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

Difference between revisions of "2020 AMC 12B Problems/Problem 18"
(→‎Solution 3(HARD Calculation)(Not Completed yet))
Line 55: Line 55:
 
Solving for <math>y^2</math> gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math> ~DrB
 
Solving for <math>y^2</math> gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math> ~DrB
  
−
==Solution 3(HARD Calculation)(Not Completed yet)==
+
==Solution 3(HARD Calculation)==
−
 
 
  
 
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.  
 
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.  
 
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>
 
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>
 
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>  
 
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>  
 +
Let <math>CG=GF=m</math>, then <math>BF=DG=2-m</math>
 +
Also notice that <math>KB=BE=2-m</math>, thus <math>KE=KB-BE=2-m-(2-\sqrt{2})=\sqrt{2}-m</math>
 +
Now use the condition that the area of quadrilateral <math>BFIE</math> is 1, we can set up the following equation:
 +
<math>\frac{1}{2}(2-m)^2-\frac{1}{4}(\sqrt{2}-m)^2=1</math>
 +
We solve the equation and yield <math>m=\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}</math>
 +
Now notice that
 +
<math>FI=AC-AL=2\sqrt{2}-1-\frac{\sqrt{2}}{2}*\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}</math>
 +
<math>=2\sqrt{2}-1-\frac{8\sqrt{2}-4-\sqrt{128-64\sqrt2}}{4}</math>
 +
<math>=\frac{\sqrt{128-64\sqrt{2}}}{4}</math>
 +
Hence <math>FI^2=\frac{128-64\sqrt{2}}{16}=8-4\sqrt{2}</math>  -HarryW
 +
 +
  
 
{{AMC12 box|year=2020|ab=B|num-b=17|num-a=19}}
 
{{AMC12 box|year=2020|ab=B|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:54, 9 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 1

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 pentagon $EIFF'E'$ is $2$. Length $AE=\sqrt{2}$, thus $EB=2-\sqrt{2}$. Triangle $EB'E'$ is isosceles, and the area of this triangle is $\frac{1}{2}(4-2\sqrt{2})(2-\sqrt{2})=6-4\sqrt{2}$. Adding these two areas, we get \[2+6-4\sqrt{2}=8-4\sqrt{2}\rightarrow \boxed{\mathrm{(B)}}\]. --OGBooger

Solution 2

Draw the auxiliary line $AC$. Denote by $M$ the point it intersects with $HE$, and by $N$ the point it intersects with $GF$. Last, denote by $x$ the segment $FN$, and by $y$ the segment $FI$. We will find two equations for $x$ and $y$, and then solve for $y^2$.

Since the overall area of $ABCD$ is $4 \;\; \Longrightarrow \;\; AB=2$, and $AC=2\sqrt{2}$. In addition, the area of $\bigtriangleup AME = \frac{1}{2} \;\; \Longrightarrow \;\; AM=1$.

The two equations for $x$ and $y$ are then:

$\bullet$ Length of $AC$: $1+y+x = 2\sqrt{2} \;\; \Longrightarrow \;\; x = (2\sqrt{2}-1) - y$

$\bullet$ Area of CMIF: $\frac{1}{2}x^2+xy = \frac{1}{2} \;\; \Longrightarrow \;\; x(x+2y)=1$.

Substituting the first into the second, yields $\left[\left(2\sqrt{2}-1\right)-y\right]\cdot \left[\left(2\sqrt{2}-1\right)+y\right]=1$

Solving for $y^2$ gives $\boxed{\textbf{(B)}\ 8-4\sqrt{2}}$ ~DrB

Solution 3(HARD Calculation)

We can easily observe that the area of square $ABCD$ is 4 and its side length is 2 since all four regions that build up the square has area 1. Extend $FI$ and let the intersection with $AB$ be $K$. Connect $AC$, and let the intersection of $AC$ and $HE$ be $L$ Notice that since the area of triangle $AEH$ is 1 and $AE=AH$ , $AE=AH=\sqrt{2}$, therefore $BE=HD=2-\sqrt{2}$ Let $CG=GF=m$, then $BF=DG=2-m$ Also notice that $KB=BE=2-m$, thus $KE=KB-BE=2-m-(2-\sqrt{2})=\sqrt{2}-m$ Now use the condition that the area of quadrilateral $BFIE$ is 1, we can set up the following equation: $\frac{1}{2}(2-m)^2-\frac{1}{4}(\sqrt{2}-m)^2=1$ We solve the equation and yield $m=\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}$ Now notice that $FI=AC-AL=2\sqrt{2}-1-\frac{\sqrt{2}}{2}*\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}$ $=2\sqrt{2}-1-\frac{8\sqrt{2}-4-\sqrt{128-64\sqrt2}}{4}$ $=\frac{\sqrt{128-64\sqrt{2}}}{4}$ Hence $FI^2=\frac{128-64\sqrt{2}}{16}=8-4\sqrt{2}$ -HarryW


2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_18&oldid=117546"