Difference between revisions of "2004 AMC 12A Problems/Problem 13"

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== Solution ==
 
== Solution ==
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[[Image:2004_AMC12A-13.png|center]]
 
=== Solution 1 ===
 
=== Solution 1 ===
 
Let's count them by cases:
 
Let's count them by cases:
  
{{image}}
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[[Image:2004_AMC12A-13b.png|center]]
  
 
*'''Case 1''': The line is horizontal or vertical, clearly <math>3 \cdot 2 = 6</math>.
 
*'''Case 1''': The line is horizontal or vertical, clearly <math>3 \cdot 2 = 6</math>.

Revision as of 19:32, 3 December 2007

Problem

Let $S$ be the set of points $(a,b)$ in the coordinate plane, where each of $a$ and $b$ may be $- 1$, $0$, or $1$. How many distinct lines pass through at least two members of $S$?

$\text {(A)}\ 8 \qquad \text {(B)}\ 20 \qquad \text {(C)}\ 24 \qquad \text {(D)}\ 27\qquad \text {(E)}\ 36$

Solution

2004 AMC12A-13.png

Solution 1

Let's count them by cases:

2004 AMC12A-13b.png
  • Case 1: The line is horizontal or vertical, clearly $3 \cdot 2 = 6$.
  • Case 2: The line has slope $\pm 1$, with $2$ through $(0,0)$ and $4$ additional ones one unit above or below those. These total $6$.
  • Case 3: The only remaining lines pass through two points, a vertex and a non-vertex point on the opposite side. Thus we have each vertex pairing up with two points on the two opposites sides, giving $4 \cdot 2 = 8$ lines.

These add up to $6+6+8=20\ \mathrm{(B)}$.

Solution 2

There are ${9 \choose 2} = 36$ ways to pick two points, but we've clearly overcounted all of the lines which pass through three points. In fact, each line which passes through three points will have been counted ${3 \choose 2} = 3$ times, so we have to subtract $2$ for each of these lines. Quick counting yields $3$ horizontal, $3$ vertical, and $2$ diagonal lines, so the answer is $36 - 2(3+3+2) = 20$ distinct lines.

See also

2004 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions