# 2020 AMC 10B Problems/Problem 21

The following problem is from both the 2020 AMC 10B #21 and 2020 AMC 12B #18, so both problems redirect to this page.

## Problem

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 $FI^2$?

$[asy] real x=2sqrt(2); real y=2sqrt(16-8sqrt(2))-4+2sqrt(2); real z=2sqrt(8-4sqrt(2)); pair A, B, C, D, E, F, G, H, I, J; A = (0,0); B = (4,0); C = (4,4); D = (0,4); E = (x,0); F = (4,y); G = (y,4); H = (0,x); I = F + z * dir(225); J = G + z * dir(225); draw(A--B--C--D--A); draw(H--E); draw(J--G^^F--I); draw(rightanglemark(G, J, I), linewidth(.5)); draw(rightanglemark(F, I, E), linewidth(.5)); dot("A", A, S); dot("B", B, S); dot("C", C, dir(90)); dot("D", D, dir(90)); dot("E", E, S); dot("F", F, dir(0)); dot("G", G, N); dot("H", H, W); dot("I", I, SW); dot("J", J, SW); [/asy]$

$\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2$

## Solutions

### Solution 1

Since the total area is $4$, the side length of square $ABCD$ is $2$. We see that since triangle $HAE$ is a right isosceles triangle with area 1, we can determine sides $HA$ and $AE$ both to be $\sqrt{2}$. Now, consider extending $FB$ and $IE$ until they intersect. Let the point of intersection be $K$. We note that $EBK$ is also a right isosceles triangle with side $2-\sqrt{2}$ and find it's area to be $3-2\sqrt{2}$. Now, we notice that $FIK$ is also a right isosceles triangle and find it's area to be $\frac{1}{2}$$FI^2$. This is also equal to $1+3-2\sqrt{2}$ or $4-2\sqrt{2}$. Since we are looking for $FI^2$, we want two times this. That gives $\boxed{\textbf{(B)}\ 8-4\sqrt{2}}$.~TLiu

### Solution 2

Since this is a geometry problem involving sides, and we know that $HE$ is $2$, we can use our ruler and find the ratio between $FI$ and $HE$. Measuring(on the booklet), we get that $HE$ is about $1.8$ inches and $FI$ is about $1.4$ inches. Thus, we can then multiply the length of $HE$ by the ratio of $\frac{1.4}{1.8},$ of which we then get $FI= \frac{14}{9}.$ We take the square of that and get $\frac{196}{81},$ and the closest answer to that is $\boxed{\textbf{(B)}\ 8-4\sqrt{2}}$. ~Celloboy (Note that this is just a strategy I happened to use that worked. Do not press your luck with this strategy, for it was a lucky guess)

### Solution 3

Draw the auxiliary line $AC$. Denote by $M$ the point it intersects with $HE$, and by $N$ the point it intersects with $GF$. Last, denote by $x$ the segment $FN$, and by $y$ the segment $FI$. We will find two equations for $x$ and $y$, and then solve for $y^2$.

Since the overall area of $ABCD$ is $4 \;\; \Longrightarrow \;\; AB=2$, and $AC=2\sqrt{2}$. In addition, the area of $\bigtriangleup AME = \frac{1}{2} \;\; \Longrightarrow \;\; AM=1$.

The two equations for $x$ and $y$ are then:

$\bullet$ Length of $AC$: $1+y+x = 2\sqrt{2} \;\; \Longrightarrow \;\; x = (2\sqrt{2}-1) - y$

$\bullet$ Area of CMIF: $\frac{1}{2}x^2+xy = \frac{1}{2} \;\; \Longrightarrow \;\; x(x+2y)=1$.

Substituting the first into the second, yields $\left[\left(2\sqrt{2}-1\right)-y\right]\cdot \left[\left(2\sqrt{2}-1\right)+y\right]=1$

Solving for $y^2$ gives $\boxed{\textbf{(B)}\ 8-4\sqrt{2}}$ ~DrB

### Solution 4

Plot a point $F'$ such that $F'I$ and $AB$ are parallel 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 pentagon $EIFF'E'$ is $2$. Length $AE=\sqrt{2}$, thus $EB=2-\sqrt{2}$. Triangle $EB'E'$ is isosceles, and the area of this triangle is $\frac{1}{2}(4-2\sqrt{2})(2-\sqrt{2})=6-4\sqrt{2}$. Adding these two areas, we get $$2+6-4\sqrt{2}=8-4\sqrt{2}\rightarrow \boxed{\mathrm{(B)}}$$. --OGBooger

### Solution 5 (HARD Calculation)

We can easily observe that the area of square $ABCD$ is 4 and its side length is 2 since all four regions that build up the square has area 1. Extend $FI$ and let the intersection with $AB$ be $K$. Connect $AC$, and let the intersection of $AC$ and $HE$ be $L$. Notice that since the area of triangle $AEH$ is 1 and $AE=AH$ , $AE=AH=\sqrt{2}$, therefore $BE=HD=2-\sqrt{2}$. Let $CG=CF=m$, then $BF=DG=2-m$. Also notice that $KB=2-m$, thus $KE=KB-BE=2-m-(2-\sqrt{2})=\sqrt{2}-m$. Now use the condition that the area of quadrilateral $BFIE$ is 1, we can set up the following equation: $\frac{1}{2}(2-m)^2-\frac{1}{4}(\sqrt{2}-m)^2=1$ We solve the equation and yield $m=\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}$. Now notice that $FI=AC-AL-\frac{m}{\sqrt{2}}=2\sqrt{2}-1-\frac{\sqrt{2}}{2}*\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}$ $=2\sqrt{2}-1-\frac{8\sqrt{2}-4-\sqrt{128-64\sqrt2}}{4}$ $=\frac{\sqrt{128-64\sqrt{2}}}{4}$. Hence $FI^2=\frac{128-64\sqrt{2}}{16}=8-4\sqrt{2}$. -HarryW

-edit: annabelle0913