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Difference between revisions of "1980 AHSME Problems/Problem 13"

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== Solution ==
 
== Solution ==
 
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>
 
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>
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== Solution ==
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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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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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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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~ jaspersun
  
 
== See also ==
 
== See also ==

Latest revision as of 23:03, 3 September 2024

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

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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