Two strips of width 1 overlap at an angle of α as shown. The area of the overlap (shown shaded) is

1989 AMC 12 Problem 13

Attempt Problem 13 of the 1989 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1989 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

13.

Two strips of width

1

overlap at an angle of

\alpha

as shown. The area of the overlap (shown shaded) is

\sin\alpha

\frac1{\sin\alpha}

\frac1{1-\cos\alpha}

\frac1{\sin^2\alpha}

\frac1{(1-\cos\alpha)^2}

The overlap is a parallelogram with altitude

1

If its slanted side has length

L,

then

L\sin\alpha=1

The overlap is a parallelogram. Taking a slanted side as its base, the perpendicular height is the width

1

of the slanted strip. If that base has length

L,

its vertical component is the width of the horizontal strip, so

L\sin\alpha=1.

Hence

L=\frac{1}{\sin\alpha},

and the area is

L\cdot1=\frac{1}{\sin\alpha}.

Thus the correct answer is B.