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Difference between revisions of "1980 AHSME Problems/Problem 13"
(→‎Solution 2 (Complex Plane))
 
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== Solution 2 (Complex Plane) ==
 
== Solution 2 (Complex Plane) ==
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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>.
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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 <math>1</math> and the second step being <math>\frac{1}{2}i</math>.  
  
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Thus, applying the infinite geometric series formula: <math>\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}</math>
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Since each step is half as long as and rotated <math>90^{\circ}</math> counterclockwise from the previous step, the complex number representing a step must be <math>\frac{1}{2}i</math> times the complex number representing the previous step.
  
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This is equivalent to the <math>x</math> coordinate being <math>4/5</math> and the <math>y</math> coordinate being <math>2/5</math>, so the answer is <math>\fbox{B}</math>
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Therefore, the bug's path can be represented by an infinite geometric series with first term <math>1</math> and common ratio <math>\frac{i}{2}</math>.  The sum of this series is <math>\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</math>, which corresponds to <math>\boxed{(\textbf{B}) \left(\frac{4}{5}, \frac{2}{5}\right)}</math>
  
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~ jaspersun
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-jaspersun, edited by j314andrews
  
 
== See also ==
 
== See also ==

Latest revision as of 17:29, 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)

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

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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