Prove a length of 6 in a triangle diagram. A puzzle:  Three equilateral triangles of size 3, 4, and 7 touch at a corner. The other corners of the size 4 triangle are 3 away from a 3 corner, and 7 away from a 7 corner. How far apart are the other 3 and 7 corners?
The answer is 6.  I have an unreasonably complicated proof.  Can anyone find an elegant proof?

The Propeller Theorem doesn't seem to help.
 A: I have a solution. It may not be very elegant, but it is doable and not too difficult. 
One of the equal angles of the triangle with sides $(7,7,4)$ is $\alpha=\cos^{-1}(2/7)$ and one of the equal angles of the triangle with sides $(3,3,4)$ is $\beta=\cos^{-1}(2/3)$. If $\gamma$ be the angle opposite to the unknown side of the triangle with sides $(7,3,a)(a$ unknown), then, $\gamma=180-(\alpha+\beta)$. Then, $$a^2=7^2+3^2-2\times7\times 3\cos \gamma=58+42\cos(\alpha+\beta)=$$ Now $\cos (\alpha+\beta)=2/7\times 2/3-15/21=-11/21\implies a^2=36\implies a=6$
A: Here's an overly-complicated answer compared to @Samrat's, but it speaks to some interesting underlying structure in the configuration:

A little angle chasing shows that $\angle APW \cong \angle ORY$; since the hypotenuses are congruent, in fact $\triangle APW \cong \triangle ORY$, so that $\overline{RY} \cong \overline{PW}$. Likewise, $\overline{SZ} \cong \overline{QX}$.
Importantly, a little more angle chasing shows that $\overline{AR}\parallel\overline{BS}$ (with common perpendicular $\overline{AB}$), and we can calculate as follows:
$$\begin{align}
|\overline{RS}|^2 &= |\overline{RT}|^2 + |\overline{ST}|^2 \\
&= |\overline{AB}|^2 + \left(\;|\overline{SZ}| - |\overline{RY}|\;\right)^2\\
&= c^2 + \left(\;|\overline{QX}| - |\overline{PW}|\;\right)^2 \\
&= c^2 + \left(\;\sqrt{b^2-\left(\frac{c}{2}\right)^2} - \sqrt{a^2-\left(\frac{c}{2}\right)^2}\;\right)^2 \qquad (\star)
\end{align}$$
At this point, we might as well substitute $a=3$, $b=7$, $c=4$ to conclude that $|\overline{RS}| = 6$.

When working this out, I had hoped to discern what's so special about the values $3$-$7$-$4$ that causes the result to be an integer. The fact that $a = b - c$ doesn't seem particularly key in itself. What appears to matter is that, while $a^2-(c/2)^2$ and $b^2-(c/2)^2$ are not themselves squares, they have the form $m^2 p$ and $n^2 p$ (specifically with $m = 1$, $n= 3$, $p = 5$ (Hmmmm...)) so that all of the radicals and fractions in the expanded form of $(\star)$ go away nicely ... and then magically combine with $c^2$ so as to give a perfect square.
So, the question becomes: What's so special about the values $1$-$3$-$5$ for $m$-$n$-$p$? For now, I'll leave that as an exercise for the reader.
