Mathmaticians please!!!?

Question

There are six lines so that each line intersects each other line and no three intersect at the same point. How many triangles are there? Come up with an empirical pattern/theoretical understanding.

Please communicate your solutions neatly and fully.

Answer 1

If I'm not mistaken, for every n lines, there are (n-2)^2 triangles.

If you draw three lines, you'll see you only have 1 triangle.

3 - 2 = 1... and 1^2 = 1

If you draw a fourth line, you'll see you can now make 4 triangles.  (Triangles are allowed to overlap).  4-2 = 2... 2^2 = 4

If you draw a 5th line, you can make 9 triangles!

5-2 = 3... 3^2 = 9

You might try it out yourself for 6 lines, you should get 16 triangles.

Once you draw your lines that no 3 lines intersect at the same point, you should label all the intersection points.  Maybe A, B, C, etc.

That way it will help you keep track of all the different triangles so you don't count the same one twice.

What I gave you is only a conjecture, not a proof!  A proof would be much more difficult and I don't think I'm up for the task.  I'm just not that good.

Take care,

David

Answer 2

Draw yourself a "Star of David"  for openers.  There are six.  However, I would not speculate on whether that holds true for all other possible solutions.