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 $FI^2?$

$\dfrac73$

$8-4\sqrt2$

$1+\sqrt2$

$\dfrac74\sqrt2$

$2\sqrt2$

Video solution:

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Written solution:

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 $K$ . Then $BE=2-\sqrt2$ , and triangle $BEK$ is right isosceles, so its area is $\frac12(2-\sqrt2)^2=3-2\sqrt2.$ The region $BFIE$ has area $1$ , so triangle $FIK$ has area $1+(3-2\sqrt2)=4-2\sqrt2.$ Triangle $FIK$ is also right isosceles, so its area is $\frac12 FI^2$ . Hence $\frac12 FI^2=4-2\sqrt2,\qquad FI^2=8-4\sqrt2.$

Thus, the correct answer is B .

2020 AMC 10B Problem 21

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Problem 21 in Other Years