2021 AMC 10B Spring Problem 21

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 AEC'?

2

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

\dfrac{13}{6}

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

\dfrac{7}{3}

Answer: A

Concepts:paper folding coordinate geometry

Difficulty rating: 2230

Small hint:

A fold is a reflection, so the crease is the perpendicular bisector of

CC'

Big hint:

Find

AE,

AC',

and

EC'

from the folded right triangle

Video solution:

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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+\frac56=2.

Thus, the answer is A .