Results 1 to 8 of 8

Math Help - 10 random points

  1. #1
    lpd
    lpd is offline
    Member
    Joined
    Sep 2009
    Posts
    100

    10 random points

    Suppose ten points are placed at random inside a triangle whose sides are all of length three. Show that there must be at least two points no more than a distance one apart. Be sure to consider all possible cases.

    So what i did was...
    dividing the triangle into 4 smaller equilateral triangles like this:
    A
    AVA
    (hope you understand what i was trying to draw out) and each length has length 1.

    And I used the pigeohole priciplae. which says if there are n+1 pigeons in n pigeon holes, at least one hall must contain 2 or more pigeon.

    so isn't it obvious that if there are 10 points placed in the triangle, obviously there will be at least 2 points which has a length less than one??

    Am I being tricked here, and what does it mean by "be consider to show all cases"???
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by lpd View Post
    Suppose ten points are placed at random inside a triangle whose sides are all of length three. Show that there must be at least two points no more than a distance one apart. Be sure to consider all possible cases.

    So what i did was...
    dividing the triangle into 4 smaller equilateral triangles like this:
    A
    AVA
    (hope you understand what i was trying to draw out) and each length has length 1.

    And I used the pigeohole priciplae. which says if there are n+1 pigeons in n pigeon holes, at least one hall must contain 2 or more pigeon.

    so isn't it obvious that if there are 10 points placed in the triangle, obviously there will be at least 2 points which has a length less than one??

    Am I being tricked here, and what does it mean by "be consider to show all cases"???
    Maybe it means don't forget to address the possibility that one or more points lie on the edges/vertices of your sub-triangles.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by lpd View Post
    suppose ten points are placed at random inside a triangle whose sides are all of length three. Show that there must be at least two points no more than a distance one apart. Be sure to consider all possible cases.

    So what i did was...
    Dividing the triangle into 4 smaller equilateral triangles like this:
    A.A
    AVA
    (hope you understand what i was trying to draw out) and each length has length 1.

    And i used the pigeonhole priciplae. Which says if there are n+1 pigeons in n pigeon holes, at least one hall must contain 2 or more pigeon.

    So isn't it obvious that if there are 10 points placed in the triangle, obviously there will be at least 2 points which has a length less than one??

    Am i being tricked here, and what does it mean by "be consider to show all cases"???
    Presumably you meant to divide the triangle into 9 smaller triangles:
    AVA
    AAVA
    AVAVA

    It is clear that none of the smaller triangles can contain more than one of the ten points in its interior, but as undefined points out there is nothing to stop one of the smaller triangles containg a point at more than one of its vertices. That is where you need to consider all possible cases carefully.

    Edit. Looking at the question more carefully, I don't think that there is anything wrong with simply using the pigeonhole principle. The question does not ask for two points to have distance less than 1, but for two points with distance no more than 1. If two points are in the same smaller triangle then their distance apart certainly cannot be more than 1.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by Opalg View Post
    Presumably you meant to divide the triangle into 9 smaller triangles:
    AVA
    AAVA
    AVAVA

    It is clear that none of the smaller triangles can contain more than one of the ten points in its interior, but as undefined points out there is nothing to stop one of the smaller triangles containg a point at more than one of its vertices. That is where you need to consider all possible cases carefully.
    That's a clever trick of changing the font color to white to make the letters line up! I drew nine subtriangles on paper but didn't really notice the OP said 4 haha.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,799
    Thanks
    685

    The subdivided triangle looks like this:


    Code:
    
                *
               / \
              *---*
             / \ / \
            *---*---*
           / \ / \ / \
          *---*---*---*

    And I agree with opalg's afterthought.


    FYI: I find that [color=beige] creates invisible characters.

    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by Opalg View Post
    Edit. Looking at the question more carefully, I don't think that there is anything wrong with simply using the pigeonhole principle. The question does not ask for two points to have distance less than 1, but for two points with distance no more than 1. If two points are in the same smaller triangle then their distance apart certainly cannot be more than 1.
    Well I think it's easy to accidentally require the points to be in the interiors of the smaller triangles, thus missing some cases; I wasn't suggesting that a simple application of the pidgeonhole principle did not apply.

    A simple solution is to arbitrarily assign each horizontal edge to the triangle above it, every other edge to the triangle to its left, and the middle vertex to the northwest triangle connected to it.
    Last edited by undefined; September 22nd 2010 at 09:48 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    lpd
    lpd is offline
    Member
    Joined
    Sep 2009
    Posts
    100
    Sorry to bring this topic up again... Just looking through the question again...

    but i did not quite understand undefined's solution:
    "
    A simple solution is to arbitrarily assign each horizontal edge to the triangle above it, every other edge to the triangle to its left, and the middle vertex to the northwest triangle connected to it."

    Does that mean, you are placing the "points" on each of the horizontal edge to the traingle above it? So, if i do so, I have used up 6 points, hence 4 points left, thus, there will be at least 2 points no more than a distance one apart.


    But I don't get this case "the middle vertex to the northwest triangle connected to it?"

    Also, how many possible cases will there there be? Just 3 i pressume?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by lpd View Post
    Sorry to bring this topic up again... Just looking through the question again...

    but i did not quite understand undefined's solution:
    "
    A simple solution is to arbitrarily assign each horizontal edge to the triangle above it, every other edge to the triangle to its left, and the middle vertex to the northwest triangle connected to it."

    Does that mean, you are placing the "points" on each of the horizontal edge to the traingle above it? So, if i do so, I have used up 6 points, hence 4 points left, thus, there will be at least 2 points no more than a distance one apart.


    But I don't get this case "the middle vertex to the northwest triangle connected to it?"

    Also, how many possible cases will there there be? Just 3 i pressume?
    I was just referring to a partition of the original triangle into subtriangles. A partition divides a set into parts whose intersection is empty and whose union is the original set. We don't really need a partition here, but I thought it easiest to define a partition rather than dealing with cases of points that belong to more than one subtriangle. So I was describing a scheme of assigning edges/vertices to certain subtriangles, so that for any point within the original triangle, we can identify which subtriangle it belongs to. See what I mean?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: September 25th 2010, 03:12 AM
  2. Distance between two random Points in a square
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: July 21st 2010, 09:05 AM
  3. Average distance between 2 random points
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 9th 2010, 11:17 PM
  4. Random Points on Convex Sets (Superellipsoids)
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: August 7th 2009, 08:08 AM
  5. N random points in a circle
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: March 20th 2009, 10:00 AM

Search Tags


/mathhelpforum @mathhelpforum