2
,\\
k_3&=1/r_3
,\\
k_4&=0
,
\end{align} 
since we can consider a line $DC$
as the fourth circle with zero curvature (infinite radius).
Thus, given that you've found values for $r_1,r_2$ ,
\begin{align} 
r_3&=
\frac{r_1\,r_2}{r_1+r_2+2\sqrt{r_1\,r_2}}
.
\end{align}
