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$", J, SW); [/asy]$

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

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 (Lucky Measuring)

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) This cannot work if the problem says not to scale - awu2014

Note that this will only work if the diagram is to scale, and at the start of the test, they mention that all diagrams are not necessarily to scale (whether or not the problem states that). Therefore, if you are to use this strategy on a problem, you are betting on the fact that this diagram IS to scale, so only use it as a last resort.

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

Solution 6

$[asy] real x=2sqrt(2); real y=2sqrt(16-8sqrt(2))-4+2sqrt(2); real z=2sqrt(8-4sqrt(2)); real k= 8-2sqrt(2); real l= 2sqrt(2)-4; pair A, B, C, D, E, F, G, H, I, J, L, M, K; A = (0,0); B = (4,0); C = (4,4); D = (0,4); E = (x,0); F = (4,y); G = (y,4); H = (0,x); I = F + z * dir(225); J = G + z * dir(225); L = (k,0); M = F + z * dir(315); K = (4,l); draw(A--B--C--D--A); draw(H--E); draw(J--G^^F--I); draw(F--M); draw(M--L); draw(E--K,dashed+linewidth(.5)); draw(K--L,dashed+linewidth(.5)); draw(B--L); draw(rightanglemark(G, J, I), linewidth(.5)); draw(rightanglemark(F, I, E), linewidth(.5)); draw(rightanglemark(F, M, L), linewidth(.5)); fill((4,0)--(k,0)--M--(4,y)--cycle, gray); dot("$A$", A, S); dot("$C$", C, dir(90)); dot("$D$", D, dir(90)); dot("$E$", E, S); dot("$G$", G, N); dot("$H$", H, W); dot("$I$", I, SW); dot("$J$", J, SW); dot("$K$", K, S); dot("$F(G)$", F, E); dot("$J'$", M, dir(90)); dot("$H'$", L, S); dot("$B(D)$", B, S); [/asy]$ Easily, we can find that: quadrilateral $BFIE$ and $DHJG$ are congruent with each other, so we can move $DHJG$ to the shaded area ( $F$ and $G$ , $B$ and $D$ overlapping) to form a square $FIKJ'$ ( $DG$ = $FB$ , $CG$ = $FC$ , ${\angle} CGF$ = ${\angle}CFG$ = $45^{\circ}$ so ${\angle} IFJ'= 90^{\circ}$ ). Then we can solve $AH$ = $AE$ = $\sqrt{2}$ , $EB$ = $2-\sqrt{2}$ , $EK$ = $2\sqrt{2}-2$ .

$FI^2$ = $area$ of $BFIE$ $+$ $area$ of $FJ'H'B$ $+$ $area$ of $EH'K$ = $1 + 1 + \frac{1}{2}(2\sqrt{2}-2)^2=8-4\sqrt{2}\rightarrow \boxed{\mathrm{(B)}}$

--Ryan Zhang @BRS

Video Solution 1

https://www.youtube.com/watch?v=AKJXB07Sat0&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=7 ~ MathEx

Video Solution 2 by the Beauty of Math

https://youtu.be/VZYe3Hu88OA?t=189

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

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