If $x$ is a positive integer, then the

$$
10 x^{2}+5 x, \quad 8 x^{2}+4 x+1, \quad 6 x^{2}+7 x+1
$$

are the lengths of the sides of a triangle, all of whose angles are less than $90^{\circ}$, and whose area is an integer.
Let $10 x^{2}+5 x=a, 8 x^{2}+4 x+1=b, 6 x^{2}+7 x+1=c$, then

$$
\begin{gathered}
s=12 x^{2}+8 x+1=(6 x+1)(2 x+1) \\
s-a=2 x^{2}+3 x+1=(2 x+1)(x+1) \\
s-b=4 x^{2}+4 x=4 x(x+1) \\
s-c=6 x^{2}+x=x(6 x+1)
\end{gathered}
$$

and

$$
\begin{gathered}
\tan \frac{\alpha}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}=\frac{2 x}{2 x+1} \\
\tan \frac{\beta}{2}=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}=\frac{1}{2} \\
\tan \frac{\gamma}{2}=\sqrt{\frac{(s-b)(s-a)}{s(s-c)}}=\frac{2(x+1)}{6 x+1}
\end{gathered}
$$

If $x$ is a positive integer, then each of these fractions is less than $1$, so

$$
\frac{\alpha}{2}<45^{\circ}, \quad \frac{\beta}{2}<45^{\circ}, \quad \frac{\gamma}{2}<45^{\circ}
$$

thus

$$
\alpha<90^{\circ}, \quad \beta<90^{\circ}, \quad \gamma<90^{\circ} .
$$

The area of the triangle

$$
t=\sqrt{s(s-a)(s-b)(s-c)}=2 x(x+1)(2 x+1)(6 x+1)
$$

is an integer expression.

(Kornél Miklóssy, Arad.)