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>...") |
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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>? | ||
+ | <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> | ||
+ | |||
+ | <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 21:30, 7 February 2020
In square , points and lie on and , respectively, so that Points and lie on and , respectively, and points and lie on so that and . See the figure below. Triangle , quadrilateral , quadrilateral , and pentagon each has area What is ?
Solution
Plot a point such that and are collinear and extend line to point such that forms a square. Extend line to meet line F'B' and point is the intersection of the two. The area of this square is equivalent to . We see that the area of square is , meaning each side is of length 2. The area of the quadrilateral is . Length , thus . Triangle is isosceles, and the area of this triangle is . Adding these two areas, we get $2+6-4\sqrt{2}=\boxed{\textbf{B) }8-4\sqrt{2}}