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

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==Solution==
 
==Solution==
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Drawing <math>4</math> lines, you see that the maximum points of intersection is <math>6</math>. Then, we see that we can not make <math>5</math> intersection points and can just make a quadrilateral, triangle, two T's, and one pair of coordinate axis to make <math>4,3,2,</math> and <math>1</math> points of intersection. Thus the answer is <math>1+2+3+4+6</math> to get <math>16, B</math>
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-Lcz
  
 
==See Also==
 
==See Also==

Revision as of 17:35, 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

Drawing $4$ lines, you see that the maximum points of intersection is $6$. Then, we see that we can not make $5$ intersection points and can just make a quadrilateral, triangle, two T's, and one pair of coordinate axis to make $4,3,2,$ and $1$ points of intersection. Thus the answer is $1+2+3+4+6$ to get $16, B$

-Lcz

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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