1980 AMC 12 Problem 9

Attempt Problem 9 of the 1980 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 1980 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

9.

A man walks xx miles due west, turns 150∘150^\circ to his left and walks 33 miles in the new direction. If he finishes at a point 3\sqrt3 miles from his starting point, then xx is
3\sqrt3
232\sqrt3
32\frac32
33
not uniquely determined by the given information
Answer: E
Concepts:vectordistance formulaquadraticcasework
Small hint:
Resolve the second walk into horizontal and vertical components
Big hint:
The distance equation is quadratic in xx; check whether both positive roots are valid
Solution:
Take east as the positive horizontal direction. After the turn, the 33-mile displacement has components (332,−32).(\frac{3\sqrt3}{2},-\frac{3}{2}). Therefore (332−x)2+(32)2=3. \left(\frac{3\sqrt3}{2}-x\right)^2 +\left(\frac32\right)^2=3. Hence 332−x=±32,\frac{3\sqrt3}{2}-x=\pm\frac{\sqrt3}{2}, giving x=3x=\sqrt3 or x=23.x=2\sqrt3. The value is not unique.
Therefore, the correct answer is E.
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