Difference between revisions of "2006 AIME A Problems/Problem 10"

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== See also ==
 
== See also ==
*[[2006 AIME II Problems/Problem 9 | Previous problem]]
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{{AIME box|year=2006|n=II|num-b=9|num-a=11}}
*[[2006 AIME II Problems/Problem 11 | Next problem]]
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*[[2006 AIME II Problems]]
 
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]

Revision as of 14:57, 25 September 2007

Problem

Eight circles of diameter 1 are packed in the first quadrant of the coordinte plane as shown. Let region $\mathcal{R}$ be the union of the eight circular regions. Line $l,$ with slope 3, divides $\mathcal{R}$ into two regions of equal area. Line $l$'s equation can be expressed in the form $ax=by+c,$ where $a, b,$ and $c$ are positive integers whose greatest common divisor is 1. Find $a^2+b^2+c^2.$

2006AimeI10.PNG

Solution

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 (proof needed). Denote the intersection of the line and the x-axis as $(x, 0)$.

The line divides the region into 2 sections. The left piece is a trapezoid, with its area $\frac{1}{2}((x) + (x+1))(3) = 3x + \frac{3}{2}$. The right piece is the addition of a trapezoid and a rectangle, and the areas are $\frac{1}{2}((1-x) + (2-x))(3)$ and $2 \cdot 1 = 2$, totaling $\frac{13}{2} - 3x$. Since we want the two regions to be equal, we find that $3x + \frac 32 = \frac {13}2 - 3x$, so $x = \frac{5}{6}$.

We have that $\left(\frac 56, 0\right)$ is a point on the line of slope 3, so $0 = 3 \cdot \frac 56 + b$ and $b = -\frac{5}{2}$. In y-intercept form, the equation of the line is $y = 3x - \frac{5}{2}$, and in the form for the answer, the line’s equation is $2y + 5 = 6x$. Thus, our answer is $2^2 + 5^2 + 6^2 = 065$.

See also

2006 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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