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>. |
| − | {{ | ||
== See also == | == See also == | ||
Revision as of 15:24, 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 | ||
