Difference between revisions of "2004 AMC 12A Problems/Problem 13"
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=== Solution 2 === | === Solution 2 === | ||
There are <math>{9 \choose 2} = 36</math> 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 <math>{3 \choose 2} = 3</math> times, so we have to subtract <math>2</math> for each of these lines. Quick counting yields <math>3</math> horizontal, <math>3</math> vertical, and <math>2</math> diagonal lines, so the answer is <math>36 - 2(3+3+2) = 20</math> distinct lines. | There are <math>{9 \choose 2} = 36</math> 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 <math>{3 \choose 2} = 3</math> times, so we have to subtract <math>2</math> for each of these lines. Quick counting yields <math>3</math> horizontal, <math>3</math> vertical, and <math>2</math> diagonal lines, so the answer is <math>36 - 2(3+3+2) = 20</math> distinct lines. | ||
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| + | === Solution 3 === | ||
| + | First consider how many lines go through <math>(-1, -1)</math> and hit two points in <math>S</math>. You can see that there are <math>5</math> such lines. Now, we cross out <math>(-1, -1)</math> and make sure to never consider consider lines that go through it anymore (As doing so would be double counting). Repeat for <math>(-1, 0)</math>, making sure not to count the vertical line as it goes through the crossed out <math>(-1, -1)</math>. Then cross out <math>(-1, 0)</math>. Repeat for the rest, and count <math>20</math> lines in total. | ||
== See also == | == See also == | ||
Revision as of 23:10, 4 November 2018
Problem
Let 
be the set of points 
in the coordinate plane, where each of 
and 
may be 
, 
, or 
. How many distinct lines pass through at least two members of ?
Solution
Solution 1
Let's count them by cases:
- Case 1: The line is horizontal or vertical, clearly
. - Case 2: The line has slope
, with 
through 
and 
additional ones one unit above or below those. These total 
. - 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
lines.
.
, with 
and 
additional ones one unit above or below those. These total 
.
lines.These add up to 
.
Solution 2
There are 
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 
times, so we have to subtract 
for each of these lines. Quick counting yields 
horizontal, vertical, and diagonal lines, so the answer is 
distinct lines.
Solution 3
First consider how many lines go through 
and hit two points in . You can see that there are 
such lines. Now, we cross out and make sure to never consider consider lines that go through it anymore (As doing so would be double counting). Repeat for 
, making sure not to count the vertical line as it goes through the crossed out . Then cross out . Repeat for the rest, and count 
lines in total.
See also
| 2004 AMC 12A (Problems • Answer Key • Resources) | |
| 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 | |
