10. Find the area of the triangle if it is known that the radius of the inscribed circle is 1, and the lengths of all three altitudes are expressed as integers.
Answer: $3 \sqrt{3}$.

Solution: Let $S-$ be the area, $a, b, c-$ be the sides of the triangle, and $h_{a}, h_{b}, h_{c}$ be the corresponding heights. Then $S=\frac{1}{2} a h_{a}=\frac{1}{2} b h_{b}=\frac{1}{2} c h_{c}$, from which we get $a=\frac{2 S}{h_{a}}, b=\frac{2 S}{h_{b}}$ and $c=\frac{2 S}{h_{c}}$. Substituting into the formula $S=p r=\frac{a+b+c}{2}$, we obtain $1=\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}$.

Without loss of generality, we can assume $h_{a} \leqslant h_{b} \leqslant h_{c}$. If we assume $h_{a}>3$, we get $\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}<\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$. Therefore, $h_{a} \leqslant 3$. Let's consider the cases:

(a) Case $h_{a}=1$. Then $\frac{1}{h_{b}}+\frac{1}{h_{c}}=0$, which is impossible.

(b) Case $h_{a}=2$. Then $\frac{1}{h_{b}}+\frac{1}{h_{c}}=\frac{1}{2}$. Then $a=\frac{2 S}{h_{a}}=S$ and $b+c=$ $\frac{2 S}{h_{b}}+\frac{2 S}{h_{c}}=2 S \cdot\left(\frac{1}{h_{b}}+\frac{1}{h_{c}}\right)=S$, which contradicts the triangle inequality.

(c) Case $h_{a}=3$. Then $\frac{1}{h_{b}}+\frac{1}{h_{c}}=\frac{2}{3}$, and $h_{b}, h_{c} \geqslant 3$, which is possible only in the case $h_{a}=h_{b}=h_{c}=3$. We get an equilateral triangle with side $a=b=c=2 \sqrt{3}$, the area of which is $3 \sqrt{3}$.