# Thread: Visualization Problem: Dots and Lines

1. ## Visualization Problem: Dots and Lines

Five.

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!
How.

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
AE: ABE, ACE, ADE
BC: ABC BCD BCE
BD: ABD BCD BDE
BE: ABE BCE BDE
CD: ACD BCD CDE
CE: ACE BCE DCE
DE: ADE BDE CDE

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
AE: ABE, ACE, ADE
BC: ABC BCD BCE
BD: ABD BCD BDE
BE: ABE BCE BDE
CD: ACD BCD CDE
CE: ACE BCE DCE
DE: ADE BDE CDE
Great reasoning.

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
AE: ABE, ACE, ADE
BC: ABC BCD BCE
BD: ABD BCD BDE
BE: ABE BCE BDE
CD: ACD BCD CDE
CE: ACE BCE DCE
DE: ADE BDE CDE
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.

10. Is.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•