1980 AHSME Problems/Problem 13

Problem

A bug (of negligible size) starts at the origin on the coordinate plane. First, it moves one unit right to $(1,0)$ . Then it makes a $90^\circ$ counterclockwise and travels $\frac 12$ a unit to $\left(1, \frac 12 \right)$ . If it continues in this fashion, each time making a $90^\circ$ degree turn counterclockwise and traveling half as far as the previous move, to which of the following points will it come closest?

$\text{(A)} \ \left(\frac 23, \frac 23 \right) \qquad \text{(B)} \ \left( \frac 45, \frac 25 \right) \qquad \text{(C)} \ \left( \frac 23, \frac 45 \right) \qquad \text{(D)} \ \left(\frac 23, \frac 13 \right) \qquad \text{(E)} \ \left(\frac 25, \frac 45 \right)$

Solution

Writing out the change in $x$ coordinates and then in $y$ coordinates gives the infinite sum $1-\frac{1}{4}+\frac{1}{16}-\dots$ and $\frac{1}{2}-\frac{1}{8}+\dots$ respectively. Using the infinite geometric sum formula, we have $\frac{1}{1+\frac{1}{4}}=\frac{4}{5}$ and $\frac{\frac{1}{2}}{1+\frac{1}{4}}=\frac{2}{5}$ , thus the answer is $\left( \frac{4}{5}, \frac{2}{5} \right)$

Solution

We can represent the bug's position on the coordinate plane using complex numbers. The first move the bug makes is $1$ , the second $i/2$ , the third $-1/4$ , and so on. It becomes clear that the distance the bug travels is an infinite geometric series with initial term 1, and common ratio $i/2$ .

Thus, applying the infinite geometric series formula: $\frac{1}{1-\frac{i}{2}} = \frac{1}{\frac{2-i}{2}} = \frac{2}{2-i} = \frac{2(2+i)}{(2-i)(2+i)} = \frac{4 + 2i}{5}$

This is equivalent to the $x$ coordinate being $4/5$ and the $y$ coordinate being $2/5$ , so the answer is $\fbox{B}$

~ jaspersun

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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1980 AHSME Problems/Problem 13

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