2020 AMC 10B Problems/Problem 21

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$", J, SW); [/asy]$ $\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2$

Solution

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 $3-\sqrt{2}$ . 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 $F'$ 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 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 4 (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 $CG=GF=m$ , then $BF=DG=2-m$ . Also notice that $KB=BE=2-m$ , 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 $FI=AC-AL=2\sqrt{2}-1-\frac{\sqrt{2}}{2}*\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}$ $=2\sqrt{2}-1-\frac{8\sqrt{2}-4-\sqrt{128-64\sqrt2}}{4}$ $=\frac{\sqrt{128-64\sqrt{2}}}{4}$ . Hence $FI^2=\frac{128-64\sqrt{2}}{16}=8-4\sqrt{2}$ . -HarryW

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search