## Task $1 / 71$

Determine the triangle of maximum area that satisfies the conditions $0<a \leq 2 \leq b \leq 4<$ $c<5$.
We use the formula $A=\frac{1}{2} a b \sin \gamma$. Clearly, $A$ is maximized when both $a$ and $b$ and $\sin \gamma$ are maximized - assuming that $c$ then meets the required condition:

$$
A_{\max }=\frac{1}{2} a_{\max } b_{\max }(\sin \gamma)_{\max }=\frac{1}{2} \cdot 2 \cdot 4 \cdot \sin \frac{\pi}{2}
$$

Thus, $\gamma=\frac{\pi}{2}$, meaning the triangle must be a right triangle with $c$ as the hypotenuse. Therefore,

$$
c^{2}=a^{2}+b^{2}=4+16=20 \rightarrow c=2 \sqrt{5}
$$

Since $4<c=2 \sqrt{5}<5$ holds, the only solution to this problem is $a=2, b=4, c=2 \sqrt{5}$.