## Task 18/79

The measures of the inradius and the sides of a triangle are to be integral members of an arithmetic sequence of the first order. How large are they if they are as small as possible?
Assume there is a triangle where the measures of the inradius $r$ and the sides $a, b, c$ are consecutive terms of an arithmetic sequence of the first order. Then, the inradius is certainly the smallest term (the inradius is smaller than any side).

W.L.O.G., let $a < b < c$. (Here and in the following, $r, a, b, c$ denote the measures of the corresponding segments.)

Then,

$$
a = r + d \quad ; \quad b = r + 2d \quad ; \quad c = r + 3d
$$

where $d$ is the difference of the arithmetic sequence. It is known that for the inradius,

$$
r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}} \quad \text{with} \quad s = \frac{a+b+c}{2}
$$

thus,

$$
\begin{aligned}
s = \frac{3r + 6d}{2} & = \frac{3}{2}(r + 2d) & ; & s - a = \frac{1}{2}(r + 4d) \\
s - b & = \frac{1}{2}(r + 2d) & ; & s - c = \frac{1}{2}r
\end{aligned}
$$

Therefore,

$$
r = \sqrt{\frac{\frac{r + 4d}{2} \cdot \frac{r + 2d}{2} \cdot \frac{r}{2}}{3 \frac{r + 2d}{2}}} = \frac{1}{2} \sqrt{\frac{r^2 + 4rd}{3}}
$$

and $12r^2 = r^2 + 4rd, \frac{11}{4}r = d$.

For $a, b,$ and $c$ to be integers when $r$ is an integer, it is necessary and sufficient that $d$ is an integer. This is the case if and only if $r \equiv 0 \pmod{4}$. The smallest possible $r$ is then $r = 4$, for which $d = 11$, and for the sides, $a = 15, b = 26, c = 37$. Indeed, these values also satisfy the triangle inequalities.