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]]}} | ||
+ | |||
+ | == 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 6: | Line 8: | ||
==Solution== | ==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. 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 <math>A</math> and <math>B</math>. Consider two cases: | ||
+ | |||
+ | '''Case 1''': No line passes through both <math>A</math> and <math>B</math> | ||
+ | |||
+ | 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 2''': There is a line passing through <math>A</math> and <math>B</math> | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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 distinct points that lie on two or more of the lines. What is the sum of all possible values of ?
Solution
It is possible to obtain , , , , , and points of intersection, as demonstrated in the following figures:
It is clear that the maximum number of possible intersections is , 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 and . Consider two cases:
Case 1: No line passes through both and
Then, since an intersection is obtained by an intersection between at least two lines, two lines pass through each of and . Then, since there can be no additional intersections, no line that passes through can intersect a line that passes through , and so each line that passes through must be parallel to every line that passes through . Then the two lines passing through are parallel to each other by transitivity of parallelism, so they coincide, contradiction.
Case 2: There is a line passing through and
Then there must be a line passing through , and a line passing through . These lines must be parallel. The fourth line must pass through either or . Without loss of generality, suppose passes through . Then since and cannot coincide, they cannot be parallel. Then and 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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.