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

(Solution 1(David C))
m (added "Problem" to the top)
 
(8 intermediate revisions by 4 users not shown)
Line 1: Line 1:
 
{{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>?
Line 5: Line 7:
 
<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 1(David C)==
+
==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));
  
We do casework to find values that work
+
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));
  
Case 1: Four Parallel Lines= 0 Intersections
+
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>
  
Case 2: Three Parallel Lines and One Line Intersecting the Three Lines= 3 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.
  
Case 3: Two Parallel Lines with another Two Parallel Lines= 4 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 <math>A</math> and <math>B</math>.  Consider two cases:
  
Case 4: Two Parallel Lines with Two Other Non-Parallel Lines=5 Intersections
+
'''Case 1''': No line passes through both <math>A</math> and <math>B</math>
  
Case 5: Four Non-Parallel Lines All Intersecting Each Other at different points = 6 Intersections
+
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.
  
Case 6: Four Non-Parallel Lines All Intersecting At One Point= 1 Intersection
+
'''Case 2''': There is a line passing through <math>A</math> and <math>B</math>
  
You can find out that you cannot have 2 Intersections
+
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.
  
Sum= <math>1+3+4+5+6</math>= <math>19 \boxed{D}</math>
+
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{\textbf{(D) } 19}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 01:32, 5 December 2019

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. 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 $0+1+3+4+5+6=\boxed{\textbf{(D) } 19}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS