Question

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 $\bigtriangleup AEC' ?$

$2$

$1+\dfrac{2}{3}\sqrt{3}$

$\dfrac{13}{6}$

$1 + \dfrac{3}{4}\sqrt{3}$

$\dfrac{7}{3}$

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Written solution:

Use coordinates with $A=(0,1)$ , $B=(0,0)$ , $C=(1,0)$ , and $D=(1,1)$ . Since $C'D=\frac13$ , we have $C'=(\frac23,1)$ , so $AC'=\frac23$ .

The fold reflects $C$ to $C'$ , so the image of side $BC$ is the line through $C'$ and $E$ . Reflecting $B=(0,0)$ across the perpendicular bisector of $CC'$ gives $(-\frac{2}{15},\frac25)$ . The line through this point and $C'$ meets $AB$ at $E=(0,\frac12)$ .

Thus $AE=\frac12$ , and

$EC'=\sqrt{\left(\frac23\right)^2+\left(\frac12\right)^2}=\frac56.$

The perimeter of $\triangle AEC'$ is

$\frac12+\frac23+rac56=2.$

Thus, the answer is A .

2021 AMC 10B Spring Problem 21

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Problem 21 in Other Years