What can be the maximum area of a triangle if the length of any of its sides is not greater than 2?
Solution. The area of the triangle does not decrease if we scale it so that the largest side is 2. Let the height corresponding to this side be $m$, which divides the side into segments of length $x$ and $2-x$. Since the largest angle of the triangle is opposite the largest side, the angles opposite the 2-unit side are acute, so $0 < x < 2$. By the Pythagorean theorem, the other two sides are $\sqrt{m^{2}+x^{2}} \leq 2$ and $\sqrt{m^{2}+(2-x)^{2}} \leq 2$.

![](https://cdn.mathpix.com/cropped/2024_05_02_bef846f3f910dc5b7eabg-1.jpg?height=247&width=366&top_left_y=375&top_left_x=869)

The triangle exists if these two inequalities, and the equivalent inequalities obtained by squaring and rearranging, $x^{2} \leq 4-m^{2}$ and $x^{2}-4 x+m^{2} \leq 0$, have a common solution for $x$. Solving these inequalities separately:

$$
0 < x \leq \sqrt{4-m^{2}} \quad \text{and} \quad 2-\sqrt{4-m^{2}} \leq x \leq 2+\sqrt{4-m^{2}}
$$

A common solution exists precisely when $2-\sqrt{4-m^{2}} \leq \sqrt{4-m^{2}}$, which simplifies to $2 \leq 2 \sqrt{4-m^{2}}, 1 \leq \sqrt{4-m^{2}}, 1 \leq 4-m^{2}$, or $m \leq \sqrt{3}$. Thus, the area of the triangle $t=\frac{1}{2} \cdot 2 m=m \leq \sqrt{3}$, and the maximum value of $\sqrt{3}$ is achieved when $m=\sqrt{3}$, $x=1$. In this case, the triangle is equilateral, with the other two sides also being 2 units long.