2020 AMC 10B Problem 21

1. Side of the square. The four regions fill the square and each has area $1$, so the square has area $4$ and side $2$.
2. Triangle $AEH$. It has a right angle at $A$ and $AE = AH$, so it is a $45$-$45$-$90$ triangle (half of a square). Area $\tfrac12 \times AE \times AH = 1$ gives $AE = AH = \sqrt2$. Its long side is $EH = \sqrt2 \times \sqrt2 = 2$.
3. Use the diagonal $AC$. The figure is symmetric about the diagonal $AC$, and $EH$ crosses it at a right angle. The distance from $A$ to $EH$ is half of $EH$, which is $1$ (in a half-square triangle the height to the long side is half that side). The diagonal is $AC = 2\sqrt2$. So the distance from $C$ to line $EH$ is
$$h = 2\sqrt2 - 1.$$
4. Make a big triangle. Extend sides $CB$ and $CD$ past $F$ and $G$ until they hit line $EH$ at points $P$ and $Q$. Triangle $CPQ$ has a right angle at $C$ and is symmetric about $AC$, so it is also a half-square triangle. Its height from $C$ is $h$, so its long side is $PQ = 2h$ and its area is
$$\frac12 \times 2h \times h = h^2.$$
5. The pentagon is this triangle with two corners cut off. Corner triangle $FIP$ has a right angle at $I$ (since $FI \perp EH$) and a $45^\circ$ angle at $P$, because line $EH$ makes a $45^\circ$ angle with side $BC$. So it is a half-square triangle too, with $IP = FI$. Call $FI = x$. Its area is $\tfrac12 x^2$.
6. By the symmetry about $AC$, corner triangle $GJQ$ is the same size. (Quadrilaterals $BFIE$ and $DHJG$ have equal area, which forces $G$ to be the mirror image of $F$.)
7. Set up the equation.
$$\text{pentagon} = h^2 - \tfrac12 x^2 - \tfrac12 x^2 = h^2 - x^2 = 1.$$
8. Solve.
$$x^2 = h^2 - 1 = (2\sqrt2 - 1)^2 - 1 = (8 - 4\sqrt2 + 1) - 1 = 8 - 4\sqrt2.$$

The answer is $\boxed{\textbf{(B)}\ 8-4\sqrt2}$.

Everything is symmetric about the diagonal $AC$, and $EH$ is perpendicular to it, so every slanted line in the picture is at $45^\circ$ to the sides of the square. That makes every triangle in sight a half-square triangle, whose area depends on just one length. Instead of finding $F$ and $G$, grow the pentagon into a simple triangle and subtract the two corners. The unknown $FI$ then shows up on its own.

Coordinates also work, just more slowly. Put $A = (0,0)$ and $C = (2,2)$, so line $EH$ is $x + y = \sqrt2$. Put $F = (2, f)$ and, by symmetry, $G = (f, 2)$. Split the pentagon into rectangle $FIJG$ plus triangle $FCG$, set the total area to $1$, and solve for $f$. The distance from $F$ to line $EH$ is $FI = \frac{2 + f - \sqrt2}{\sqrt2}$, and squaring it again gives $8 - 4\sqrt2$.