Minimal

My solution assumes we don't know the side lengths of the triangle; though, the heuristic approach for these kinds of problems would be setting the legs of the triangle equal to each other and simply solving for the area.

A well known property of triangles and inradii is that $A = rs$ where $A$ , $r$ , and $s$ are the area, the inradius, and the semi-perimeter of the triangle respectively. To minimize the area of the triangle, we must minimize its semi-perimeter.

The problem also specifies that the triangle is a right triangle. Let $a$ and $b$ be the length of the legs that triangle and $c$ be the length of the hypotenuse. These properties hold for right triangles:

$A = \frac{1}{2}ab$

$a^2 + b^2 = c^2$

Now we'll use the root-mean square and arithmetic mean inequality (Cauchy-Schwarz Inequality) on $a$ and $b$ .

$\sqrt{\frac{a^2+b^2}{2}} \geq \frac{a+b}{2} \Rightarrow \sqrt{2}c \geq a+b$

We know that $\sqrt{2}c$ is minimized when it is equal to $a+b$ .

$\sqrt{2}c = a+b$

$\rightarrow 2c^2 = a^2+b^2+2ab$

$\rightarrow a^2 - 2ab + b^2 = 0$

$\rightarrow (a-b)^2 = 0$

$\rightarrow a=b$

Setting both area equations equal to each other yields us this:

$A = \frac{1}{2}ab = \frac{1}{2}r(a+b+c)$

Now we just have do some substitution and solve for $a$ :

$\frac{1}{2}a^2 = \frac{1}{2}r(2a+\frac{2a}{\sqrt{2}}) \Rightarrow a = r(2+\sqrt{2})$

$A = \frac{1}{2}a^2 = \frac{1}{2}(r(2+\sqrt{2}))^2 = r^2(3+2\sqrt{2}) = \boxed{3+2\sqrt{2}}$

The area of ΔABC = ab/2 = (a+b+c)/2, Pythagoras : a^2 + b^2 = c^2 , and also c = a+b – 2 . So the area of ΔABC = a+b – 1 . To minimize A, we must minimize a+b. We know that a+b ≥ 2√ab, minimum when a=b. So a^2/2 = 2a – 1 , or a^2 – 4a + 2 = 0. a= 2 + √2, and the area of ΔABC = a^2/2 = 5.83.

Lets do a little thought experiment.

Let $a$ , $b$ and $c$ be the sides of the triangle with side $c$ being the hypotenuse and the inradius ( $r$ ) kept constant.

Now, the bigger the difference between $a$ and $b$ , the larger the area., eventually approaching infinity as $a$ or $b$ approaches $2r$ . Of course you can do a more mathematical method by using calculus, but this is RMO, there's time constrain. (I've never joined a maths competition but through my experience in physics competitions...)

So it becomes clear that the minimum is when $a=b$ . So do a little bit more calculations to obtain $\boxed{5.83...}$