Difference between revisions of "2019 AMC 10A Problems/Problem 14"

(See Also)
(Solution 1(David C))
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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</math>
+
Sum= <math>1+3+4+5+6</math>= <math>19 \boxed{D}</math>
+
 
 
==See Also==
 
==See Also==
  

Revision as of 19:05, 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 $N$ distinct points that lie on two or more of the lines. What is the sum of all possible values of $N$?

$\textbf{(A) } 14 \qquad \textbf{(B) } 16 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 21$

Solution 1(David C)

We do casework to find values that work

Case 1: Four Parallel Line= 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= $1+3+4+5+6$= $19 \boxed{D}$

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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