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

Revision as of 18:46, 11 February 2021

Problem

[url=https://aops.com/community/p20334805][size=150][b]Problem 21[/b][/size][/url] 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 $\bigtriangleup AEC' ?$

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

Solution (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

@@ Line 26: / Line 26: @@
 label("C",CC,N);
 </asy>
-==Solution==
+==Solution (Quicksolve) ==
-A
+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

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

Revision as of 18:46, 11 February 2021

Problem

Solution (Quicksolve)