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I got an interesting resut

  1. Apr 28, 2007 #1

    daniel_i_l

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    If you have a chain of links of equal lengths whose ends are connected (so that it forms a loop) then by holding three points and "pulling" them outwards different triangles can be formed. For example, with a chain that has three links 1 triangle can be made. With 4 links no triangles can be made. With 9 links three can be made {(3,3,3),(2,3,4),(1,4,4)} ... I made a program to search through all numbers of links up to 1000000 and look for chains that give the same amount of triangles as there are links. The result was that there's only one - a chain of 48 links can make 48 triangles. And interestingly, numbers lower than 48 give less triangles than links and numbers over 48 give more. By numbers like 1000 you get about 10^6 triangles... also if you take some number of links and keep on multiplying it by the same number, then the number of triangles changes in a fixed pattern. And for every chain, if you either add or subtract 3 links you get the same number of triangles...
    Is there any significance to all this or is it just silly number games?
    Thanks.
     
  2. jcsd
  3. Apr 28, 2007 #2
    I really can't see what you mean...
     
  4. Apr 28, 2007 #3

    daniel_i_l

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    The general idea is just to find triangles with sides of integer length where the sum of the sides is some number. The "number" is the number of links. So for the number three, the only triangle is (1,1,1) cause the only three integers whose sum is 3 is 1+1+1. For 9 you get the 3 results posted above... Is that what you where asking?
     
  5. Apr 28, 2007 #4
    Oh wow, that's really interesting. I can't think of any use, but I'm sure that it has a lot of implications for discrete math in general.
     
  6. Apr 29, 2007 #5

    AKG

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    3 numbers x, y, z form a chain that can be made into a triangle of perimeter n iff x+y+z = n and max{x,y,z} < [n/2] where [.] is the "round down" function. The reason for the condition "x+y+z = n" is obvious. The second condition basically says that the triangle inequality is satisfied. So given an n, how many (x,y,z) satisfy this? To count this, condition on the size of the smallest side, which has length k anywhere between 1 and [n/3]. So the remaining two sides have to add up to n-k, and both must have length greater than or equal to k, and less than or equal to [n/2]. The next smallest side has length m, so we need [n/2] > m > k, [n/2] > n-(m+k) > m, since n-(m+k) will be the length of the largest side. So this means:

    max{n-k-[n/2], k} < m < min{[n/2], [(n-k)/2]}

    which is just

    max{n-k-[n/2], k} < m < [(n-k)/2]

    For each m satisfying the above, we get one and only one possibility, so we just have to count the number of m satisfying this, and that is exactly

    [(n-k)/2] - max{n-k-[n/2], k}

    So the total number of triangles of perimeter n and integer length sides is:

    [tex]\sum _{k=1} ^{\lfloor n/3 \rfloor}\left \lfloor \frac{n-k}{2} \right \rfloor - \max \left (n-k-\left \lfloor \frac{n}{2} \right \rfloor ,\, k\right )[/tex]

    When is n-k-[n/2] > k? When k < [{n/2}/2], where {.} is the round up function. So the above sum is:

    [tex]\sum _{k=1} ^{\lfloor \lceil (n/2) \rceil /2 \rfloor}\left \lfloor \frac{n-k}{2} \right \rfloor - \left (n-k-\left \lfloor \frac{n}{2} \right \rfloor \right ) + \sum _{k=\lfloor \lceil (n/2) \rceil /2 \rfloor + 1} ^{\lfloor n/3 \rfloor}\left \lfloor \frac{n-k}{2} \right \rfloor - k[/tex]

    [tex]= -\left \lfloor \frac{\lceil n/2 \rceil}{2} \right \rfloor \left \lceil \frac{n}{2} \right \rceil + \sum _{k=1} ^{\lfloor \lceil (n/2) \rceil /2 \rfloor}\left \lfloor \frac{n-k}{2} \right \rfloor + k + \sum _{k=\lfloor \lceil (n/2) \rceil /2 \rfloor + 1} ^{\lfloor n/3 \rfloor}\left \lfloor \frac{n-k}{2} \right \rfloor - k[/tex]

    [tex] = -\left \lfloor \frac{\lceil n/2 \rceil}{2} \right \rfloor \left \lceil \frac{n}{2} \right \rceil + \left \lfloor \frac{\lceil n/2 \rceil}{2}\right \rfloor ^ 2 + \left \lfloor \frac{\lceil n/2 \rceil}{2} \right \rfloor - \frac{1}{2} \left ( \left \lfloor \frac{n}{3}\right \rfloor ^ 2 + \left \lfloor \frac{n}{3} \right \rfloor \right ) + \sum _{k=1} ^{\lfloor n/3 \rfloor}\left \lfloor \frac{n-k}{2} \right \rfloor[/tex]
     
    Last edited: Apr 29, 2007
  7. Apr 29, 2007 #6

    AKG

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    Actually, figure out this number for the first, say, 8 n. You get 0, 0, 1, 0, 1, 1, 2, 1. Go to the On-Line Encyclopedia of Integer Sequences, type in that sequence of numbers, and hit "Search". You'll see your sequence has already been discovered, it's called Alcuin's sequence, and a formula much better than mine is already given for it. It's actually 12 different formulas because it depends on the value of n (mod 12). From that site:

    For n=0..11 (mod 12), a(n) is respectively
    n^2/48
    (n^2 + 6n - 7)/48
    (n^2 - 4)/48
    (n^2 + 6n + 21)/48
    (n^2 - 16)/48
    (n^2 + 6n - 7)/48
    (n^2 + 12)/48
    (n^2 + 6n + 5)/48
    (n^2 - 16)/48
    (n^2 + 6n + 9)/48
    (n^2 - 4)/48
    (n^2 + 6n + 5)/48


    48 certainly does have some significance! By the way, something went wrong in the formula I derived, because when I plug in n=48 I get 52 as the result (unless I just computed wrong).
     
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