Difference between revisions of "1995 AIME Problems/Problem 1"
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Latest revision as of 18:29, 4 July 2013
Problem
Square 
is 
For 
the lengths of the sides of square 
are half the lengths of the sides of square 
two adjacent sides of square 
are perpendicular bisectors of two adjacent sides of square 
and the other two sides of square are the perpendicular bisectors of two adjacent sides of square 
The total area enclosed by at least one of 
can be written in the form 
where 
and 
are relatively prime positive integers. Find 
Solution
The sum of the areas of the squares if they were not interconnected is a geometric sequence:
Then subtract the areas of the intersections, which is 
:
![$1^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{16}\right)^2 - \left[\left(\frac{1}{4}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{16}\right)^2 + \left(\frac{1}{32}\right)^2\right]$](http://latex.artofproblemsolving.com/6/0/9/609fafb8171705bc071b28b2302550e4282c0eab.png)
The majority of the terms cancel, leaving 
, which simplifies down to 
. Thus, 
.
Alternatively, take the area of the first square and add 
of the areas of the remaining squares. This results in ![$1+ \frac{3}{4}\left[\left(\frac{1}{2}\right)^2 + \ldots + \left(\frac{1}{16}^2\right)\right]$](http://latex.artofproblemsolving.com/f/1/a/f1ad25f7cc94b2582dca91b9ebdd23b3bec63068.png)
, 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
