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| ==Problem==
| | #redirect [[2020 AMC 10B Problems/Problem 21]] |
| 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 1 ==
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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>3-2\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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| == Solution 4 ==
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| 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
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| == Solution 5 (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=CF=m</math>, then <math>BF=DG=2-m</math>.
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| Also notice that <math>KB=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-\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>.
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| Hence <math>FI^2=\frac{128-64\sqrt{2}}{16}=8-4\sqrt{2}</math>. -HarryW
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| -edit: annabelle0913
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| == Solution 6 ==
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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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| 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;
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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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| 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);
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| 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);
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| </asy>
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| 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>
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| --Ryan Zhang @BRS
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| ==Video Solution 1==
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| https://www.youtube.com/watch?v=AKJXB07Sat0&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=7 ~ MathEx
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| ==Video Solution 2 by the Beauty of Math==
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| Solution starts at 3:09: https://youtu.be/VZYe3Hu88OA
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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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| {{AMC12 box|year=2020|ab=B|num-b=17|num-a=19}}
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| [[Category:Intermediate Geometry Problems]]
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| {{MAA Notice}}
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