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

$\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{13}{6} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}$

\* 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);
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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

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2021 AMC 10B Problems/Problem 21

Problem

Solution (Quicksolve)