Smallest inradius in a triangle

Solution 1:

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 $r$).

Solution 2:

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.