How many triangles can you form with 21 evenly distributed dots?

[davee123]

So, being inundated with the "How many triangles?" questions on Facebook, I noticed this one which is actually more difficult than I expect the question author intended:
http://creative.ak.facebook.com/ads3/flyers/36/28/6002237517496_1_992e4bd8.jpg
Assuming you have 21 dots evenly distributed in an equilateral triangular pattern (like bowling pins), how many distinct triangles can be formed by connecting the dots?

Of course, I'll bet they expect people to interpret "equilateral triangles" rather than simply "triangles", but it does make for a more interesting challenge.

And on that note, how many are possible with 3 dots, 6 dots, 10 dots, and 15 dots? Is there a nice formula for progression as the number of available dots increases?

DaveE

 

[Rogerio]

When the "starting triangle's base" has N dots, then the number of equilateral triangles is

Spoiler

(N-1)*N*(N+1)*(N+2)/24

which means

Spoiler

70

equilateral triangles in the figure.

 

[Gokul43201]

Of course, I'll bet they expect people to interpret "equilateral triangles" rather than simply "triangles", but it does make for a more interesting challenge.

Which way do you want us to interpret it: only equilaterals or all triangles?

 

[davee123]

I suppose both questions are interesting-- answer what you will, I guess! But I couldn't think of a formula that expressed the number of "triangles" given N dots. I honestly never even tried to answer the equilateral triangle question, since it seemed more trivial. But there are some interesting caveats to it.

DaveE

 

[Gokul43201]

I get a different answer from rogerio for equilateral triangles. Maybe I am overlooking some triangles somewhere...

For N even:

Spoiler

n(2n^2-n-2)/8 = 48 triangles for n=6 base dots

For N odd:

Spoiler

(n-1)(n+1)(2n-1)/8 triangles.

 

[davee123]

I get a different answer from rogerio for equilateral triangles. Maybe I am overlooking some triangles somewhere...

I got the same answer as Rogerio for the equilateral triangles-- there are some non-standard orientations, don't forget!

DaveE

 

[Gokul43201]

I got the same answer as Rogerio for the equilateral triangles-- there are some non-standard orientations, don't forget!

DaveE

Oops, yes! I was missing those.

 

[DaveC426913]

I'm assuming your formulae simply count all possible 3-dot combinations and assumes a triangle joins them. Is that correct?

Do these formula eliminate "degenerate shapes"? i.e. three dots in a straight line does not a triangle make, so some combos of 3 dots are not valid.

Oh, I see you guys are pursing only equilateral triangles so far, so my point is moot.

 

[davee123]

I see you guys are pursing only equilateral triangles so far, so my point is moot.

Well, moot in terms of equilateral triangles, but not all triangles, which is what I was more interested in. Figuring out all the sets of 3 points is pretty simple, but figuring out which sets of 3 are in straight lines was tougher-- at least in terms of trying to get a formula.

DaveE

 

[DaveC426913]

Well, moot in terms of equilateral triangles, but not all triangles, which is what I was more interested in. Figuring out all the sets of 3 points is pretty simple, but figuring out which sets of 3 are in straight lines was tougher-- at least in terms of trying to get a formula.

DaveE

I suppose it would be easier to write an algorithm (where you have access to loops and decision trees) than a formula. I wonder if all algorithms are transposable into formulae...

 

[Rogerio]

Figuring out all the sets of 3 points is pretty simple, but figuring out which sets of 3 are in straight lines was tougher-- at least in terms of trying to get a formula.

Agreed.

I got 114 degenerated triangles (3 points in line).
So the number of general triangles in the figure is
21*20*19/6 - 114 = 1216.

But it was a very ugly way...

 

[davee123]

Yep, that matches what I got, although I just did 21 choose 3 - 114. Same difference, though.

DaveE

 

[Gokul43201]

General formula for the 114? Now that should be fun!