Visualization Problem: Dots and Lines

bennyJ

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Five.
 
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stapel

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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! ;)
 

yma16

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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
 
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JeffM

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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.

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

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

ABC
ABD
ABE
ACD
ACE
ADE
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.

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

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

AB: ABC ABD ABE
AC: ABC ACD ACE
AD: ABD, ACD, ADE
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
 
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Dr.Peterson

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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.

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

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

AB: ABC ABD ABE
AC: ABC ACD ACE
AD: ABD, ACD, ADE
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.
 

JeffM

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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.
Clearly you are correct, and I was wrong. If two planes share a point, they are not parallel and so must intersect. So my previous answer was wrong.

Each of the ten planes is defined by three of the five points points so each plane must intersect with each of the others. That validates yma's computation of 45 intersections to be considered before eliminating duplicates:

\(\displaystyle \dbinom{10}{2} = \dfrac{10 * 9}{2} = 45.\)

Now we must eliminate the duplicates. But I do not see how yma calculates that there are 20 duplicates. We definitely have ten lines that represent a common intersection of three planes. I do not see where the other ten come from, but perhaps I am again missing something.
 

Dr.Peterson

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Agreed. Could anyone enlighten this thread's participants about the easiest way to eliminate the duplicates from the count?
Yma16 did so, correctly. They just didn't explain how to get the 20 (which may have been through mere counting).

Here is one way to get it: We have 5 points, which form 10 lines and 10 planes. The 10 planes determine 45 pairs of planes. We want to subtract from the 45, all the "extra" pairs that do not form distinct lines. Each of the 10 lines is the intersection of 3 of the 45 pairs, so we need to subtract 2 of those 3 pairs for each of the 10 lines, that is, 20 "extra" lines. So we get 45 - 20 = 25 distinct lines of intersection.

Here is another way to get the same end result: We have 5 points, which form 10 lines and 10 planes. The 10 planes determine 45 pairs of planes. But each of the 10 lines is the intersection of 3 pairs of planes, so together they account for 3*10 = 30 of those pairs. This leave 45 - 30 = 15 pairs, which (assuming we are convinced that none of these intersect in duplicate lines) give us a total of 10 + 15 lines of intersection: Yma16's 25.

Here's another: We have 10 lines formed by pairs of points; any other intersection of planes must be the intersection of two planes that share only one of the 5 points, like my ABC and CDE. So to count these additional lines, we can count the shared points (5 choices), and choose 2 of the other 4 to form one plane (4C2 = 6 choices), but divide by 2 because we get the same pair of planes if the two we chose are on the other plane. This yields 5 * 6 / 2 = 15 additional lines; adding these to the 10 we know, the total is again 25.

When I can get the same answer three ways, I start to be confident.
 

JeffM

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Yma16 did so, correctly. They just didn't explain how to get the 20 (which may have been through mere counting).

Here is one way to get it: We have 5 points, which form 10 lines and 10 planes. The 10 planes determine 45 pairs of planes. We want to subtract from the 45, all the "extra" pairs that do not form distinct lines. Each of the 10 lines is the intersection of 3 of the 45 pairs, so we need to subtract 2 of those 3 pairs for each of the 10 lines, that is, 20 "extra" lines. So we get 45 - 20 = 25 distinct lines of intersection.
I am feeling abashed.

Planes ABC and ABD intersect in line AB as do planes ABC and ABE as well as planes ABD and ABE. I actually pointed that out in my first post. So the one line AB counts as three of the forty-five intersections. So each of the ten lines I specified is duplicated twice in terms of intersections.

I'll go stand in the corner. Maybe denis will come visit me.
 

yma16

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As I said in my last post, I counted the duplicated line

For example, the following 3 planes intersect each other. They all have a line that pass through the point 1 and 2. So I consider there is an extra line I need to subtract.

123 (the plane contain 123)

124
125

Do the same for all the combinations of the planes, I got the duplicated lines are 20.
 

Dr.Peterson

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For example, the following 3 planes intersect each other. They all have a line that pass through the point 1 and 2. So I consider there is an extra line I need to subtract.

123 (the plane contain 123)

124
125

Do the same for all the combinations of the planes, I got the duplicated lines are 20.
I think you are essentially using my first method, though you may have been thinking a little differently. Did you actually list and count, or use combinations to get the 10 lines and subtract two duplicates for each, as I did?

In any case, good work.
 

yma16

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I really counted them since 10 planes is a small quantity. If

I think you are essentially using my first method, though you may have been thinking a little differently. Did you actually list and count, or use combinations to get the 10 lines and subtract two duplicates for each, as I did?

In any case, good work.
the number of planes is large, we may need to develop a formula to figure out the duplicated lines. In that case, more than three planes can share a line, so it is more complicated.
 

Dr.Peterson

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the number of planes is large, we may need to develop a formula to figure out the duplicated lines. In that case, more than three planes can share a line, so it is more complicated.
You're right about that. I used a little bit of formula, based on an obvious observation; with more, it would become less obvious.
 
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