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

(Added a complete solution with full diagrams and a proof)
(→‎Solution 1)
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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.
 
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.
  
−
Let the intersection points be <math>A</math> and <math>B</math>.  Consider two cases:
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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:
  
 
Case 1:  No line passes through both <math>A</math> and <math>B</math>
 
Case 1:  No line passes through both <math>A</math> and <math>B</math>

Revision as of 20:20, 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

It is possible to obtain 0, 1, 3, 4, 5, and 6 intersections, 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. We now prove that it is impossible to obtain two intersections.

We proceed by contradiction. Assume a configuration of four lines exists such that there exist only two intersection points. Let these intersection points be $A$ and $B$. Consider two cases:

Case 1: No line passes through both $A$ and $B$

Then, since an intersection is obtained by an intersection between at least two lines, two lines pass through each of $A$ and $B$. Then, since there can be no additional intersections, no line that passes through $A$ can intersect a line that passes through $B$, and so each line that passes through $A$ must be parallel to every line that passes through $B$. Then the two lines passing through $B$ are parallel to each other by transitivity of parallelism, so they coincide, contradiction.

Case 2: There is a line passing through $A$ and $B$

Then there must be a line $l_a$ passing through $A$, and a line $l_b$ passing through $B$. These lines must be parallel. The fourth line $l$ must pass through either $A$ or $B$. Without loss of generality, suppose $l$ passes through $A$. Then since $l$ and $l_a$ cannot coincide, they cannot be parallel. Then $l$ and $l_b$ cannot be parallel either, so they intersect, contradiction.

All possibilities have been exhausted, and thus we can conclude that two intersections is impossible. Our answer is given by the sum $0+1+3+4+5+6=\boxed{19}$, or $\boxed{\text{D}}$.

(Thomas Lam)

Solution 2(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= $1+3+4+5+6$= $19 \boxed{D}$

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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