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{Combinatorics} Finding the number of triangles in a plane

  1. Aug 16, 2015 #1
    1. The problem statement, all variables and given/known data
    In the coordinate plane let ##A_i=(i,1)## for ##l\leq i\leq15##, and let ##A_i=(i-15,4)## for ##16\leq i \leq 30##. Find the number of all isosceles triangles, where all three vertices belong to the set ##\{A_1,A_2, \cdots,A_{30}\}##

    2. Relevant equations
    Knowing the number of vertices that are equal distance?

    3. The attempt at a solution

    I have really no idea what to do since I don't really know how to reword the question or of similar problems. I know that the vertices will all be equal distance apart from each other and however many vertices I have in the plane I will divide that by 3 to get the left over number of triangles. Is this some what in the right direction? Or any relevant wikipedia articles I can look at that might give me some insight. I know this is a weak attempt but I don't have background to know what to try really.
    Last edited: Aug 16, 2015
  2. jcsd
  3. Aug 16, 2015 #2


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    An equilateral triangle has all three sides of equal length.

    An isosceles triangle has two sides of equal length; some consider an equilateral triangle to be a special case of an isosceles triangle.

  4. Aug 16, 2015 #3
    So I plot the points do I have to worry about using a point more then once such as 123 345

    I thought it read equilateral for some reason so yeah I know two points will be equal distance to another point. Where do I go from that thought? Plot them all would be a guess.
  5. Aug 16, 2015 #4


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    You have a set of 30 points. There may be more than one isosceles triangle which has a vertex at one of these thirty points.
  6. Aug 16, 2015 #5
    I have to account for all the repeats how would I use the choose function so I don't count them all because I feel it will be quite a large number. Like what are my constraints is what i'm having trouble seeing. I can't count them by hand I see now.

    Like with those 30 points can I use them as many times as I want to make triangles or once its on the plane I can't generate it again.

    If I can't generate it again I feel it should just be ##14## triangles going down the line 123, 345 etc.
  7. Aug 16, 2015 #6
    So if know that the triangle will either have 2 vertices in the top row and 1 in the bottom row, or vice versa. If I have 2 vertices in the top row, how many ways can I choose a vertex in the bottom row so the triangle is isosceles?
  8. Aug 17, 2015 #7


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    The most apparent cases are those like: (2,1),(1,4),(3,4) or (8,1), (3,4), (13,4) having two vertices in the upper set and one in the lower as well as their mirror images.
    Next, you will want to consider triangles like (1,1), (1,4), (4,4) where one pair of vertices are on the same x coordinate and the other vertex is three away.
    Finally, you would look for any distances where the diagonal could be a whole number, eg (1,1), (1,6), (5,4) where the diagonal from (1,1) to (5,4) is length 5.
    Once you account for those three types of triangles, you will need to find a method to count how many are possible.

    This seems like it might take some time. I recommend only looking at the set with one vertex on one side, then double the count, since there is symmetry.
  9. Aug 18, 2015 #8


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    Since they're symmetric about a vertical axis, the mirror images are the same triangles.
    6 is not a valid value for y, so I'm not sure what you meant here.
    youngstudent, you need to do a bit of geometry here. Suppose there is an isosceles triangle with two vertices on the lower line snd one on the upper line, and the side along the lower line is equal to one of the others. The Ai point on the lower line directly below the vertex on the upper line lies between the two vertices on the lower line. If the distances from it to those two vertices are a and b, what equation can you write?
  10. Aug 18, 2015 #9


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    Sorry, I meant (1,1) (6,1) (5,4).
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