Difference between revisions of "2019 AMC 10A Problems/Problem 14"
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You can find out that you cannot have 2 Intersections | You can find out that you cannot have 2 Intersections | ||
| − | Sum= <math>1+3+4+5+6</math>= <math>19 \boxed{D}</math> | + | Sum= <math>1+3+4+5+6</math>= <math>19 \boxed{D}</math> |
==See Also== | ==See Also== | ||
Revision as of 18:06, 9 February 2019
- The following problem is from both the 2019 AMC 10A #14 and 2019 AMC 12A #8, so both problems redirect to this page.
For a set of four distinct lines in a plane, there are exactly 
distinct points that lie on two or more of the lines. What is the sum of all possible values of ?
Solution 1(David C)
We do casework to find values that work
Case 1: Four Parallel Lines= 0 Intersections
Case 2: Three Parallel Lines and One Line Intersecting the Three Lines= 3 Intersections
Case 3: Two Parallel Lines with another Two Parallel Lines= 4 Intersections
Case 4: Two Parallel Lines with Two Other Non-Parallel Lines=5 Intersections
Case 5: Four Non-Parallel Lines All Intersecting Each Other at different points = 6 Intersections
Case 6: Four Non-Parallel Lines All Intersecting At One Point= 1 Intersection
You can find out that you cannot have 2 Intersections
Sum= 
= 
See Also
| 2019 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 10 Problems and Solutions | ||
| 2019 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
