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2019 AMC 10A Problems/Problem 14: Difference between revisions

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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]]}}
== 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>?


<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>
==Solution==
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:
<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 <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>.
==Video Solution 1==
https://youtu.be/-0s2xGhU8wM
~Education, the Study of Everything


==See Also==
==See Also==
Line 10: Line 78:
{{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}}
[[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

These problems are copyrighted © by the Mathematical Association of America.

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