Different ways of solving a contest problem (Part 1)

The following problem appeared on the American High School Mathematics Examination (now called the AMC 12) in 1988:

If 3 \sin \theta = \cos \theta, what is \sin \theta \cos \theta?

When I presented this problem to a group of students, I was pleasantly surprised by the amount of creativity shown when solving this problem.

Here’s the first solution that I received: draw the appropriate triangles for the angle \theta:

3 \sin \theta = \cos \theta

\tan \theta = \displaystyle \frac{1}{3}

Therefore, the angle \theta must lie in either the first or third quadrant, as shown. (Of course, \theta could be coterminal with either displayed angle, but that wouldn’t affect the values of \sin \theta or \cos \theta.)

AHSME problem

In Quadrant I, \sin \theta = \displaystyle \frac{1}{\sqrt{10}} and \cos \theta = \displaystyle \frac{3}{\sqrt{10}}. Therefore,

\sin \theta \cos \theta = \displaystyle \frac{1}{\sqrt{10}} \times \frac{3}{\sqrt{10}} = \displaystyle \frac{3}{10}.

In Quadrant III, \sin \theta = \displaystyle -\frac{1}{\sqrt{10}} and \cos \theta = -\displaystyle \frac{3}{\sqrt{10}}. Therefore,

\sin \theta \cos \theta = \displaystyle \left( - \frac{1}{\sqrt{10}} \right) \times \left( -\frac{3}{\sqrt{10}} \right) = \displaystyle \frac{3}{10}.

Either way, we can be certain that \sin \theta \cos \theta = \displaystyle \frac{3}{10}.

3 thoughts on “Different ways of solving a contest problem (Part 1)

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