Difference between revisions of "2021 AMC 10B Problems/Problem 21"

Revision as of 16:19, 12 February 2021

Problem

A square piece of paper has side length $1$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C'$ , and edge $\overline{AB}$ at point $E$ . Suppose that $C'D = \frac{1}{3}$ . What is the perimeter of triangle $\bigtriangleup AEC' ?$

$\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{13}{6} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}$ $[asy] /* Made by samrocksnature */ pair A=(0,1); pair CC=(0.666666666666,1); pair D=(1,1); pair F=(1,0.62); pair C=(1,0); pair B=(0,0); pair G=(0,0.25); pair H=(-0.13,0.41); pair E=(0,0.5); dot(A^^CC^^D^^C^^B^^E); draw(E--A--D--F); draw(G--B--C--F, dashed); fill(E--CC--F--G--H--E--CC--cycle, gray); draw(E--CC--F--G--H--E--CC); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,NE); label("E",E,NW); label("C'",CC,N); [/asy]$

Solution 1

We can set the point on $CD$ where the fold occurs as point $F$ . Then, we can set $FD$ as $x$ , and $CF$ as $1-x$ because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for $x$ , we get,

$x^2 + (\frac{1}{3})^2 = (1-x)^2 \rightarrow x^2 + \frac{1}{9} = x^2 - 2x + 1 \rightarrow x=\frac{4}{9}$

We know this is a 3-4-5 triangle because the side lengths are $\frac{3}{9}, \frac{4}{9}, \frac{5}{9}$ . We also know that $EAC'$ is similar to $C'DF$ because angle $C'$ is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of $C'DF * \frac{AC'}{DF}$ . Thats just $\frac{4}{3} * \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} * \frac{3}{2} = 2$ . Therefore, the final answer is $\boxed{A}$

~Tony_Li2007

Solution 2

Let line we're reflecting over be $\ell$ , and let the points where it hits $AB$ and $CD$ , be $M$ and $N$ , respectively. Notice, to reflect over a line we find the perpendicular passing through the midpoint of the two points (which are the reflected and the original). So, we first find the equation of the line $\ell$ . The segment $CC'$ has slope $\frac{0 - 1}{1 - 2/3} = -3$ , implying line $\ell$ has a slope of $\frac{1}{3}$ . Also, the midpoint of segment $CC'$ is $\left( \frac{5}{6}, \frac{1}{2} \right)$ , so line $\ell$ passes through this point. Then, we get the equation of line $\ell$ is simply $y = \frac{1}{3} x + \frac{2}{9}$ . Then, if the point where $B$ is reflected over line $\ell$ is $B'$ , then we get $BB'$ is the line $y = -3x$ . The intersection of $\ell$ and segment $BB'$ is $\left( - \frac{1}{15}, \frac{1}{5} \right)$ . So, we get $B' = \left(- \frac{2}{15}, \frac{2}{5} \right)$ . Then, line segment $B'C'$ has equation $y = \frac{3}{4} x + \frac{1}{2}$ , so the point $E$ is the $y$ -intercept, or $\left(0, \frac{1}{2} \right)$ . This implies that $AE = \frac{1}{2}$ , AC' = \frac{2}{3} $, and by the Pythagorean Theorem,$ EC' = \frac{5}{6} $(or you could notice$ \triangle AEC' $is a$ 3-4-5 $right triangle). Then, the perimeter is$ \frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2 $, so our answer is$ \boxed{\textbf{(A)} ~2}$. ~rocketsri

==Solution 3 (Fakesolve):== Assume that E is the midpoint of$ (Error compiling LaTeX. Unknown error_msg)\overline{AB} $. Then,$ \overline{AE}=\frac{1}{2} $and since$ C'D=\frac{1}{3} $,$ \overline{AC'}=\frac{2}{3} $. By the Pythagorean Theorem,$ \overline{EC'}=\frac{5}{6} $. It easily follows that our desired perimeter is$ 2 \rightarrow \boxed{A}$ ~samrocksnature

Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles)

https://youtu.be/cagzLmdbqYQ

~ pi_is_3.14

2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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 10 Problems and Solutions

@@ Line 38: / Line 38: @@
 ==Solution 2==
-Let line we're reflecting over be <math>\ell</math>, and let the points where it hits <math>AB</math> and <math>CD</math>, be <math>M</math> and <math>N</math>, respectively. Notice, to reflect over a line we find the perpendicular passing through the midpoint of the two points (which are the reflected and the original). So, we first find the equation of the line <math>\ell</math>. The segment <math>CC'</math> has slope <math>\frac{0 - 1}{1 - 2/3} = \frac{-3}</math>, implying line <math>\ell</math> has a slope of <math>\frac{1}{3}</math>. Also, the midpoint of segment <math>CC'</math> is <math>\left( \frac{5}{6}, \frac{1}{2} \right)</math>, so line <math>\ell</math> passes through this point. Then, we get the equation of line <math>\ell</math> is simply <math>y = \frac{1}{3} x + \frac{2}{9}</math>. Then, if the point where <math>B</math> is reflected over line <math>\ell</math> is <math>B'</math>, then we get <math>BB'</math> is the line <math>y = -3x</math>. The intersection of <math>\ell</math> and segment <math>BB'</math> is <math>\left( - \frac{1}{15}, \frac{1}{5} \right)</math>. So, we get <math>B' = \left(- \frac{2}{15}, \frac{2}{5} \right)</math>. Then, line segment <math>B'C'</math> has equation <math>y = \frac{3}{4} x + \frac{1}{2}</math>, so the point <math>E</math> is the <math>y</math>-intercept, or <math>\left(0, \frac{1}{2} \right)</math>. This implies that <math>AE = \frac{1}{2}</math>, AC' = \frac{2}{3}<math>, and by the Pythagorean Theorem, </math>EC' = \frac{5}{6}<math> (or you could notice </math>\triangle AEC'<math> is a </math>3-4-5<math> right triangle). Then, the perimeter is </math>\frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2<math>, so our answer is </math>\boxed{\textbf{(A)} ~2}<math>. ~rocketsri
+Let line we're reflecting over be <math>\ell</math>, and let the points where it hits <math>AB</math> and <math>CD</math>, be <math>M</math> and <math>N</math>, respectively. Notice, to reflect over a line we find the perpendicular passing through the midpoint of the two points (which are the reflected and the original). So, we first find the equation of the line <math>\ell</math>. The segment <math>CC'</math> has slope <math>\frac{0 - 1}{1 - 2/3} = -3</math>, implying line <math>\ell</math> has a slope of <math>\frac{1}{3}</math>. Also, the midpoint of segment <math>CC'</math> is <math>\left( \frac{5}{6}, \frac{1}{2} \right)</math>, so line <math>\ell</math> passes through this point. Then, we get the equation of line <math>\ell</math> is simply <math>y = \frac{1}{3} x + \frac{2}{9}</math>. Then, if the point where <math>B</math> is reflected over line <math>\ell</math> is <math>B'</math>, then we get <math>BB'</math> is the line <math>y = -3x</math>. The intersection of <math>\ell</math> and segment <math>BB'</math> is <math>\left( - \frac{1}{15}, \frac{1}{5} \right)</math>. So, we get <math>B' = \left(- \frac{2}{15}, \frac{2}{5} \right)</math>. Then, line segment <math>B'C'</math> has equation <math>y = \frac{3}{4} x + \frac{1}{2}</math>, so the point <math>E</math> is the <math>y</math>-intercept, or <math>\left(0, \frac{1}{2} \right)</math>. This implies that <math>AE = \frac{1}{2}</math>, AC' = \frac{2}{3}<math>, and by the Pythagorean Theorem, </math>EC' = \frac{5}{6}<math> (or you could notice </math>\triangle AEC'<math> is a </math>3-4-5<math> right triangle). Then, the perimeter is </math>\frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2<math>, so our answer is </math>\boxed{\textbf{(A)} ~2}<math>. ~rocketsri
 ==Solution 3 (Fakesolve):==

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