, 2018

Solution:

Without loss of generality, suppose that $AB$ is the hypotenuse.

If $F$ is the right angle, then $F$ must be the midpoint of $AB$. To prove this, let $X$ and $Y$ be the feet from $F$ to $BC$ and $AC$. Since $\angle XFY = \angle DFE = 90^{\circ}$, we have $\angle XFD = \angle YFE$ so
$$XF = DF \cos \angle XFD = EF \cos \angle YFE = YF$$
Hence $F$ is equidistant from $AC$ and $BC$ so it is the midpoint of $AB$. Then the minimum area is achieved by minimizing $DF$; this occurs when $DF$ is perpendicular to $BC$. The triangle $DEF$ then becomes the medial triangle of $ABC$, so its area is $\frac{1}{4}$.

If $F$ is not the right angle, without loss of generality, let the right angle be $D$. Place this triangle in the complex plane such that $C$ is the origin, $B = \sqrt{2}$, and $A = \sqrt{2}i$.

Now since $D$ is on the real axis and $E$ is on the imaginary axis, $D = x$ and $E = yi$, and we can obtain $F$ by a $90$ degree counterclockwise rotation of $D$ around $E$: this evaluates to $F = y + (x + y)i$. For $F$ to be on $AB$, the only constraint is to have $y + (x + y) = \sqrt{2} \Longrightarrow x = \sqrt{2} - 2y$.

To minimize the area, we minimize
$$\frac{DE^2}{2} = \frac{x^2 + y^2}{2} = \frac{(\sqrt{2} - 2y)^2 + y^2}{2} = \frac{5y^2 - 4\sqrt{2}y + 2}{2}$$
which has a minimum of $\frac{1}{5}$ at $y = \frac{2\sqrt{2}}{5}$. Since this is between $0$ and $\sqrt{2}$, this is indeed a valid configuration.

Finally, we take the smallest area of $\frac{1}{5}$.