1987 AMC 12 Problem 14

Attempt Problem 14 of the 1987 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 1987 AMC 12 solutions, or check the answer key.

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14.

ABCD

is a square and

M

and

N

are the midpoints of

BC

and

CD,

respectively. Then

\sin\theta=

\frac{\sqrt5}{5}

\frac35

\frac{\sqrt{10}}5

\frac45

none of these

Assign coordinates to the square and write vectors

\overrightarrow{AM}

and

\overrightarrow{AN}

Use the determinant formula for the sine of the angle between two vectors

Take the square to have side

2,

with

A=(0,0),\ M=(1,2),

and

N=(2,1).

Then

\begin{aligned} \sin\theta &=\frac{|\det((1,2),(2,1))|} {\sqrt5\sqrt5}\\ &=\frac{|1-4|}{5} =\frac35. \end{aligned}

Thus the correct answer is B.

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