Difference between revisions of "2006 AIME I Problems/Problem 10"
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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> | 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> | ||
+ | [[Image:2006AimeI10.PNG]] | ||
− | [[ | + | == Solution == |
+ | 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>. | ||
+ | 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>\displaystyle \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>. | ||
− | == | + | 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 == | == See also == |
Revision as of 16:24, 13 March 2007
Problem
Eight circles of diameter 1 are packed in the first quadrant of the coordinte 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
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 .
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
and
. In y-intercept form, the equation of the line is
, and in the form for the answer, the line’s equation is
. Thus, our answer is
.
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 |