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 1

Each step the bug takes is half as long as its previous step. So each horizontal step is $\frac{1}{4}$ as long as and in the opposite direction of the previous horizontal step. Similarly, each vertical step is $\frac{1}{4}$ as long as and in the opposite direction the previous vertical step.

Therefore, the $x$ -coordinate is the sum of an infinite geometric series with first term $1$ and common ratio $-\frac{1}{4}$ , so it must be $\frac{1}{1-\left(-\frac{1}{4}\right)} = \frac{4}{5}$ . Similarly, the $y$ -coordinate is the sum of an infinite geometric series with first term $\frac{1}{2}$ and common ratio $-\frac{1}{4}$ , so it must be $\frac{\frac{1}{2}}{1-\left(-\frac{1}{4}\right)} = \frac{2}{5}$ . Therefore, the bug's path approaches $\boxed{(\textbf{B}) \left(\frac{4}{5}, \frac{2}{5}\right)}$

Solution 2 (Complex Plane)

Consider the bug's path as an infinite sequence of vectors in the complex plane. Then each step can be represented by a complex number, with the first step being $1$ and the second step being $\frac{1}{2}i$ .

Since each step is half as long as and rotated $90^{\circ}$ counterclockwise from the previous step, the complex number representing a step must be $\frac{1}{2}i$ times the complex number representing the previous step.

Therefore, the bug's path can be represented by an infinite geometric series with first term $1$ and common ratio $\frac{i}{2}$ . The sum of this series is $\frac{1}{1-\frac{1}{2}i} = \frac{2}{2-i} = \frac{2(2+i)}{(2-i)(2+i)} = \frac{4}{5} + \frac{2}{5}i$ , which corresponds to $\boxed{(\textbf{B}) \left(\frac{4}{5}, \frac{2}{5}\right)}$

-jaspersun, edited by j314andrews

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

Contents

Problem

Solution 1

Solution 2 (Complex Plane)

See also