1980 AMC 12 Problem 13

Attempt Problem 13 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).

13.

A bug (of negligible size) starts at the origin on the coordinate plane. First it moves 11 unit right to (1,0).(1,0). Then it makes a 90∘90^\circ turn counterclockwise and travels 12\frac12 a unit to (1,12).(1,\frac12). If it continues in this fashion, each time making a 90∘90^\circ turn counterclockwise and traveling half as far as in the previous move, to which of the following points will it come closest?
(23,23)(\frac23,\frac23)
(45,25)(\frac45,\frac25)
(23,45)(\frac23,\frac45)
(23,13)(\frac23,\frac13)
(25,45)(\frac25,\frac45)
Answer: B
Concepts:geometric sequencecomplex numbersummation
Small hint:
Represent a 90∘90^\circ counterclockwise turn by multiplication by ii
Big hint:
The displacement vectors form an infinite geometric series with ratio i2\frac{i}{2}
Solution:
As complex numbers, the successive displacement vectors are 1, i2, (i2)2,…. 1,\ \frac i2,\ \left(\frac i2\right)^2,\ldots. Their sum is 11−i2=1+i21+14=45+25i. \frac1{1-\frac{i}{2}} =\frac{1+\frac{i}{2}}{1+\frac{1}{4}} =\frac45+\frac25i. Thus the bug approaches (45,25).(\frac{4}{5},\frac{2}{5}).
Therefore, the correct answer is B.
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