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| − | ==Problem==
| + | #redirect [[2020 AMC 12B Problems/Problem 18]] |
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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>?
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| − | <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==
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| − | 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
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| − | ==Solution 2==
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| − | 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)
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| − | ==Solution 3==
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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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| − | ==See Also==
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| − | {{AMC10 box|year=2020|ab=B|num-b=20|num-a=22}}
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| − | {{MAA Notice}}
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