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| − | == Problem ==
| + | #REDIRECT [[2006 AIME I Problems/Problem 10]] |
| − | Eight circles of diameter 1 are packed in the first quadrant of the coordinte plane as shown. Let region <math> \mathcal{R} </math> be the union of the eight circular regions. Line <math> l, </math> with slope 3, divides <math> \mathcal{R} </math> into two regions of equal area. Line <math> l </math>'s equation can be expressed in the form <math> ax=by+c, </math> where <math> a, b, </math> and <math> c </math> are positive integers whose greatest common divisor is 1. Find <math> a^2+b^2+c^2. </math>
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| − | [[Image:2006AimeI10.PNG]]
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| − | == Solution ==
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| − | Assume that if unit [[square]]s 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 <math>(x, 0)</math>.
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| − | The line divides the region into 2 sections. The left piece is a [[trapezoid]], with its area <math>\frac{1}{2}((x) + (x+1))(3) = 3x + \frac{3}{2}</math>. The right piece is the addition of a [[trapezoid]] and a [[rectangle]], and the areas are <math>\frac{1}{2}((1-x) + (2-x))(3)</math> and <math>2 \cdot 1 = 2</math>, totaling <math>\frac{13}{2} - 3x</math>. Since we want the two regions to be equal, we find that <math>3x + \frac 32 = \frac {13}2 - 3x</math>, so <math>x = \frac{5}{6}</math>.
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| − | We have that <math>\left(\frac 56, 0\right)</math> is a point on the line of slope 3, so <math>0 = 3 \cdot \frac 56 + b</math> and <math>b = -\frac{5}{2}</math>. In [[y-intercept]] form, the equation of the line is <math>y = 3x - \frac{5}{2}</math>, and in the form for the answer, the line’s equation is <math>2y + 5 = 6x</math>. Thus, our answer is <math>2^2 + 5^2 + 6^2 = 065</math>.
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| − | == See also ==
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| − | *[[2006 AIME II Problems/Problem 9 | Previous problem]]
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| − | *[[2006 AIME II Problems/Problem 11 | Next problem]]
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| − | *[[2006 AIME II Problems]]
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| − | [[Category:Intermediate Combinatorics Problems]]
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