1988 AMC 12 Problem 20

Question

In one of the adjoining figures a square of side

2

is dissected into four pieces so that

E

and

F

are the midpoints of opposite sides and

AG

is perpendicular to

BF.

These four pieces can then be reassembled into a rectangle as shown in the second figure. The ratio of height to base,

\frac{XY}{YZ},

in this rectangle is

4

1+2\sqrt3

2\sqrt5

\frac{8+4\sqrt3}{3}

5

Find

BF

and

DE

from the side length and midpoint conditions

Use the unchanged area to determine the rectangle’s base

Both

BF

and

DE

have horizontal and vertical changes

1

and

2,

so each has length

\sqrt5.

Thus

XY=BF+DE=2\sqrt5.

The rectangle has area

4,

so

YZ=\frac{4}{2\sqrt5}=\frac{2}{\sqrt5}.

Therefore

\frac{XY}{YZ}=5.

Thus the correct answer is E.

Problem 20 in other years