2024 CGMO

Proof. Let the right triangle be $△ABC$ 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 $△ABC$ 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 $△BMN$, and the circle with diameter $MB$ covers the quadrilateral $ACMN$. Thus, these two disks together cover $△ABC$. 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, $△ABC$ 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 $\frac{3\sqrt{3}}{2}$. $\square$