Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (rv)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
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]]
−
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a <math> 50\% </math> chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumilated to decide the ranks of the teams. In the first game of the tournament, team <math> A </math> beats team <math> B. </math> The probability that team <math> A </math> finishes with more points than team <math> B </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers. Find <math> m+n. </math>
 
  
 
== Solution ==
 
== Solution ==
−
You can break this into cases based on how many rounds A wins out of the remaining 5 games.
+
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>.  
−
 
 
−
If A wins 0 games, then B must win 0 games and the probability of this is <math> \frac{{0 \choose 5}}{2^5} \frac{{0 \choose 5}}{2^5} = \frac{1}{1024} </math>.
 
−
 
 
−
If A wins 1 games, then B must win 1 or less games and the probability of this is <math> \frac{{1 \choose 5}}{2^5} \frac{{0 \choose 5}+{1 \choose 5}}{2^5} = \frac{5}{1024} </math>.
 
−
 
 
−
If A wins 2 games, then B must win 2 or less games and the probability of this is <math> \frac{{2 \choose 5}}{2^5} \frac{{0 \choose 5}+{1 \choose 5}+{2 \choose 5}}{2^5} = \frac{160}{1024} </math>.
 
−
 
 
−
If A wins 3 games, then B must win 3 or less games and the probability of this is <math> \frac{{3 \choose 5}}{2^5} \frac{{0 \choose 5}+{1 \choose 5}+{2 \choose 5}+{3 \choose 5}}{2^5} = \frac{260}{1024} </math>.
 
−
 
 
−
If A wins 4 games, then B must win 4 or less games and the probability of this is <math> \frac{{4 \choose 5}}{2^5} \frac{{0 \choose 5}+{1 \choose 5}+{2 \choose 5}+{3 \choose 5}+{4 \choose 5}}{2^5} = \frac{155}{1024} </math>.
 
  
−
If A wins 5 games, then B must win 5 or less games and the probability of this is <math> \frac{{5 \choose 5}}{2^5} \frac{{0 \choose 5}+{1 \choose 5}+{2 \choose 5}+{3 \choose 5}+{4 \choose 5}+{5 \choose 5}}{2^5} = \frac{32}{1024} </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>\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>.  
  
−
Summing these 6 cases, we get <math> \frac{638}{1024} </math>, which simplifies to <math> \frac{319}{512} </math>, so out answer is <math>319 + 512 = 831</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 19:33, 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 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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_10&oldid=16998"