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

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{{AMC10 box|year=2021|ab=B|num-b=20|num-a=22}}

Revision as of 23:53, 11 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, $x^2 + (\frac{1}{3})^2 = (1-x)^2 \rightarrow x^2 + \frac{1}{9} = x^2 - 2x + 1 \rightarrow x=4/9$. 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 (Outlined)

double angle tangent to find angle ACE then trig

Solution (Quicksolve)

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

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
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All AMC 10 Problems and Solutions
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