, 2024

Proof. Let the right triangle be $\triangle ABC$ with $\angle C = 90^\circ$. In this solution, the interior of a triangle or a circle includes its boundary.
Let $\omega_1$ and $\omega_2$ be the circumferences of two closed disks with radius $1$, such that every point inside $\triangle ABC$ lies in the interior of $\omega_1$ or in the interior of $\omega_2$. The distance between any two points within the same disk does not exceed $2$.
Consider the assignment of the three vertices $A$, $B$, and $C$.
If $A$ and $B$ lie in the same disk, then the length of the hypotenuse $|AB| \le 2$, and the area $S \le \frac{1}{4}|AB|^2 = 1$.
If $A$ and $B$ do not lie in the same disk, assume $A$ lies inside $\omega_1$ and $B$ lies inside $\omega_2$. Without loss of generality, let the right-angled vertex $C$ lie inside $\omega_1$. Since $B$ lies outside $\omega_1$, the line segment $CB$ intersects the circumference $\omega_1$ at a unique point $M$. The point $M$ divides $CB$ into two segments $CM$ and $MB$, where $CM$ lies inside $\omega_1$, and $MB$ (excluding the endpoint $M$) lies outside $\omega_1$ and must therefore lie inside $\omega_2$. Consequently, the endpoint $M$ also lies inside $\omega_2$.
Since both $M$ and $A$ lie inside $\omega_1$, we have $|MA| \le 2$; similarly, since both $M$ and $B$ lie inside $\omega_2$, we have $|MB| \le 2$.

On the other hand, let $N$ be the foot of the perpendicular from $M$ to $AB$. The circle with diameter $MA$ covers $\triangle BMN$, and the circle with diameter $MB$ covers the quadrilateral $ACMN$. Thus, these two disks together cover $\triangle ABC$. Therefore, the problem reduces to finding a point $M$ on the side $BC$ such that $|MA|, |MB| \le 2$.
Let $|MC| = x$ and $|AC| = y$. Then $x^2 + y^2 = |MA|^2 \le 4$, and
$$2S = |BC| \cdot |AC| \le (2+x)y \le (2+x)\sqrt{4-x^2}.$$
$$4S^2 \le (2+x)^2(4-x^2) = (2+x)^3(2-x) = 27 \times \left(\frac{2+x}{3}\right)^3 (2-x) \le 27 \times \left(\frac{4}{4}\right)^4 = 27.$$

Thus, $S \le \frac{3\sqrt{3}}{2}$. Equality holds when $x = 1$ and $y = \sqrt{3}$, corresponding to the side lengths $AC = \sqrt{3}$, $BC = 3$, and $AB = 2\sqrt{3}$, with the point $M$ on $BC$ satisfying $MC = 1$ and $MB = MA = 2$. In this case, $\triangle ABC$ can be covered by two closed disks of radius $1$ (i.e., the disks with diameters $MB$ and $MA$), fulfilling the requirement.
Therefore, the maximum possible area of the right triangle is $\frac{3\sqrt{3}}{2}$. $\square$