2006 AMC 12A Problem 15

Attempt Problem 15 of the 2006 AMC 12A 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 2006 AMC 12A solutions, or check the answer key.

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15.

Suppose cosx=0\cos x = 0 and cos(x+z)=12.\cos(x + z) = \tfrac{1}{2}. What is the smallest possible positive value of z?z?

π6\dfrac{\pi}{6}

π3\dfrac{\pi}{3}

π2\dfrac{\pi}{2}

5π6\dfrac{5\pi}{6}

7π6\dfrac{7\pi}{6}

Answer: A
Concepts:trigonometry
Difficulty rating: 1590
Solution:

Since cosx=0,\cos x = 0, we have x=π2+kπ.x = \tfrac{\pi}{2} + k\pi. Since cos(x+z)=12,\cos(x + z) = \tfrac{1}{2}, we have x+z=2nπ±π3.x + z = 2n\pi \pm \tfrac{\pi}{3}.

On the unit circle, the smallest positive angular separation between an odd multiple of π2\tfrac{\pi}{2} and an angle congruent to ±π3\pm\tfrac{\pi}{3} is π6.\tfrac{\pi}{6}. It is attained by taking x=π2x = -\tfrac{\pi}{2} and x+z=π3,x + z = -\tfrac{\pi}{3}, which gives z=π3+π2=π6.z = -\tfrac{\pi}{3} + \tfrac{\pi}{2} = \tfrac{\pi}{6}.

Thus, the correct answer is A.

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