A side of triangle $H_{1}$ is equal to the arithmetic mean of the other two sides. The sides of triangle $H_{2}$ are each 10 units smaller, and the sides of triangle $H_{3}$ are each 14 units larger than the sides of $H_{1}$. The radius of the inscribed circle of $H_{2}$ is 5 units smaller than that of $H_{1}$, and the radius of the inscribed circle of $H_{3}$ is 5 units larger. Calculate the sides of $H_{1}$.
In the triangle $H_{1}$, we denote the sides in the usual way such that $a \leq b \leq c$. According to the assumption, $b=(a+c) / 2$, which implies $b-a=c-b$. Denoting this non-negative difference by $d$, we have $a=b-d, c=b+d, 2s=a+b+c=3b$, and for the radius of the inscribed circle

\[
\begin{aligned}
& \varrho=\frac{t}{s}=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}=\sqrt{\frac{(b-2d)(b+2d)}{12}} \\
& 12 \varrho^{2}=b^{2}-4d^{2}
\end{aligned}
\]

The sides of $H_{2}$ and $H_{3}$ are obtained from those of $H_{1}$ by decreasing and increasing $b$ by the desired amount, while keeping $d$ unchanged. According to the problem's conditions,

\[
\begin{aligned}
& 12(\varrho-5)^{2}=(b-10)^{2}-4d^{2} \\
& 12(\varrho+5)^{2}=(b+14)^{2}-4d^{2}
\end{aligned}
\]

Expanding the parentheses in (2) and (3), and subtracting (1) from them,

\[
\begin{aligned}
-120 \varrho+300 & =-20b+100 \\
120 \varrho+300 & =28b+196
\end{aligned}
\]

From here, $b=38, \varrho=8$ follows, and substituting these into (1) (since $d \geq 0$) we get $d=13$. Thus, the sides of $H_{1}$ are $a=25, b=38$, $c=51$ units, the sides of $H_{2}$ are 15, 28, and 41, and the sides of $H_{3}$ are 39, 52, 65 (a right triangle); the radius of the inscribed circle in $H_{2}$ is 3, and in $H_{3}$ it is 13 units, meeting the requirements of the problem.

Lóránt (Pannonhalma, Benedekrendi g. IV. o. t.)