2020 AMC 12B Problems/Problem 18

Solution

Plot a point $F'$ such that $F'$ and $I$ are collinear and extend line $FB$ to point $B'$ such that $FIB'F'$ forms a square. Extend line $AE$ to meet line F'B' and point $E'$ is the intersection of the two. The area of this square is equivalent to $FI^2$ . We see that the area of square $ABCD$ is $4$ , meaning each side is of length 2. The area of the quadrilateral $EIFF'E'$ is $2$ . Length $AE=\sqrt{2}$ , thus $EB=2-\sqrt{2}$ . Triangle $EB'E'$ is isosceles, and the area of this triangle is $\frac{1}{2}*(4-2\sqrt{2})*(2-\sqrt{2})=6-4sqrt(2)$ . Adding these two areas, we get $2+6-4\sqrt{2}=\boxed{\textbf{B) }8-4\sqrt{2}}

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2020 AMC 12B Problems/Problem 18

Solution