Difference between revisions of "2020 AMC 10B Problems/Problem 21"

Latest revision as of 16:52, 3 November 2024

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

1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4 (HARD Calculation)
6 Solution 5 (Basically Same as Solution 3)
7 Solution 6
8 Solution 7 (Easy to See)
9 Solution 8
10 Video Solution (HOW TO THINK CREATIVELY!!!)
11 Video Solution by MathEx
12 Video Solution by TheBeautyOfMath
13 Really Good Vid Explanation
14 See Also

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 $FI=AC-AL-\frac{m}{\sqrt{2}}=2\sqrt{2}-1-\frac{\sqrt{2}}{2} \cdot \frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}$ $=2\sqrt{2}-1-\frac{8\sqrt{2}-4-\sqrt{128-64\sqrt2}}{4}$ $=\frac{\sqrt{128-64\sqrt{2}}}{4}$ . 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 $AH$ = $AE$ = $\sqrt{2}$ , $EB$ = $2-\sqrt{2}$ , $EK$ = $2\sqrt{2}-2$ .

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

--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 $x = NC$ , which means $FG=2x$ . The area of the pentagon is then $[FIJG]+[GCF]=GF \cdot FI + x^2 = (2x)(2\sqrt{2}-1-x)+x^2=1$ 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

Solution 8

We extend $\overline{FB}$ and $\overline{IE}$ to meet at a point $X$ . Since $\angle AEI = \angle BEX$ and $FX$ is parallel to $DA$ , we know that $\triangle{BEX} \sim \triangle{AEH}$ , and because $\angle BXE = \angle IXF$ and $\angle XBE = \angle XIF$ , we can conclude that $\triangle{BEX} \sim \triangle{AEH} \sim \triangle{IFX}$ .

Now, because $\triangle{AEH}$ is isosceles, right, and has an area of 1, we can conclude that $AE = AH = \sqrt{2}$ and that $BE = 2-\sqrt{2}$ . Armed with this knowledge, and setting $IF = a$ and the area of $\triangle{BEX} = b$ , we can use similarity to say that $(\frac{a}{2-\sqrt{2}})^2 = \frac{1+b}{b}$ Since we know the side lengths of $\triangle{BEX}$ due to the fact that it is also an isosceles right triangle, we know that the area is $\frac{(2-\sqrt{2})^2}{2}$ . Simplifying further and plugging in values, we have $\frac{a^2}{(2-\sqrt{2})^2} = 1 + \frac{2}{(2-\sqrt{2})^2)}$ Multiplying by $(2-\sqrt{2})^2$ on both sides, we get $a^2 = (2-\sqrt{2})^2 + 2 = \boxed{\textbf{(B)}\ 8-4\sqrt{2}}$ ~yingkai_0_

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

@@ Line 61: / Line 61: @@
 == Solution 3 ==
-Plot a point <math>F'</math> such that <math>F'I</math> and <math>AB</math> are parallel and extend line <math>FB</math> to point <math>B'</math> such that <math>FIB'F'</math> forms a square. Extend line <math>AE</math> to meet line <math>F'B'</math> and point <math>E'</math> is the intersection of the two. The area of this square is equivalent to <math>FI^2</math>. We see that the area of square <math>ABCD</math> is <math>4</math>, meaning each side is of length 2. The area of the pentagon <math>EIFF'E'</math> is <math>2</math>. Length <math>AE=\sqrt{2}</math>, thus <math>EB=2-\sqrt{2}</math>. Triangle <math>EB'E'</math> is isosceles, and the area of this triangle is <math>\frac{1}{2}(4-2\sqrt{2})(2-\sqrt{2})=6-4\sqrt{2}</math>. Adding these two areas, we get <cmath>2+6-4\sqrt{2}=8-4\sqrt{2}\rightarrow \boxed{\mathrm{(B)}}</cmath>. --OGBooger
+Plot a point <math>F'</math> such that <math>F'I</math> and <math>AB</math> are parallel and extend line <math>FB</math> to point <math>B'</math> such that <math>FIB'F'</math> forms a square. Extend line <math>AE</math> to meet line <math>F'B'</math> and point <math>E'</math> is the intersection of the two. The area of this square is equivalent to <math>FI^2</math>. We see that the area of square <math>ABCD</math> is <math>4</math>, meaning each side is of length 2. The area of the pentagon <math>EIFF'E'</math> is <math>2</math>. Length <math>AE=\sqrt{2}</math>, thus <math>EB=2-\sqrt{2}</math>. Triangle <math>EB'E'</math> is isosceles, and the area of this triangle is <math>\frac{1}{2}(4-2\sqrt{2})(2-\sqrt{2})=6-4\sqrt{2}</math>. Adding these two areas, we get <cmath>2+6-4\sqrt{2}=8-4\sqrt{2}\rightarrow \boxed{\textbf{(B)}\ 8-4\sqrt{2}}</cmath>. --OGBooger
 == Solution 4 (HARD Calculation) ==
@@ Line 73: / Line 73: @@
 We solve the equation and yield <math>m=\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}</math>.
 Now notice that
-<math>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}</math>
+<math>FI=AC-AL-\frac{m}{\sqrt{2}}=2\sqrt{2}-1-\frac{\sqrt{2}}{2} \cdot \frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}</math>
 <math>=2\sqrt{2}-1-\frac{8\sqrt{2}-4-\sqrt{128-64\sqrt2}}{4}</math>
 <math>=\frac{\sqrt{128-64\sqrt{2}}}{4}</math>.
 Hence <math>FI^2=\frac{128-64\sqrt{2}}{16}=8-4\sqrt{2}</math>.  -HarryW
--edit: annabelle0913
+== Solution 5 (Basically Same as Solution 3)==
-== Solution 6 ==
 <asy>
@@ Line 133: / Line 131: @@
 Easily, we can find that: quadrilateral <math>BFIE</math> and <math>DHJG</math> are congruent with each other, so we can move <math>DHJG</math> to the shaded area (<math>F</math> and <math>G</math>, <math>B</math> and <math>D</math> overlapping) to form a square <math>FIKJ'</math> (<math>DG</math> = <math>FB</math>, <math>CG</math> = <math>FC</math>, <math>{\angle} CGF</math> = <math>{\angle}CFG</math> = <math>45^{\circ}</math> so <math>{\angle} IFJ'= 90^{\circ}</math>). Then we can solve <math>AH</math> = <math>AE</math> = <math>\sqrt{2}</math>, <math>EB</math> = <math>2-\sqrt{2}</math>, <math>EK</math> = <math>2\sqrt{2}-2</math>.
-<math>FI^2</math> = <math>area</math> of  <math>BFIE</math> <math>+</math> <math>area</math> of <math>FJ'H'B</math> <math>+</math> <math>area</math> of <math>EH'K</math> = <math>1 + 1 + \frac{1}{2}(2\sqrt{2}-2)^2=8-4\sqrt{2}\rightarrow \boxed{\mathrm{(B)}}</math>
+<math>FI^2=\text{area of} \: BFIE+\text{area of} \:FJ'H'B+\text{area of} \:EH'K \\= 1 + 1 + \frac{1}{2}(2\sqrt{2}-2)^2=8-4\sqrt{2}\rightarrow \boxed{\mathrm{(B)}}</math>
 --Ryan Zhang @BRS
-==Video Solution 1==
+== Solution 6 ==
-https://www.youtube.com/watch?v=AKJXB07Sat0&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=7 ~ MathEx
+<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));
-==Video Solution 2 by the Beauty of Math==
+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>
+<math>[ABCD] = 4</math>, <math>AB = 2</math>, <math>[AHE] = 1</math>, <math>AH = AE = \sqrt{2}</math>, <math>DH = 2 - \sqrt{2}</math>, <math>JL = HK = \sqrt{2} \cdot DH = 2 \sqrt{2} - 2</math>
+Because <math>ABCD</math> is a square and <math>AH = AE</math>, <math>AC</math> is the line of symmetry of pentagon <math>CDHEB</math>. Because <math>[DHJG] = [BFIE]</math>, <math>DHJG</math> is the reflection of <math>BFIE</math> about line <math>AC</math>
+Let <math>FI = GJ = x</math>, <math>KL = LG = GJ - LJ = x - 2 \sqrt{2} + 2</math>
+<math>[DHK] = \frac{(2 - \sqrt{2})^2}{2} = 3 - 2 \sqrt {2}</math>
+<math>[GKL] = \frac{(x - 2 \sqrt{2} + 2)^2}{2} = \frac{x^2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6</math>
+<math>[HKJL] = (x - 2 \sqrt{2} + 2) \cdot (2 \sqrt{2} - 2) = 2x \sqrt{2} - 2x + 8 \sqrt{2} -12</math>
+<cmath>[DHK] + [GKL] + [HKLJ] = [DHJG]</cmath>
+<cmath>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</cmath>
+<cmath>\frac{x^2}{2} + 2 \sqrt{2} - 4 = 0</cmath>
+<cmath>x^2 = 8 - 4 \sqrt{2}</cmath>
+<cmath>FI^2 = \boxed{\textbf{(B)}\ 8-4\sqrt{2}}</cmath>
+~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
+== Solution 7 (Easy to See) ==
+Note that the side length of <math>ABCD</math> is 2 and thus the diagonal is of length <math>2\sqrt{2}</math>. However, the height to side <math>HE</math> in triangle <math>HAE</math> is 1, implying that <math>CM = 2\sqrt{2}-1</math> where <math>M</math> is the midpoint of <math>JI</math>. From here suppose that <math>N</math> is the midpoint of <math>\overline{FG}</math> and let <math>x = NC</math>, which means <math>FG=2x</math>. The area of the pentagon is then
+<cmath>[FIJG]+[GCF]=GF \cdot FI + x^2 = (2x)(2\sqrt{2}-1-x)+x^2=1</cmath>
+Solving this quadratic for <math>x</math> yields <math>x=2\sqrt{2}-1 \pm \sqrt{8-4\sqrt{2}}</math> (technically the smaller value is the correct one but it doesn’t matter for our purposes). We can then calculate <math>FI^2 = (2\sqrt{2} -1 -x)^2 = \boxed{\textbf{(B) } 8-4\sqrt{2}}</math>.
+~Dhillonr25
+== Solution 8 ==
+We extend <math>\overline{FB}</math> and <math>\overline{IE}</math> to meet at a point <math>X</math>. Since <math>\angle AEI = \angle BEX</math> and <math>FX</math> is parallel to <math>DA</math>, we know that <math>\triangle{BEX} \sim \triangle{AEH}</math>, and because <math>\angle BXE = \angle IXF</math> and <math>\angle XBE = \angle XIF</math>, we can conclude that <math>\triangle{BEX} \sim \triangle{AEH} \sim \triangle{IFX}</math>.
+Now, because <math>\triangle{AEH}</math> is isosceles, right, and has an area of 1, we can conclude that <math>AE = AH = \sqrt{2}</math> and that <math>BE = 2-\sqrt{2}</math>. Armed with this knowledge, and setting <math>IF = a</math> and the area of <math>\triangle{BEX} = b</math>, we can use similarity to say that
+<cmath>(\frac{a}{2-\sqrt{2}})^2 = \frac{1+b}{b}</cmath>
+Since we know the side lengths of <math>\triangle{BEX}</math> due to the fact that it is also an isosceles right triangle, we know that the area is <math>\frac{(2-\sqrt{2})^2}{2}</math>.
+Simplifying further and plugging in values, we have
+<cmath>\frac{a^2}{(2-\sqrt{2})^2} = 1 + \frac{2}{(2-\sqrt{2})^2)}</cmath>
+Multiplying by <math>(2-\sqrt{2})^2</math> on both sides, we get
+<cmath>a^2 = (2-\sqrt{2})^2 + 2 = \boxed{\textbf{(B)}\ 8-4\sqrt{2}}</cmath>
+~yingkai_0_
+==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
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All AMC 12 Problems and Solutions

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