1989 AMC 12 Problema 13

Question

Dos franjas de ancho

1

se superponen formando un ángulo

\alpha

. El área de la región sombreada es

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}

La región es un paralelogramo de altura

1

The overlap is a parallelogram with altitude

1

Si su lado inclinado mide

L,

entonces

L\sin\alpha=1

If its slanted side has length

L,

then

L\sin\alpha=1

La región es un paralelogramo. Con el lado inclinado como base, la altura es el ancho

1

de la franja inclinada. Si la base mide

L,

su componente vertical da

L\sin\alpha=1.

Así,

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

y el área es

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

Por lo tanto, la respuesta correcta es B.

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.

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