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

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{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #14]] and [[2019 AMC 12A Problems|2019 AMC 12A #8]]}}
 
{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #14]] and [[2019 AMC 12A Problems|2019 AMC 12A #8]]}}
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== Problem ==
  
 
For a set of four distinct lines in a plane, there are exactly <math>N</math> distinct points that lie on two or more of the lines. What is the sum of all possible values of <math>N</math>?
 
For a set of four distinct lines in a plane, there are exactly <math>N</math> distinct points that lie on two or more of the lines. What is the sum of all possible values of <math>N</math>?
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<math>\textbf{(A) } 14 \qquad \textbf{(B) } 16 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 21</math>
 
<math>\textbf{(A) } 14 \qquad \textbf{(B) } 16 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 21</math>
  
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==Solution 1==
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==Solution==
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It is possible to obtain <math>0</math>, <math>1</math>, <math>3</math>, <math>4</math>, <math>5</math>, and <math>6</math> points of intersection, as demonstrated in the following figures:
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It is possible to obtain 0, 1, 3, 4, 5, and 6 intersections, as demonstrated in the following figures:
 
 
<asy>
 
<asy>
 
unitsize(2cm);
 
unitsize(2cm);
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draw((-1/3,1-d)--(-1/3,-1-d),Arrows);
 
draw((-1/3,1-d)--(-1/3,-1-d),Arrows);
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draw((1/3,1-d)--(1/3,-1-d),Arrows);
 
draw((1/3,1-d)--(1/3,-1-d),Arrows);
 
draw((-1,-1/3-d)--(1,-1/3-d),Arrows);
 
draw((-1,-1/3-d)--(1,-1/3-d),Arrows);
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</asy>
 
</asy>
  
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It is clear that the maximum number of possible intersections is <math>{4 \choose 2} = 6</math>, since each pair of lines can intersect at most once. We now prove that it is impossible to obtain two intersections.
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It is clear that the maximum number of possible intersections is <math>{4 \choose 2} = 6</math>, since each pair of lines can intersect at most once. In addition, by looking at the answer choices, we know that we cannot have 7 points of intersection or else our answer would be greater than the given answer choices. Our answer is given by the sum <math>0+1+3+4+5+6=\boxed{\textbf{(D)} 19}</math>.
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We proceed by contradiction.  Assume a configuration of four lines exists such that there exist only two intersection points.  Let these intersection points be <math>A</math> and <math>B</math>.  Consider two cases:
 
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Case 1:  No line passes through both <math>A</math> and <math>B</math>
 
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Then, since an intersection is obtained by an intersection between at least two lines, two lines pass through each of <math>A</math> and <math>B</math>.  Then, since there can be no additional intersections, no line that passes through <math>A</math> can intersect a line that passes through <math>B</math>, and so each line that passes through <math>A</math> must be parallel to every line that passes through <math>B</math>.  Then the two lines passing through <math>B</math> are parallel to each other by transitivity of parallelism, so they coincide, contradiction.
 
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Case 2:  There is a line passing through <math>A</math> and <math>B</math>
 
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Then there must be a line <math>l_a</math> passing through <math>A</math>, and a line <math>l_b</math> passing through <math>B</math>.  These lines must be parallel.  The fourth line <math>l</math> must pass through either <math>A</math> or <math>B</math>.  Without loss of generality, suppose <math>l</math> passes through <math>A</math>.  Then since <math>l</math> and <math>l_a</math> cannot coincide, they cannot be parallel. Then <math>l</math> and <math>l_b</math> cannot be parallel either, so they intersect, contradiction.
 
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All possibilities have been exhausted, and thus we can conclude that two intersections is impossible.  Our answer is given by the sum <math>0+1+3+4+5+6=\boxed{19}</math>, or <math>\boxed{\text{D}}</math>.  
 
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(Thomas Lam)
 
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==Solution 2(David C)==
 
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We do casework to find values that work
 
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Case 1: Four Parallel Lines= 0 Intersections
 
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Case 2: Three Parallel Lines and One Line Intersecting the Three Lines= 3 Intersections
 
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Case 3: Two Parallel Lines with another Two Parallel Lines= 4 Intersections
 
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Case 4: Two Parallel Lines with Two Other Non-Parallel Lines=5 Intersections
 
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Case 5: Four Non-Parallel Lines All Intersecting Each Other at different points = 6 Intersections
 
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Case 6: Four Non-Parallel Lines All Intersecting At One Point= 1 Intersection
 
  
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You can find out that you cannot have 2 Intersections
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==Video Solution 1==
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https://youtu.be/-0s2xGhU8wM
  
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Sum= <math>1+3+4+5+6</math>= <math>19  \boxed{D}</math>
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~Education, the Study of Everything
  
 
==See Also==
 
==See Also==
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{{AMC12 box|year=2019|ab=A|num-b=7|num-a=9}}
 
{{AMC12 box|year=2019|ab=A|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category: Introductory Combinatorics Problems]]

Latest revision as of 18:55, 26 October 2025

The following problem is from both the 2019 AMC 10A #14 and 2019 AMC 12A #8, so both problems redirect to this page.

Problem

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

It is possible to obtain $0$, $1$, $3$, $4$, $5$, and $6$ points of intersection, as demonstrated in the following figures:


[asy] unitsize(2cm); real d = 2.5; draw((-1,.6)--(1,.6),Arrows); draw((-1,.2)--(1,.2),Arrows); draw((-1,-.2)--(1,-.2),Arrows); draw((-1,-.6)--(1,-.6),Arrows); draw((-1+d,0)--(1+d,0),Arrows); draw((0+d,1)--(0+d,-1),Arrows); draw(dir(45)+(d,0)--dir(45+180)+(d,0),Arrows); draw(dir(135)+(d,0)--dir(135+180)+(d,0),Arrows); dot((0+d,0)); draw((-1+2*d,sqrt(3)/3)--(1+2*d,sqrt(3)/3),Arrows); draw((-1/4-1/2+2*d, sqrt(3)/12-sqrt(3)/2)--(-1/4+1/2+2*d,sqrt(3)/12+sqrt(3)/2),Arrows); draw((1/4+1/2+2*d, sqrt(3)/12-sqrt(3)/2)--(1/4-1/2+2*d,sqrt(3)/12+sqrt(3)/2),Arrows); draw((-1+2*d,-sqrt(3)/6)--(1+2*d,-sqrt(3)/6),Arrows); dot((0+2*d,sqrt(3)/3)); dot((-1/2+2*d,-sqrt(3)/6)); dot((1/2+2*d,-sqrt(3)/6)); draw((-1/3,1-d)--(-1/3,-1-d),Arrows); draw((1/3,1-d)--(1/3,-1-d),Arrows); draw((-1,-1/3-d)--(1,-1/3-d),Arrows); draw((-1,1/3-d)--(1,1/3-d),Arrows); dot((1/3,1/3-d)); dot((-1/3,1/3-d)); dot((1/3,-1/3-d)); dot((-1/3,-1/3-d)); draw((-1+d,sqrt(3)/12-d)--(1+d,sqrt(3)/12-d),Arrows); draw((-1/4-1/2+d, sqrt(3)/12-sqrt(3)/2-d)--(-1/4+1/2+d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw((1/4+1/2+d, sqrt(3)/12-sqrt(3)/2-d)--(1/4-1/2+d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw((-1+d,-sqrt(3)/6-d)--(1+d,-sqrt(3)/6-d),Arrows); dot((0+d,sqrt(3)/3-d)); dot((-1/2+d,-sqrt(3)/6-d)); dot((1/2+d,-sqrt(3)/6-d)); dot((-1/4+d,sqrt(3)/12-d)); dot((1/4+d,sqrt(3)/12-d)); draw((-1/4-1/2+2*d, sqrt(3)/12-sqrt(3)/2-d)--(-1/4+1/2+2*d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw((1/4+1/2+2*d, sqrt(3)/12-sqrt(3)/2-d)--(1/4-1/2+2*d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw(dir(30)+(2*d,-d)--dir(30+180)+(2*d,-d),Arrows); draw(dir(150)+(2*d,-d)--dir(-30)+(2*d,-d),Arrows); dot((0+2*d,0-d)); dot((0+2*d,sqrt(3)/3-d)); dot((-1/2+2*d,-sqrt(3)/6-d)); dot((1/2+2*d,-sqrt(3)/6-d)); dot((-1/4+2*d,sqrt(3)/12-d)); dot((1/4+2*d,sqrt(3)/12-d)); [/asy]

It is clear that the maximum number of possible intersections is ${4 \choose 2} = 6$, since each pair of lines can intersect at most once. In addition, by looking at the answer choices, we know that we cannot have 7 points of intersection or else our answer would be greater than the given answer choices. Our answer is given by the sum $0+1+3+4+5+6=\boxed{\textbf{(D)} 19}$.

Video Solution 1

https://youtu.be/-0s2xGhU8wM

~Education, the Study of Everything

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

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