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2020 AMC 10B Problems/Problem 21

Revision as of 15:47, 28 October 2023 by Scinderella220 (talk | contribs) (→‎Solution 5)
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 its area to be $3-2\sqrt{2}$. Now, we notice that $FIK$ is also a right isosceles triangle (because $\angle EKB=45^\circ$) 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

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 3

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 \[2+6-4\sqrt{2}=8-4\sqrt{2}\rightarrow \boxed{\textbf{(B)}\ 8-4\sqrt{2}}\]. --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 $CG=CF=m$, 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

Solution 5 (Basically Same as Solution 3)

[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

Solution 6

[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, K, L; 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); K = (4-x,4); L = J + 1.68 * dir(45); draw(A--B--C--D--A); draw(H--E); draw(J--G^^F--I); draw(H--K,dashed+linewidth(.5)); draw(L--K,dashed+linewidth(.5)); draw(rightanglemark(G, J, I), linewidth(.5)); draw(rightanglemark(F, I, E), linewidth(.5)); draw(rightanglemark(H, K, L), linewidth(.5)); draw(rightanglemark(K, L, G), 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); dot("$K$", K, N); dot("$L$", L, S); [/asy]

$[ABCD] = 4$, $AB = 2$, $[AHE] = 1$, $AH = AE = \sqrt{2}$, $DH = 2 - \sqrt{2}$, $JL = HK = \sqrt{2} \cdot DH = 2 \sqrt{2} - 2$

Because $ABCD$ is a square and $AH = AE$, $AC$ is the line of symmetry of pentagon $CDHEB$. Because $[DHJG] = [BFIE]$, $DHJG$ is the reflection of $BFIE$ about line $AC$

Let $FI = GJ = x$, $KL = LG = GJ - LJ = x - 2 \sqrt{2} + 2$

$[DHK] = \frac{(2 - \sqrt{2})^2}{2} = 3 - 2 \sqrt {2}$

$[GKL] = \frac{(x - 2 \sqrt{2} + 2)^2}{2} = \frac{x^2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6$

$[HKJL] = (x - 2 \sqrt{2} + 2) \cdot (2 \sqrt{2} - 2) = 2x \sqrt{2} - 2x + 8 \sqrt{2} -12$

\[[DHK] + [GKL] + [HKLJ] = [DHJG]\]

\[3 - 2 \sqrt {2} + \frac{x^2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6 + 2x \sqrt{2} - 2x + 8 \sqrt{2} -12= 1\]

\[\frac{x^2}{2} + 2 \sqrt{2} - 4 = 0\]

\[x^2 = 8 - 4 \sqrt{2}\]

\[FI^2 = \boxed{\textbf{(B)}\ 8-4\sqrt{2}}\]

~isabelchen

Solution 7 (Easy to See)

Note that the side length of $ABCD$ is 2 and thus the diagonal is of length $2\sqrt{2}$. However, the height to side $HE$ in triangle $HAE$ is 1, implying that $CM = 2\sqrt{2}-1$ where $M$ is the midpoint of $JI$. From here suppose that $N$ is the midpoint of $\overline{FG}$ and let $x = NC$. The area of the pentagon is then \[2 \cdot FI + x^2 = 2(2\sqrt{2}-1-x)+x^2=1\] Solving this quadratic for $x$ yields $x=2\sqrt{2}-1 \pm \sqrt{8-4\sqrt{2}}$ (technically the smaller value is the correct one but it doesn’t matter for our purposes). We can then calculate $FI^2 = (2\sqrt{2} -1 -x)^2 = \boxed{\textbf{(B) } 8-4\sqrt{2}}$.

~Dhillonr25

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/oRvHHywcw4w

~Education, the Study of Everything

Video Solution by MathEx

https://www.youtube.com/watch?v=AKJXB07Sat0

Video Solution by TheBeautyOfMath

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

Really Good Vid Explanation

https://www.youtube.com/watch?v=AUndgrOH8U8&ab_channel=ReachTheStars

See Also

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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