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Thread: Visualization Problem: Dots and Lines

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    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?

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    Quote Originally Posted by bennyJ View Post
    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!

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    Quote Originally Posted by stapel View Post
    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?

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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
    Last edited by yma16; 01-26-2018 at 12:35 PM. Reason: I made a mistake

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    Quote Originally Posted by yma16 View Post
    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?

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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.

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

    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.

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

    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
    Last edited by JeffM; 02-13-2018 at 12:56 AM.

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    Quote Originally Posted by JeffM View Post
    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.

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

    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.

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

    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
    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?

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    Quote Originally Posted by JeffM View Post
    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.

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

    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.

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    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?

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    Quote Originally Posted by bennyJ View Post
    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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