
1. Find the least possible area of a triangle ABC whose altitudes satisfy the inequalities $h_{a} \geqslant 3 \mathrm{~cm}, h_{b} \geqslant 4 \mathrm{~cm}$ and $h_{c} \geqslant 5 \mathrm{~cm}$.


Solution. Denote by $a, b, c$ the lengths of the sides of the triangle $A B C$. Its altitude $h_{b}$ satisfies the inequality

$$
c \geqslant h_{b}
$$

since $h_{b}$ is the length of the shortest segment connecting the vertex $B$ with a point on the line $A C$. The area $S$ of the triangle $A B C$ therefore satisfies

$$
S=\frac{c h_{c}}{2} \geqslant \frac{h_{b} h_{c}}{2} \geqslant 10 \mathrm{~cm}^{2}
$$

If there exists a triangle $A B C$ satisfying the conditions of the problem whose area is exactly $10 \mathrm{~cm}^{2}$, then both inequalities $S=\frac{1}{2} c h_{c} \geqslant \frac{1}{2} h_{b} h_{c} \geqslant 10 \mathrm{~cm}^{2}$ must become equalities. This means that $c=h_{b}=4 \mathrm{~cm}$ and at the same time $h_{c}=5 \mathrm{~cm}$. The first equality means that the triangle is right, with the right angle at the vertex $A$. The length of its cathetus $A C$ then satisfies $b=h_{c}=5 \mathrm{~cm}$, while the length $A$ of its hypotenuse $B C$ equals $\sqrt{41} \mathrm{~cm}$. From the formula $S=\frac{1}{2} a h_{a}$ we obtain for the altitude $h_{a}$

$$
h_{a}=\frac{2 S}{a}=\frac{20}{\sqrt{41}} \mathrm{~cm}>3 \mathrm{~cm}
$$

This means that the right triangle $A B C$ with catheti of lengths $b=5 \mathrm{~cm}$ and $c=4 \mathrm{~cm}$ satisfies the conditions of the problem.

The least possible area of the triangle $A B C$ whose altitudes have the requested properties is thus $10 \mathrm{~cm}^{2}$.
