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Difference between revisions of "2020 AMC 12B Problems/Problem 18"

(Created page with "==Solution== 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>...")
 
Line 1: Line 1:
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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));
 +
 +
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>
 +
 +
<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==
  
 
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 F'B' 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 quadrilateral <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})=6-4sqrt(2)</math>. Adding these two areas, we get $2+6-4\sqrt{2}=\boxed{\textbf{B) }8-4\sqrt{2}}
 
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 F'B' 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 quadrilateral <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})=6-4sqrt(2)</math>. Adding these two areas, we get $2+6-4\sqrt{2}=\boxed{\textbf{B) }8-4\sqrt{2}}

Revision as of 22:30, 7 February 2020

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

Plot a point $F'$ such that $F'$ and $I$ are collinear 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 quadrilateral $EIFF'E'$ is $2$. Length $AE=\sqrt{2}$, thus $EB=2-\sqrt{2}$. Triangle $EB'E'$ is isosceles, and the area of this triangle is $\frac{1}{2}*(4-2\sqrt{2})*(2-\sqrt{2})=6-4sqrt(2)$. Adding these two areas, we get $2+6-4\sqrt{2}=\boxed{\textbf{B) }8-4\sqrt{2}}

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