Difference between revisions of "2006 AIME I Problems/Problem 10"
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Revision as of 19:05, 4 July 2013
Problem
Eight circles of diameter 1 are packed in the first quadrant of the coordinate plane as shown. Let region 
be the union of the eight circular regions. Line 
with slope 3, divides into two regions of equal area. Line 
's equation can be expressed in the form 
where 
and 
are positive integers whose greatest common divisor is 1. Find 
|
Solution
Solution 1
The line passing through the tangency point of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the four aforementioned circles splitting into congruent areas, and there are an additional two circles on each side. The line passes through 
and 
, which can be easily solved to be 
. Thus, 
.
Solution 2
Assume that if unit squares are drawn circumscribing the circles, then the line will divide the area of the concave hexagonal region of the squares equally (as of yet, there is no substantiation that such would work, and definitely will not work in general). Denote the intersection of the line and the x-axis as 
.
The line divides the region into 2 sections. The left piece is a trapezoid, with its area 
. The right piece is the addition of a trapezoid and a rectangle, and the areas are 
and 
, totaling 
. Since we want the two regions to be equal, we find that 
, so 
.
We have that 
is a point on the line of slope 3, so 
. Our answer is 
.
We now assess the validity of our starting assumption. We can do that by seeing that our answer passes through the tangency of the two circles, cutting congruent areas, a result explored in solution 1.
See also
| 2006 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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. 
