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

## 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 $\overline{BC}$ intersects edge $\overline{AB}$ at point $E$. Suppose that $C'D = \frac{1}{3}$. What is the perimeter of triangle $\bigtriangleup AEC' ?$

$\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{136} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}$ $[asy] /* Made by samrocksnature */ pair A=(0,1); pair CC=(0.666666666666,1); pair D=(1,1); pair F=(1,0.62); pair C=(1,0); pair B=(0,0); pair G=(0,0.25); pair H=(-0.13,0.41); pair E=(0,0.5); dot(A^^CC^^D^^C^^B^^E); draw(E--A--D--F); draw(G--B--C--F, dashed); fill(E--CC--F--G--H--E--CC--cycle, gray); draw(E--CC--F--G--H--E--CC); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,NE); label("E",E,NW); label("C'",CC,N); [/asy]$

## 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,

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

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 $\overline{AB}$. 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

## Solution 4

As described in Solution 1, we can find that $DF=\frac{4}{9}$, and $C'F = \frac{5}{9}.$

Then, we can find we can find the length of $\overline{AE}$ by expressing the length of $\overline{EF}$ in two different ways, in terms of $AE$. If let $AE = a$, by the Pythagorean Theorem we have that $EC = \sqrt{a^2 + \left(\frac{2}{3}\right)^2} = \sqrt{a^2 + \frac{4}{9}}.$ Therefore, since we know that $\angle EC'F$ is right, by Pythagoras again we have that $EF = \sqrt{\left(\sqrt{a^2+\frac{4}{9}}\right)^2 + \left(\frac{5}{9}\right)^2} = \sqrt{a^2 + \frac{61}{81}}.$

Another way we can express $EF$ is by using Pythagoras on $\triangle XEF$, where $X$ is the foot of the perpendicular from $F$ to $\overline{AE}.$ We see that $ADFX$ is a rectangle, so we know that $AD = 1 = FX$. Secondly, since $FD = \frac{4}{9}, EX = a - \frac{4}{9}$. Therefore, through the Pythagorean Theorem, we find that $EF = \sqrt{\left(a-\frac{4}{9}\right)^2 + 1^2} = \sqrt{a^2 - \frac{8}{9}a +\frac{97}{81}}.$

Since we have found two expressions for the same length, we have the equation $\sqrt{a^2 + \frac{61}{81}} = \sqrt{a^2 - \frac{8}{9}a +\frac{97}{81}}.$ Solving this, we find that $a=\frac{1}{2}$.

Finally, we see that the perimeter of $\triangle AEC'$ is $\frac{1}{2} + \frac{2}{3} + \sqrt{\left(\frac{1}{2}\right)^2 + \frac{4}{9}},$ which we can simplify to be $2$. Thus, the answer is $\boxed{\textbf{(A)} ~2}.$ ~laffytaffy

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

~ pi_is_3.14

 2021 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 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