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Difference between revisions of "1980 AHSME Problems/Problem 13"
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<math>\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)</math>
 
<math>\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)</math>
  
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== Solution 1 ==
 +
Each step the bug takes is half as long as its previous step.  So each horizontal step is <math>\frac{1}{4}</math> as long as and in the opposite direction of the previous horizontal step.  Similarly, each vertical step is <math>\frac{1}{4}</math> as long as and in the opposite direction the previous vertical step.
  
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== Solution ==
+
Therefore, the <math>x</math>-coordinate is the sum of an infinite geometric series with first term <math>1</math> and common ratio <math>-\frac{1}{4}</math>, so it must be <math>\frac{1}{1-\left(-\frac{1}{4}\right)} = \frac{4}{5}</math>. Similarly, the <math>y</math>-coordinate is the sum of an infinite geometric series with first term <math>\frac{1}{2}</math> and common ratio <math>-\frac{1}{4}</math>, so it must be <math>\frac{\frac{1}{2}}{1-\left(-\frac{1}{4}\right)}  
−
Writing out the change in <math>x</math> coordinates and then in <math>y</math> coordinates gives the infinite sum <math>1-\frac{1}{4}+\frac{1}{16}-\dots</math> and <math>\frac{1}{2}-\frac{1}{8}+\dots</math> respectively. Using the infinite geometric sum formula, we have <math>\frac{1}{1+\frac{1}{4}}=\frac{4}{5}</math> and <math>\frac{\frac{1}{2}}{1+\frac{1}{4}}=\frac{2}{5}</math>, thus the answer is <math>\left( \frac{4}{5}, \frac{2}{5} \right)</math>
+
= \frac{2}{5}</math>.  Therefore, the bug's path approaches <math>\boxed{(\textbf{B}) \left(\frac{4}{5}, \frac{2}{5}\right)}</math>
  
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== Solution ==
+
== Solution 2 (Complex Plane) ==
 
We can represent the bug's position on the coordinate plane using complex numbers. The first move the bug makes is <math>1</math>, the second <math>i/2</math>, the third <math>-1/4</math>, and so on. It becomes clear that the distance the bug travels is an infinite geometric series with initial term 1, and common ratio <math>i/2</math>.
 
We can represent the bug's position on the coordinate plane using complex numbers. The first move the bug makes is <math>1</math>, the second <math>i/2</math>, the third <math>-1/4</math>, and so on. It becomes clear that the distance the bug travels is an infinite geometric series with initial term 1, and common ratio <math>i/2</math>.
  

Revision as of 15:35, 14 August 2025

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)

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

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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