3. An acute triangle with sides that are consecutive natural numbers has an area that is an integer. Find at least one such triangle!
Solution. Let the sides of the triangle be $a-1$, $a$, and $a+1$. According to Heron's formula, we get

$$
\begin{gathered}
P=\sqrt{\frac{3}{2} a\left(\frac{1}{2} a+1\right) \frac{1}{2} a\left(\frac{1}{2} a-1\right)} \\
P=\frac{1}{4} a \sqrt{3(a-2)(a+2)}
\end{gathered}
$$

If $a$ is an odd number, then $a \sqrt{3(a-2)(a+2)}$ cannot be an even natural number, so $P$ cannot be a natural number. Therefore, $a$ must be an even number. Setting $a=2 k$, we get

$$
P=k \sqrt{3(k+1)(k-1)}.
$$

The value under the square root must be divisible by 9. Therefore,

$1^{\circ} k+1=3 m$, i.e., $k=3 m-1$, or

$2^{\circ} k-1=3 m$, i.e., $k=3 m+1$.

In the first case, we get

$$
P=3(3 m-1) \sqrt{m(3 m-2)}
$$

For $m=9$, we get $P=1170$ and the sides of the triangle are 51, 52, and 53, making it an acute-angled triangle.

In the second case, we get

$$
P=3(3 m+1) \sqrt{m(3 m+2)}
$$

For $m=2$, we get $P=84$ and the sides of the triangle are 13, 14, and 15, making it an acute-angled triangle.