Question

In square $ABCD,$ points $E$ and $H$ lie on $\overline{AB}$ and $\overline{DA},$ respectively, so that $AE = AH.$ Points $F$ and $G$ lie on $\overline{BC}$ and $\overline{CD},$ respectively, and points $I$ and $J$ lie on $\overline{EH}$ so that $\overline{FI} \perp \overline{EH}$ and $\overline{GJ} \perp \overline{EH}.$ See the figure below. Triangle $AEH,$ quadrilateral $BFIE,$ quadrilateral $DHJG,$ and pentagon $FCGJI$ each has area $1.$ What is $FI^2?$

$\dfrac73$

$8 - 4\sqrt{2}$

$1 + \sqrt{2}$

$\dfrac74 \sqrt{2}$

$2\sqrt{2}$

Solution:

The four regions have total area $4,$ so the square has side $2.$ Put $A = (0, 0), B = (2, 0), C = (2, 2), D = (0, 2).$ Since $\triangle AEH$ is an isosceles right triangle with area $1,$ we get $AE = AH = \sqrt2,$ so $E = (\sqrt2, 0)$ and $H = (0, \sqrt2).$ Line $EH$ is $x + y = \sqrt2.$

Let $F = (2, t).$ Its perpendicular distance to line $EH$ is $FI = \tfrac{2 + t - \sqrt2}{\sqrt2}.$ Writing $s = FI/\sqrt2 = \tfrac{2 + t - \sqrt2}{2},$ the quadrilateral $BFIE$ has area $1.$ Solving this condition gives $s^2 = 4 - 2\sqrt2.$

Then $FI^2 = 2s^2 = 8 - 4\sqrt2.$

Thus, the correct answer is B.

2020 AMC 12B Problem 18

Solution:

Problem 18 in Other Years