2021 AMC 10B Problem 21

Problem 21 of 25HarderGeometry

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'?

Show solution

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),

AC'=\frac23.

The fold reflects

C

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

E=(0,\frac12).

Thus

AE=\frac12,

and

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

The perimeter of

\triangle AEC'

\frac12+\frac23+\frac56=2.

Thus, the answer is A .

Tagged: paper folding · coordinate geometry