Smallest inradius in a triangle inradius http://sphotos-a.ak.fbcdn.net/hphotos-ak-ash4/603253_4699189150138_1902686882_n.jpg
Inside triangle ABC there are three circles with radius $r_1$, $r_2$, and $r_3$ each of which is tangent to two sides of the triangle and to its incircle with radius r. All of $r$, $r_1$, $r_2$, and $r_3$ are distinct perfect square integers. Find the smallest value of inradius $r$.
 A: For typographic convenience (and reduced visual clutter) in the following, I write "$A_2$", "$B_2$", "$C_2$" for the half-angles "$A/2$", "$B/2$", "$C/2$".

Let $P$ be the incenter of the triangle with radius $r =: s^2$; and let $P_1$, $P_2$, $P_3$ be the centers of the circles with respective radii $r_1 =: s_1^2$, $r_2 =: s_2^2$, $r_3 =: s_3^2$.
If $Q$ is the point of tangency of the incircle with $AB$, then $\triangle APQ$ has a right angle at $Q$, and we have $|AP| = \frac{r}{\sin A_2}$. With $Q_1$ the corresponding point of tangency creating $\triangle AP_1Q_1$, we have
$$\frac{r_1}{r} = \frac{|AP_1|}{|AP|} = \frac{|AP|-r-r_1}{|AP|} = \frac{r-(r+r_1)\sin A_2}{r}$$
so that
$$\sin A_2 = \frac{r-r_1}{r+r_1} = \frac{s^2 - s_1^2}{s^2+s_1^2}$$
Likewise,
$$\sin B_2 = \frac{s^2-s_2^2}{s+s_2^2} \qquad \sin C_2 = \frac{s^2-s_3^2}{s+s_3^2}$$
whence
$$\cos B_2 = \frac{2 s s_2}{s^2+s_2^2} \qquad \cos C_2 = \frac{2 s s_3}{s^2+s_3^2}$$
(Note: We know the cosines of the half-angles must be non-negative.)
Now, $A_2 + B_2 + C_2 = \pi_2$, so that
$$\begin{align}
\sin A_2 &= \cos(B_2+C_2) \\
\sin A_2 &= \cos B_2 \cos C_2 - \sin B_2 \sin C_2 \\
\frac{s^2-s_1^2}{s^2+s_1^2} &= \frac{4 s^2 s_2 s_3 - ( s^2 - s_2^2)( s^2 - s_3^2)}{(s^2+s_2^2)(s^2+s_3^2)}
\end{align}$$
Thus,
$$2 s^2 \left( s^2 - s_1 s_2 - s_2 s_3 - s_3 s_1 \right) \left( s^2 + s_1 s_2 - s_2 s_3 + s_3 s_1 \right) = 0$$
and we have
$$s^2 = s_1 s_2 + s_2 s_3 + s_3 s_1 \qquad \text{or} \qquad - s_1 s_2 + s_2 s_3 - s_3 s_1$$
As the latter option is clearly less than $\max\{s_1^2, s_2^2, s_3^2\}$, whereas $s^2$ must exceed that value (the incircle is bigger than the other three), we are left to solve the Diophantine equation
$$s^2 = s_1 s_2 + s_2 s_3 + s_3 s_1$$
in distinct positive integers dominated by $s$. One solution (found via Mathematica's FindInstance function) is
$$(s,s_1,s_2,s_3) = (9, 7, 6, 3)$$
which (approximately) corresponds to an $106.26^\circ$-$45.24^\circ$-$28.5^\circ$ triangle. No guarantees that this minimizes $s$ (and thus $r$).
A: I managed to obtain some relations between $r,r_1,r_2,r_3$ and $a,b,c$. I used only standard geometry tools and I won't give the details for now... 
The relations are: (I consider $r_1$ the circle which is not tangent to the side $a$, and the same for $r_2,b$ and $r_3,c$)
$$ p-a =2\sqrt{r_1r} \frac{r}{r-r_1}$$
(construct symilarly the other two)
From here you can construct a relatioin between $r,r_1,r_2,r_3$ in the following way:
$$ r=\frac{S}{p}=\sqrt{\frac{(p-a)(p-b)(p-c)}{p}} $$
and substituting we have
$$ r= \sqrt{\displaystyle \frac{\frac{8\sqrt{r_1r_2r_3}r^4\sqrt{r}}{(r-r_1)(r-r_2)(r-r_3)}}{2\sqrt{r_1r}\frac{r}{r-r_1}+2\sqrt{r_2r}\frac{r}{r-r_2}+2\sqrt{r_3r}\frac{r}{r-r_3}}}. $$
Here you may be able to use your hypothesis on $r,r_1,r_2,r_3$ to get something out of this.
