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

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{{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #21]] and [[2020 AMC 12B Problems|2020 AMC 12B #18]]}}
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==Problem==
 
==Problem==
 +
In square <math>ABCD</math>, points <math>E</math> and <math>H</math> lie on <math>\overline{AB}</math> and <math>\overline{DA}</math>, respectively, so that <math>AE=AH.</math> Points <math>F</math> and <math>G</math> lie on <math>\overline{BC}</math> and <math>\overline{CD}</math>, respectively, and points <math>I</math> and <math>J</math> lie on <math>\overline{EH}</math> so that <math>\overline{FI} \perp \overline{EH}</math> and <math>\overline{GJ} \perp \overline{EH}</math>. See the figure below. Triangle <math>AEH</math>, quadrilateral <math>BFIE</math>, quadrilateral <math>DHJG</math>, and pentagon <math>FCGJI</math> each has area <math>1.</math> What is <math>FI^2</math>?
  
In square <math>ABCD</math>, points <math>E</math> and <math>H</math> lie on <math>\overline{AB}</math> and <math>\overline{DA}</math>, respectively, so that <math>AE=AH.</math> Points <math>F</math> and <math>G</math> lie on <math>\overline{BC}</math> and <math>\overline{CD}</math>, respectively, and points <math>I</math> and <math>J</math> lie on <math>\overline{EH}</math> so that <math>\overline{FI} \perp \overline{EH}</math> and <math>\overline{GJ} \perp \overline{EH}</math>. See the figure below. Triangle <math>AEH</math>, quadrilateral <math>BFIE</math>, quadrilateral <math>DHJG</math>, and pentagon <math>FCGJI</math> each has area <math>1.</math> What is <math>FI^2</math>?
 
 
<asy>
 
<asy>
 
real x=2sqrt(2);
 
real x=2sqrt(2);
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</asy>
 
</asy>
 +
 
<math>\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</math>
 
<math>\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</math>
  
==Solution==
+
== Solution 1 ==
Since the total area is <math>4</math>, the side length of square <math>ABCD</math> is <math>2</math>. We see that since triangle <math>HAE</math> is a right isosceles triangle with area 1, we can determine sides <math>HA</math> and <math>AE</math> both to be <math>\sqrt{2}</math>. Now, consider extending <math>FB</math> and <math>IE</math> until they intersect. Let the point of intersection be <math>K</math>. We note that <math>EBK</math> is also a right isosceles triangle with side <math>2-\sqrt{2}</math> and find it's area to be <math>3-\sqrt{2}</math>. Now, we notice that <math>FIK</math> is also a right isosceles triangle and find it's area to be <math>\frac{1}{2}</math><math>FI^2</math>. This is also equal to <math>1+3-2\sqrt{2}</math> or <math>4-2\sqrt{2}</math>. Since we are looking for <math>FI^2</math>, we want two times this. That gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math>.~TLiu
+
Since the total area is <math>4</math>, the side length of square <math>ABCD</math> is <math>2</math>. We see that since triangle <math>HAE</math> is a right isosceles triangle with area 1, we can determine sides <math>HA</math> and <math>AE</math> both to be <math>\sqrt{2}</math>. Now, consider extending <math>FB</math> and <math>IE</math> until they intersect. Let the point of intersection be <math>K</math>. We note that <math>EBK</math> is also a right isosceles triangle with side <math>2-\sqrt{2}</math> and find its area to be <math>3-2\sqrt{2}</math>. Now, we notice that <math>FIK</math> is also a right isosceles triangle (because <math>\angle EKB=45^\circ</math>) and find it's area to be <math>\frac{1}{2}</math><math>FI^2</math>. This is also equal to <math>1+3-2\sqrt{2}</math> or <math>4-2\sqrt{2}</math>. Since we are looking for <math>FI^2</math>, we want two times this. That gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math>.~TLiu
  
==Solution 2==
+
== Solution 2 ==
Since this is a geometry problem involving sides, and we know that <math>HE</math> is <math>2</math>, we can use our ruler and find the ratio between <math>FI</math> and <math>HE</math>. Measuring(on the booklet), we get that <math>HE</math> is about <math>1.8</math> inches and <math>FI</math> is about <math>1.4</math> inches. Thus, we can then multiply the length of <math>HE</math> by the ratio of <math>\frac{1.4}{1.8},</math> of which we then get <math>FI= \frac{14}{9}.</math> We take the square of that and get <math>\frac{196}{81},</math> and the closest answer to that is <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math>. ~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)
+
Draw the auxiliary line <math>AC</math>. Denote by <math>M</math> the point it intersects with <math>HE</math>, and by <math>N</math> the point it intersects with <math>GF</math>. Last, denote by <math>x</math> the segment <math>FN</math>, and by <math>y</math> the segment <math>FI</math>. We will find two equations for <math>x</math> and <math>y</math>, and then solve for <math>y^2</math>.  
 
 
==Solution 3==
 
Draw the auxiliary line <math>AC</math>. Denote by <math>M</math> the point it intersects with <math>HE</math>, and by <math>N</math> the point it interacts with <math>GF</math>. Last, denote by <math>x</math> the segment <math>FN</math>, and by <math>y</math> the segment <math>FE</math>. We will find two equations for <math>x</math> and <math>y</math>, and then solve for <math>y^2</math>.  
 
  
 
Since the overall area of <math>ABCD</math> is <math>4 \;\; \Longrightarrow \;\;  AB=2</math>, and <math>AC=2\sqrt{2}</math>. In addition, the area of <math>\bigtriangleup AME = \frac{1}{2} \;\; \Longrightarrow \;\; AM=1</math>.
 
Since the overall area of <math>ABCD</math> is <math>4 \;\; \Longrightarrow \;\;  AB=2</math>, and <math>AC=2\sqrt{2}</math>. In addition, the area of <math>\bigtriangleup AME = \frac{1}{2} \;\; \Longrightarrow \;\; AM=1</math>.
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Solving for <math>y^2</math> gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math> ~DrB
 
Solving for <math>y^2</math> gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math> ~DrB
 +
 +
== 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{\textbf{(B)}\ 8-4\sqrt{2}}</cmath>. --OGBooger
 +
 +
== Solution 4 (HARD Calculation) ==
 +
We can easily observe that the area of square <math>ABCD</math> is 4 and its side length is 2 since all four regions that build up the square has area 1.
 +
Extend <math>FI</math> and let the intersection with <math>AB</math> be <math>K</math>. Connect <math>AC</math>, and let the intersection of <math>AC</math> and <math>HE</math> be <math>L</math>.
 +
Notice that since the area of triangle <math>AEH</math> is 1 and <math>AE=AH</math> , <math>AE=AH=\sqrt{2}</math>, therefore <math>BE=HD=2-\sqrt{2}</math>.
 +
Let <math>CG=CF=m</math>, then <math>BF=DG=2-m</math>.
 +
Also notice that <math>KB=2-m</math>, thus <math>KE=KB-BE=2-m-(2-\sqrt{2})=\sqrt{2}-m</math>.
 +
Now use the condition that the area of quadrilateral <math>BFIE</math> is 1, we can set up the following equation:
 +
<math>\frac{1}{2}(2-m)^2-\frac{1}{4}(\sqrt{2}-m)^2=1</math>
 +
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>
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<math>=2\sqrt{2}-1-\frac{8\sqrt{2}-4-\sqrt{128-64\sqrt2}}{4}</math>
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<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
 +
 +
== Solution 5 ==
 +
 +
<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);
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real l= 2sqrt(2)-4;
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pair A, B, C, D, E, F, G, H, I, J, L, M, K;
 +
A = (0,0);
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B = (4,0);
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C = (4,4);
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D = (0,4);
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E = (x,0);
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F = (4,y);
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G = (y,4);
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H = (0,x);
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I = F + z * dir(225);
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J = G + z * dir(225);
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L = (k,0);
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M = F + z * dir(315);
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K = (4,l);
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 +
draw(A--B--C--D--A);
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draw(H--E);
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draw(J--G^^F--I);
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draw(F--M);
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draw(M--L);
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draw(E--K,dashed+linewidth(.5));
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draw(K--L,dashed+linewidth(.5));
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draw(B--L);
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draw(rightanglemark(G, J, I), linewidth(.5));
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draw(rightanglemark(F, I, E), linewidth(.5));
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draw(rightanglemark(F, M, L), linewidth(.5));
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fill((4,0)--(k,0)--M--(4,y)--cycle, gray);
 +
dot("$A$", A, S);
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dot("$C$", C, dir(90));
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dot("$D$", D, dir(90));
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dot("$E$", E, S);
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dot("$G$", G, N);
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dot("$H$", H, W);
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dot("$I$", I, SW);
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dot("$J$", J, SW);
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dot("$K$", K, S);
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dot("$F(G)$", F, E);
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dot("$J'$", M, dir(90));
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dot("$H'$", L, S);
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dot("$B(D)$", B, S);
 +
 +
 +
</asy>
 +
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>.
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 +
<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>
 +
 +
--Ryan Zhang @BRS
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 +
== 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);
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D = (0,4);
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E = (x,0);
 +
F = (4,y);
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G = (y,4);
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H = (0,x);
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I = F + z * dir(225);
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J = G + z * dir(225);
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K = (4-x,4);
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L = J + 1.68 * dir(45);
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 +
draw(A--B--C--D--A);
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draw(H--E);
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draw(J--G^^F--I);
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draw(H--K,dashed+linewidth(.5));
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draw(L--K,dashed+linewidth(.5));
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draw(rightanglemark(G, J, I), linewidth(.5));
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draw(rightanglemark(F, I, E), linewidth(.5));
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draw(rightanglemark(H, K, L), linewidth(.5));
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draw(rightanglemark(K, L, G), linewidth(.5));
 +
 +
dot("$A$", A, S);
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dot("$B$", B, S);
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dot("$C$", C, dir(90));
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dot("$D$", D, dir(90));
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dot("$E$", E, S);
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dot("$F$", F, dir(0));
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dot("$G$", G, N);
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dot("$H$", H, W);
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dot("$I$", I, SW);
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dot("$J$", J, SW);
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dot("$K$", K, N);
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dot("$L$", L, S);
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</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>
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 +
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>
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 +
<math>[DHK] = \frac{(2 - \sqrt{2})^2}{2} = 3 - 2 \sqrt {2}</math>
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 +
<math>[GKL] = \frac{(x - 2 \sqrt{2} + 2)^2}{2} = \frac{x^2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6</math>
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<math>[HKJL] = (x - 2 \sqrt{2} + 2) \cdot (2 \sqrt{2} - 2) = 2x \sqrt{2} - 2x + 8 \sqrt{2} -12</math>
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 +
<cmath>[DHK] + [GKL] + [HKLJ] = [DHJG]</cmath>
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 +
<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]
 +
 +
==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
  
 
==See Also==
 
==See Also==
 +
{{AMC10 box|year=2020|ab=B|num-b=20|num-a=22}}
 +
{{AMC12 box|year=2020|ab=B|num-b=17|num-a=19}}
  
{{AMC10 box|year=2020|ab=B|num-b=20|num-a=22}}
+
[[Category:Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:38, 5 October 2022

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

[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

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

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AMC 10 Problems and Solutions
2020 AMC 12B (ProblemsAnswer KeyResources)
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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