# Smallest inradius in a triangle

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 $r$.

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Nice problem. Where is it from? – Beni Bogosel Dec 27 '12 at 12:31
Saw it on a math group. :) – John Chang Dec 27 '12 at 12:32
Is there a requirement that the circle with radius $r$ is tangent to all three sides of the triangle? – Jay Dec 27 '12 at 12:33
It is the incircle so it must be tangent to all three sides. – John Chang Dec 27 '12 at 12:38
A hint for a solution method (I could try it, but can't right now): Notice that each small circle is homothetic to the incircle so you should be able to find the ratios $r/r_i$ in terms of $a,b,c$. After that I don't know what you could do. I guess you may need some assumptions on $a,b,c$, also. – Beni Bogosel Dec 27 '12 at 12:38

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.

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Is $p$ the semi-perimeter? – John Chang Dec 28 '12 at 3:06
At this point, why not just plug in $r_1=1$, $r_2=4$, $r_3=9$ and solve the resulting quadratic? I get $r=27/2$ (the other root is $r=1$, but we need to toss that because it's impossible). – Potato Dec 28 '12 at 5:31
Yes, $p$ is the semiperimeter. @Potato: I guess that's the way we should proceed. Plug in the smallest possible $r_1,r_2,r_3$ and find $r$. note that $r$ must also be a perfect square, so your result is not good. – Beni Bogosel Dec 28 '12 at 19:05
@BeniBogosel Ah, I didn't see the requirement that $r$ be a perfect square. Anyway, if you square and multiply out, the equation simplifies considerably, and you get a quadratic in $r$... – Potato Dec 28 '12 at 19:31

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 $r$).

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The equation AP -r - r1 – Edwin Gray Jun 27 '15 at 18:06