# Thread: Visualization Problem: Dots and Lines

1. ## Visualization Problem: Dots and Lines

Five dots are arranged in space so that no more than three at a time can have a single flat surface pass through them. If each group of three dots has a flat surface pass through it and extend an infinite distance in every direction, what is the maximum number of different lines at which these surfaces may intersect one another?

2. Originally Posted by bennyJ
Five dots are arranged in space so that no more than three at a time can have a single flat surface pass through them. If each group of three dots has a flat surface pass through it and extend an infinite distance in every direction, what is the maximum number of different lines at which these surfaces may intersect one another?
What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?

When you reply, please also let us know what topics you've recently studied, including any formulas or algorithms that you think might apply. Thank you!

3. Originally Posted by stapel
What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?

When you reply, please also let us know what topics you've recently studied, including any formulas or algorithms that you think might apply. Thank you!
To be honest, I'm not entirely sure what the question is asking specifically and what the parameters are. Can anyone help shape this problem more concretely for me?

4. ## From 5 points choose 3, the combination is 10 (5x4/2)

Therefore, there are 10 different planes that go though these 5 dots. Each two planes intersect to form a line. So from 10 planes choose 2 plane to get a line, the ways is 10x9/2=45, which is the answer.

I made a mistake. After I get 10 planes, I should count the lines, because 3 plane could share the same line.
Let the 5 points be 12345. The 10 plane contain the points are
123 (the plane contain 123)
124
125
134
135
145
234
235
245
345
When two points appear on more than 2 row , it means more than two planes will share the line, therefore, we need to subtract them. So the answer should be 45-20=25

5. Originally Posted by yma16
Therefore, there are 10 different planes that go though these 5 dots. Each two planes intersect to form a line. So from 10 planes choose 2 plane to get a line, the ways is 10x9/2=45, which is the answer.

I made a mistake. After I get 10 planes, I should count the lines, because 3 plane could share the same line.
Let the 5 points be 12345. The 10 plane contain the points are
123 (the plane contain 123)
124
125
134
135
145
234
235
245
345
When two points appear on more than 2 row , it means more than two planes will share the line, therefore, we need to subtract them. So the answer should be 45-20=25
Does everyone agree with this solution or have the same approach and way of explaining the problem?

6. Well, If I understand the problem, I do not agree with yma's answer. You cannot determine 25 distinct lines from 5 points. You can at most determine 10 distinct lines. But the error in yma's approach is far from obvious.

Label the points A, B, C, D, and E. Three points define a plane surface. So what is the maximum number planes to be considered? Notice that planes ABC, ACB, BAC, BCA, CAB, and CBA are the same plane. So we are dealing with combinations rather than permutations.

$\therefore \text { maximum number of planes } = \dbinom{5}{3} = \dfrac{5!}{3! * (5 - 3)!} = \dfrac{5 * 4}{2} = 10.$

So far I agree with yma's answer. Here are the planes.

ABC
ABD
ABE
ACD
ACE
BCD
BCE
BDE
CDE

Let's consider planes ABC, ABD, and ABE. Those are three distinct planes because, by hypothesis, no more than three of the five points lie in the same plane. Moreover, none of those three planes is parallel to either of the other two because they all contain the line joining A and B. We have three planes intersecting in one line. So yma is also correct that each intersection of planes does not create a distinct line.

A line joining any two of the five points creates a line. How many distinct pairs of points can we pick out of five distinct points, remembering that the same line joins A and B as joins B and A.

$\text {number of lines of intersection } = \dbinom{5}{2} = \dfrac{5!}{2! * (5 - 2)!} = \dfrac{5 * 4 * 3}{3 * 2} = 10.$

What are those lines and in what planes are they found?

AB: ABC ABD ABE
AC: ABC ACD ACE
BC: ABC BCD BCE
BD: ABD BCD BDE
BE: ABE BCE BDE
CD: ACD BCD CDE
CE: ACE BCE DCE

7. Originally Posted by JeffM
Well, If I understand the problem, I do not agree with yma's answer. You cannot determine 25 distinct lines from 5 points. You can at most determine 10 distinct lines. But the error in yma's approach is far from obvious.

Label the points A, B, C, D, and E. Three points define a plane surface. So what is the maximum number planes to be considered? Notice that planes ABC, ACB, BAC, BCA, CAB, and CBA are the same plane. So we are dealing with combinations rather than permutations.

$\therefore \text { maximum number of planes } = \dbinom{5}{3} = \dfrac{5!}{3! * (5 - 3)!} = \dfrac{5 * 4}{2} = 10.$

So far I agree with yma's answer. Here are the planes.

ABC
ABD
ABE
ACD
ACE
BCD
BCE
BDE
CDE

Let's consider planes ABC, ABD, and ABE. Those are three distinct planes because, by hypothesis, no more than three of the five points lie in the same plane. Moreover, none of those three planes is parallel to either of the other two because they all contain the line joining A and B. We have three planes intersecting in one line. So yma is also correct that each intersection of planes does not create a distinct line.

A line joining any two of the five points creates a line. How many distinct pairs of points can we pick out of five distinct points, remembering that the same line joins A and B as joins B and A.

$\text {number of lines of intersection } = \dbinom{5}{2} = \dfrac{5!}{2! * (5 - 2)!} = \dfrac{5 * 4 * 3}{3 * 2} = 10.$

What are those lines and in what planes are they found?

AB: ABC ABD ABE
AC: ABC ACD ACE
BC: ABC BCD BCE
BD: ABD BCD BDE
BE: ABE BCE BDE
CD: ACD BCD CDE
CE: ACE BCE DCE
Great reasoning. How would you answer this one: Six dots are marked on a flat surface so that no more than two at a time may be connected by any straight line of whatever length. If a straight line is drawn through each and every pair of dots and extends beyond them any desired distance, what is the maximum number of different points at which these lines may intersect one another?

8. Originally Posted by JeffM
A line joining any two of the five points creates a line. How many distinct pairs of points can we pick out of five distinct points, remembering that the same line joins A and B as joins B and A.

$\text {number of lines of intersection } = \dbinom{5}{2} = \dfrac{5!}{2! * (5 - 2)!} = \dfrac{5 * 4 * 3}{3 * 2} = 10.$

What are those lines and in what planes are they found?

AB: ABC ABD ABE
AC: ABC ACD ACE
BC: ABC BCD BCE
BD: ABD BCD BDE
BE: ABE BCE BDE
CD: ACD BCD CDE
CE: ACE BCE DCE
I think you've wrongly assumed that the line where any two planes intersect is the line determined by two of the points. What about the intersection of the planes ABC and CDE? That's a line not in your list.

9. There definitely seem to be 10 distinct groupings of 3 dots comprised of 10 distinct lines...but is that really what the question is asking for?

10. Originally Posted by bennyJ
Five dots are arranged in space so that no more than three at a time can have a single flat surface pass through them. If each group of three dots has a flat surface pass through it and extend an infinite distance in every direction, what is the maximum number of different lines at which these surfaces may intersect one another?
Is the question how many times the ten lines that were produced from ten surfaces can intersect one another? If so, I would imagine that the answer would be higher than ten.

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