# 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 $\left( \frac{4}{5}, \frac{2}{5} \right)$

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