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

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[[Image:AIME 1995 Problem 1.png]]
 
[[Image:AIME 1995 Problem 1.png]]
 
== Solution ==
 
== Solution ==
−
The sum of the areas of the [[square]]s if they were not interconnected is a [[geometric sequence]]: <math>1^2 + (\frac{1}{2})^2 + \1dots + (\frac{1}{16})^2</math>. Then subtract the areas of the intersections: <math>(\frac{1}{4})^2 + \1dots + (\frac{1}{32})^2</math>. The majority of the terms cancel, leaving <math>1 + \frac{1}{4} - \frac{1}{1024}</math>, which simplifies down to <math>\frac{1024 + (256 - 1)}{1024}</math>. Thus, <math>m-n =  255</math>.
+
The sum of the areas of the [[square]]s if they were not interconnected is a [[geometric sequence]]:  
  
−
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}(\frac{1}{2}^2 + \1dots + \frac{1}{16}^2)</math>, which when simplified will produce the same answer.
+
:<math>1^2 + (\frac{1}{2})^2 + (\frac{1}{4})^2 + (\frac{1}{8})^2 + (\frac{1}{16})^2</math>
 +
 
 +
Then subtract the areas of the intersections (<math>(\frac{1}{4})^2 + \ldots + (\frac{1}{32})^2</math>):
 +
 
 +
:<math>1^2 + (\frac{1}{2})^2 + (\frac{1}{4})^2 + (\frac{1}{8})^2 + (\frac{1}{16})^2 - [(\frac{1}{4})^2 + (\frac{1}{8})^2 + (\frac{1}{16})^2 + (\frac{1}{32})^2]</math>
 +
:<math>= 1 + \frac{1}{2}^2 - \frac{1}{32}^2</math>
 +
 
 +
The majority of the terms cancel, leaving <math>1 + \frac{1}{4} - \frac{1}{1024}</math>, which simplifies down to <math>\frac{1024 + (256 - 1)}{1024}</math>. Thus, <math>m-n =  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}(\frac{1}{2}^2 + \ldots + \frac{1}{16}^2)</math>, which when simplified will produce the same answer.
  
 
== See also ==
 
== See also ==

Revision as of 21:56, 8 February 2007

Problem

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

AIME 1995 Problem 1.png

Solution

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

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

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

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

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

Alternatively, take the area of the first square and add $\,\frac{3}{4}$ of the areas of the remaining squares. This results in $1+ \frac{3}{4}(\frac{1}{2}^2 + \ldots + \frac{1}{16}^2)$, 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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