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

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==Problem==
 
==Problem==
−
[url=https://aops.com/community/p20334805][size=150][b]Problem 21[/b][/size][/url]
 
 
A square piece of paper has side length <math>1</math> and vertices <math>A,B,C,</math> and <math>D</math> in that order. As shown in the figure, the paper is folded so that vertex <math>C</math> meets edge <math>\overline{AD}</math> at point <math>C'</math>, and edge <math>\overline{AB}</math> at point <math>E</math>. Suppose that <math>C'D = \frac{1}{3}</math>. What is the perimeter of triangle <math>\bigtriangleup AEC' ?</math>
 
A square piece of paper has side length <math>1</math> and vertices <math>A,B,C,</math> and <math>D</math> in that order. As shown in the figure, the paper is folded so that vertex <math>C</math> meets edge <math>\overline{AD}</math> at point <math>C'</math>, and edge <math>\overline{AB}</math> at point <math>E</math>. Suppose that <math>C'D = \frac{1}{3}</math>. What is the perimeter of triangle <math>\bigtriangleup AEC' ?</math>
  
 
<math>\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}</math>
 
<math>\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}</math>
 
<asy>
 
<asy>
 +
\* Made by samrocksnature *\
 
pair A=(0,1);
 
pair A=(0,1);
 
pair CC=(0.666666666666,1);
 
pair CC=(0.666666666666,1);
Line 26: Line 26:
 
label("C",CC,N);
 
label("C",CC,N);
 
</asy>
 
</asy>
 +
 
==Solution (Quicksolve) ==
 
==Solution (Quicksolve) ==
 
Assume that E is the midpoint of <math>\overline{AB}</math>. Then, <math>\overline{AE}=\frac{1}{2}</math> and since <math>C'D=\frac{1}{3}</math>, <math>\overline{AC'}=\frac{2}{3}</math>. By the Pythagorean Theorem, <math>\overline{EC'}=\frac{5}{6}</math>. It easily follows that our desired perimeter is <math>2 \rightarrow \boxed{A}</math> ~samrocksnature
 
Assume that E is the midpoint of <math>\overline{AB}</math>. Then, <math>\overline{AE}=\frac{1}{2}</math> and since <math>C'D=\frac{1}{3}</math>, <math>\overline{AC'}=\frac{2}{3}</math>. By the Pythagorean Theorem, <math>\overline{EC'}=\frac{5}{6}</math>. It easily follows that our desired perimeter is <math>2 \rightarrow \boxed{A}</math> ~samrocksnature

Revision as of 19:46, 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 $\bigtriangleup AEC' ?$

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

\* 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);
 (Error making remote request. Unknown error_msg)

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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