2006 AMC 12A Problem 15

Question

Suppose

\cos x = 0

and

\cos(x + z) = \tfrac{1}{2}.

What is the smallest possible positive value of

z?

\dfrac{\pi}{6}

\dfrac{\pi}{3}

\dfrac{\pi}{2}

\dfrac{5\pi}{6}

\dfrac{7\pi}{6}

\cos x = 0

means

x

is an odd multiple of

\tfrac{\pi}{2}

\cos(x + z) = \tfrac{1}{2}

means

x + z = 2n\pi \pm \tfrac{\pi}{3};

minimize

z = (x+z) - x

Since

\cos x = 0,

we have

x = \tfrac{\pi}{2} + k\pi.

Since

\cos(x + z) = \tfrac{1}{2},

we have

x + z = 2n\pi \pm \tfrac{\pi}{3}.

On the unit circle, the smallest positive angular separation between an odd multiple of

\tfrac{\pi}{2}

and an angle congruent to

\pm\tfrac{\pi}{3}

is

\tfrac{\pi}{6}.

It is attained by taking

x = -\tfrac{\pi}{2}

and

x + z = -\tfrac{\pi}{3},

which gives

z = -\tfrac{\pi}{3} + \tfrac{\pi}{2} = \tfrac{\pi}{6}.

Thus, the correct answer is A.

Problem 15 in other years