math teaser, fun with triangles

[lotus776]

Given fourteen points draw equidistant to each other in the shape of a pyramid missing the top piece (5 bottom row, 4 the row above, 3 the row above that and finally two on the row above that which is the top), determine the number of triangles that can be drawn formed with the vertices at these points.
the picture should look like Pascals Triangle without the top point . I don't know how to draw the image of the dots properly on this website. Imagine 14 dots arranged in a pyramid without a top piece. Please give a brief explanation. good luck, have fun!

-Brent

[RocketManJames]

My stab at it...

Assuming you meant all triangles, then I think you need to count all sets of 3 vertices where they aren't colinear.

I get 340. Have no idea if this is right or not... but that's what I ended up with.

Edit: I just realized I'm overcounting a few triangles, because points 3, 8, and 14, for example, are colinear.

Edit #2: I am only counting 2 of these, so I guess my final answer before I head off to bed is 338.

-RMJ

[BruceZ]

[ QUOTE ]
My stab at it...

Assuming you meant all triangles, then I think you need to count all sets of 3 vertices where they aren't colinear.

I get 340. Have no idea if this is right or not... but that's what I ended up with.

Edit: I just realized I'm overcounting a few triangles, because points 3, 8, and 14, for example, are colinear.

Edit #2: I am only counting 2 of these, so I guess my final answer before I head off to bed is 338.

-RMJ

[/ QUOTE ]

There can't be any more than 329. A triangle is determined uniquely by 3 non-collinear points. C(14,3) = 364. Subtract 16 for the 4 diagonal lines containing 4 points each (each of the 4 lines has 4 sets of 3 collinear points). Subtract 2 for the 2 diagonal lines of 3 points each. The bottom horizontal line of 5 points has C(5,3) = 10 sets of 3 collinear points. The horizontal line of 4 points has 4 sets of 3 collinear points. The horizontal line of 3 points has 1 set of 3 collinear points. Then there are 2 more shallow diagonal lines emanating from each of the 2 bottom corners, and these contain 1 set of 3 collinear points each. All together that is:

C(14,3) - 16 - 2 - 10 - 4 - 1 - 2 = 329.

Could the OP have meant equilateral triangles?

[RocketManJames]

[ QUOTE ]
There can't be any more than 329.

...

All together that is:

C(14,3) - 16 - 2 - 10 - 4 - 1 - 2 = 329.

[/ QUOTE ]

I went back and looked at my numbers, and in fact I had overcounted one colinear possibility, and then had a ten's place addition error, ugh. Correcting what I had, I now get your 329 from my approach. My way is certainly more sloppy... but here are my numbers. First, I enumerate the vertices. Then I count triangles that contained a fixed vertex where it was the lowest enumerated one for that set.

C(13,2) - 6 = 72 (i.e. Fix vertex #1, remove colinear sets)
C(12,2) - 6 = 60
C(11,2) - 3 - 1 = 51
C(10,2) - 2 = 43
C(9,2) - 2 - 1 = 33
C(8,2) - 3 = 25
C(7,2) - 1 = 20
C(6,2) = 15
C(5,2) = 10

The long way to do it I guess, but after my correction, it looks like it works out.

-RMJ

[eOXevious]

23
15 Small Ones
6 with a base of 2 lengths
and 2 with a base of 3 lengths

Wait I'm seeing more