Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2020 AMC 12B Problems/Problem 18: Difference between revisions

Newer edit →
Ogbooger (talk | contribs)
24 edits
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>..."
 
N828335 (talk | contribs)
70 edits
No edit summary
Line 1: Line 1:
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 $ABCD$, 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}}

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_18&oldid=117188"