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| − | 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>?
| + | #REDIRECT [[2020 AMC 10B Problems/Problem 21]] |
| − | <asy>
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| − | real x=2sqrt(2);
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| − | real y=2sqrt(16-8sqrt(2))-4+2sqrt(2);
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| − | real z=2sqrt(8-4sqrt(2));
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| − | pair A, B, C, D, E, F, G, H, I, J;
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| − | 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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| − | 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(rightanglemark(G, J, I), linewidth(.5));
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| − | draw(rightanglemark(F, I, E), linewidth(.5));
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| − | 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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| − | </asy>
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| − | <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>
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| − | ==Solution 1==
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| − | Plot a point <math>F'</math> such that <math>F'</math> and <math>I</math> are collinear 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
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| − | ==Solution 2==
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| − | 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>.
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| − | 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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| − | The two equations for <math>x</math> and <math>y</math> are then:
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| − | <math>\bullet</math> Length of <math>AC</math>: <math>1+y+x = 2\sqrt{2} \;\; \Longrightarrow \;\; x = (2\sqrt{2}-1) - y</math>
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| − | <math>\bullet</math> Area of CMIF: <math>\frac{1}{2}x^2+xy = \frac{1}{2} \;\; \Longrightarrow \;\; x(x+2y)=1</math>.
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| − | Substituting the first into the second, yields
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| − | <math>\left[\left(2\sqrt{2}-1\right)-y\right]\cdot \left[\left(2\sqrt{2}-1\right)+y\right]=1</math>
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| − | Solving for <math>y^2</math> gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math> ~DrB
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| − | ==Solution 3(HARD Calculation)==
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| − | 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.
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| − | 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>.
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| − | 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>.
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| − | Let <math>CG=GF=m</math>, then <math>BF=DG=2-m</math>.
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| − | Also notice that <math>KB=BE=2-m</math>, thus <math>KE=KB-BE=2-m-(2-\sqrt{2})=\sqrt{2}-m</math>.
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| − | Now use the condition that the area of quadrilateral <math>BFIE</math> is 1, we can set up the following equation:
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| − | <math>\frac{1}{2}(2-m)^2-\frac{1}{4}(\sqrt{2}-m)^2=1</math>
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| − | We solve the equation and yield <math>m=\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}</math>.
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| − | Now notice that
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| − | <math>FI=AC-AL=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>.
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| − | Hence <math>FI^2=\frac{128-64\sqrt{2}}{16}=8-4\sqrt{2}</math>. -HarryW
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| − | {{AMC12 box|year=2020|ab=B|num-b=17|num-a=19}}
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| − | {{MAA Notice}}
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