Problem 3. In $\triangle A B C$ the lengths of the sides are natural numbers, and the length of the smallest side is $2 \mathrm{~cm}$. Find the area of $\triangle A B C$ if

$$
h_{c}=h_{a}+h_{b}
$$
Solution. Obviously $h_{c}$ is the largest height, so $c$ is the smallest side, i.e., $c=2 \text{ cm}$. From the formulas for the area of a triangle, we have

$$
h_{a}=\frac{2 P}{a}, h_{b}=\frac{2 P}{b}, h_{c}=\frac{2 P}{c}
$$

and if we substitute in the equation $h_{c}=h_{a}+h_{b}$, we get

$$
\frac{1}{c}=\frac{1}{a}+\frac{1}{b}
$$

and since $c=2 \text{ cm}$, we get $\frac{1}{2}=\frac{1}{a}+\frac{1}{b}$. The last equation is equivalent to the equation

$$
a b-2 a-2 b=0
$$

i.e., to the equation

$$
(a-2)(b-2)=4
$$

In the set of natural numbers, the last equation has solutions

$$
a=4, b=4 ; a=3, b=6 \text{ and } a=6, b=3 \text{. }
$$

Furthermore, $a, b, c$ are sides of a triangle, so $|a-b|<c$, which means $a=4 \text{ cm}, b=4 \text{ cm}$ and $c=2 \text{ cm}$.