2016 AMC 8 Problem 22

Question

Rectangle

DEFA

below is a

3 \times 4

rectangle with

DC=CB=BA=1.

The area of the “bat wings” (shaded area) is

2

2 \dfrac{1}{2}

3

3 \dfrac{1}{2}

4

Use the intersection point of the two diagonals that form the bat wings.

Similar triangles give the height of the small triangle cut off at the top.

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Define

I

to be the midpoint of

\overline{AD}

and

G

to be the midpoint of

\overline{EF}.

Also define

H

to be the intersection of

\overline{CF}

and

\overline{BE}.

The area of

\triangle BCE

is

\dfrac12\cdot1\cdot4=2.

By symmetry, we can see that

\triangle BCH

and

\triangle EFH

are similar. Since their bases are in a

1 : 3

ratio, so are their altitudes. This means that

3IH = HG,

which implies that

IH = 1.

Therefore,

[\triangle BCH]=\dfrac12\cdot1\cdot1=\dfrac12.

It follows that

[\triangle ECH]=2-\dfrac12=\dfrac32.

Since the figure is symmetric, the total area of the bat wings is

2\cdot\dfrac32=3.

Thus, C is the correct answer.

Problem 22 in other years