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Difference between revisions of "1995 AIME Problems/Problem 1"

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:<math>= 1 + \frac{1}{2}^2 - \frac{1}{32}^2</math>
 
:<math>= 1 + \frac{1}{2}^2 - \frac{1}{32}^2</math>
  
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The majority of the terms cancel, leaving <math>1 + \frac{1}{4} - \frac{1}{1024}</math>, which simplifies down to <math>\frac{1024 + \left(256 - 1\right)}{1024}</math>. Thus, <math>m-n =  255</math>.
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The majority of the terms cancel, leaving <math>1 + \frac{1}{4} - \frac{1}{1024}</math>, which simplifies down to <math>\frac{1024 + \left(256 - 1\right)}{1024}</math>. Thus, <math>m-n =  \boxed{255}</math>.
  
 
Alternatively, take the area of the first square and add <math>\,\frac{3}{4}</math> of the areas of  the remaining squares. This results in <math>1+ \frac{3}{4}\left[\left(\frac{1}{2}\right)^2 + \ldots + \left(\frac{1}{16}^2\right)\right]</math>, which when simplified will produce the same answer.
 
Alternatively, take the area of the first square and add <math>\,\frac{3}{4}</math> of the areas of  the remaining squares. This results in <math>1+ \frac{3}{4}\left[\left(\frac{1}{2}\right)^2 + \ldots + \left(\frac{1}{16}^2\right)\right]</math>, which when simplified will produce the same answer.
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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Latest revision as of 18:29, 4 July 2013

Problem

Square $S_{1}$ is $1\times 1.$ For $i\ge 1,$ the lengths of the sides of square $S_{i+1}$ are half the lengths of the sides of square $S_{i},$ two adjacent sides of square $S_{i}$ are perpendicular bisectors of two adjacent sides of square $S_{i+1},$ and the other two sides of square $S_{i+1},$ are the perpendicular bisectors of two adjacent sides of square $S_{i+2}.$ The total area enclosed by at least one of $S_{1}, S_{2}, S_{3}, S_{4}, S_{5}$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m-n.$

AIME 1995 Problem 1.png

Solution

The sum of the areas of the squares if they were not interconnected is a geometric sequence:

$1^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{16}\right)^2$

Then subtract the areas of the intersections, which is $\left(\frac{1}{4}\right)^2 + \ldots + \left(\frac{1}{32}\right)^2$:

$1^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{16}\right)^2 - \left[\left(\frac{1}{4}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{16}\right)^2 + \left(\frac{1}{32}\right)^2\right]$
$= 1 + \frac{1}{2}^2 - \frac{1}{32}^2$

The majority of the terms cancel, leaving $1 + \frac{1}{4} - \frac{1}{1024}$, which simplifies down to $\frac{1024 + \left(256 - 1\right)}{1024}$. Thus, $m-n = \boxed{255}$.

Alternatively, take the area of the first square and add $\,\frac{3}{4}$ of the areas of the remaining squares. This results in $1+ \frac{3}{4}\left[\left(\frac{1}{2}\right)^2 + \ldots + \left(\frac{1}{16}^2\right)\right]$, which when simplified will produce the same answer.

See also

1995 AIME (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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