2020 AMC 10B Problem 21

The four named regions fill the square and each has area $1,$ so the square has area $4$ and side length $2.$ Since triangle $AEH$ is right isosceles with area $1,$ we have $AE=AH=\sqrt2.$ Extend $FI$ to meet $AB$ at $K,$ and set $x=BF$ and $t=BE=2-\sqrt2.$ Because $EH$ has slope $-1,$ line $FK$ has slope $1,$ so $BF=BK=x.$ If $K$ were on segment $EB,$ then region $BFIE$ would lie inside triangle $BFK,$ whose area would be at most $\frac{t^2}{2}<1,$ a contradiction. Thus $K$ lies to the left of $E,$ and $EK=x-t.$ Triangle $BFK$ is right isosceles with area $\frac{x^2}{2}.$ Triangle $EIK$ is right isosceles with hypotenuse $EK=x-t,$ so its area is $\frac{(x-t)^2}{4}.$ Since their difference is region $BFIE,$ $$1=\frac{x^2}{2}-\frac{(x-t)^2}{4}.$$ Therefore $$\begin{aligned} 4&=2x^2-(x-t)^2\\ &=(x+t)^2-2t^2. \end{aligned}$$ Also, $FK=x\sqrt2$ and $KI=\frac{x-t}{\sqrt2},$ so $$FI=FK-KI=\frac{x+t}{\sqrt2}.$$ It follows that $$\begin{aligned} FI^2&=\frac{(x+t)^2}{2}\\ &=2+t^2\\ &=2+(2-\sqrt2)^2\\ &=8-4\sqrt2. \end{aligned}$$ Thus, the correct answer is B .