Counting Triangles with Polygons

[AvgStudent]

How many different triangles can be formed by selecting three vertices of a tridecagon and connecting them with a rubber band?

My intuition tells me that it's $13 \choose 3$, but I'm not so sure.

I tried with a pentagon and manually counted 10 which matches my hypothesis $\displaystyle {5 \choose 3}$

The circular shape makes it for me to think about.

Follow-up question: If my hypothesis is true, how can it prove it for n-gons?

 

[The Highlander]

How many different triangles can be formed by selecting three vertices of a tridecagon and connecting them with a rubber band?

My intuition tells me that it's $13 \choose 3$, but I'm not so sure.

I tried with a pentagon and manually counted 10 which matches my hypothesis $\displaystyle {5 \choose 3}$

The circular shape makes it for me to think about.

Follow-up question: If my hypothesis is true, how can it prove it for n-gons?

Confused by your wording: how does $\displaystyle {5 \choose 3}$ "match" your "hypothesis" of $\displaystyle {13 \choose 3}$?

Is it a regular tridecagon?
If so, then at first glance and numbering the vertices 1 to 13, I would count 11 "different" triangles unless you don't count the one between vertices 1, 2 & 3 as being the same as the one between vertices 2, 3 & 4.
If you do count those as being "different" then I would guess that the number increases to 143 but I'm sure someone else will have a more elegant solution.

 

[blamocur]

Confused by your wording: how does $\displaystyle {5 \choose 3}$ "match" your "hypothesis" of $\displaystyle {13 \choose 3}$?

Is it a regular tridecagon?
If so, then at first glance and numbering the vertices 1 to 13, I would count 11 "different" triangles unless you don't count the one between vertices 1, 2 & 3 as being the same as the one between vertices 2, 3 & 4.
If you do count those as being "different" then I would guess that the number increases to 143 but I'm sure someone else will have a more elegant solution.

If "different" means "non-congruent" then, I believe, there are 14 "different" ones. E.g.:
1 6 10
1 6 11
1 7 11
1 7 12
1 7 13
1 8 11
1 8 12
1 8 13
1 9 12
1 9 13
1 10 12
1 10 13
1 11 13
1 12 13

 

[Dr.Peterson]

How many different triangles can be formed by selecting three vertices of a tridecagon and connecting them with a rubber band?

My intuition tells me that it's $13 \choose 3$, but I'm not so sure.

I tried with a pentagon and manually counted 10 which matches my hypothesis $\displaystyle {5 \choose 3}$

The circular shape makes it for me to think about.

Follow-up question: If my hypothesis is true, how can it prove it for n-gons?

As I understand it, your "intuition" is that for an n-gon (not necessarily regular, but perhaps convex) there are C(n,3) distinct triangles that can be formed, and you've given two examples, for n=13 and n=5. This is correct, as long as "different" just means distinct triangles, and not non-congruent triangles. It's not just intuition or hypothesis; it's correct reasoning: any three vertices determine a triangle, so the number of distinct triangles is the number of distinct subsets of three vertices.

Why do you not think you've proved it? Perhaps because you haven't clearly defined the problem in your own mind. A tridecagon doesn't have to be "circular", for example.

 

[The Highlander]

As I understand it, your "intuition" is that for an n-gon (not necessarily regular, but perhaps convex) there are C(n,3) distinct triangles that can be formed, and you've given two examples, for n=13 and n=5. This is correct, as long as "different" just means distinct triangles, and not non-congruent triangles. It's not just intuition or hypothesis; it's correct reasoning: any three vertices determine a triangle, so the number of distinct triangles is the number of distinct subsets of three vertices.

Why do you not think you've proved it? Perhaps because you haven't clearly defined the problem in your own mind. A tridecagon doesn't have to be "circular", for example.

Surely C$\displaystyle {13 \choose 3}$ = 286 (which is double my 'alternative' answer because it "re-counts" in an anticlockwise direction?) includes swathes of both similar and congruent triangles!
It seems to me that any sensible answer to this problem first requires a precise definition of "different"; "distinct" (although it 'fits' with C$\displaystyle {13 \choose 3}$ as an 'answer') seems far too "free" a definition (IMHO).

PS:-

Confused by your wording: how does $\displaystyle {5 \choose 3}$ "match" your "hypothesis" of $\displaystyle {13 \choose 3}$?

Please excuse & ignore this comment; there was nothing 'confusing' about it at all! I simply failed to notice the word "pentagon" in the third line of the OP and presumed it was still referring to a tridecagon! My Bad! ?

 

[Dr.Peterson]

Surely C$\displaystyle {13 \choose 3}$ = 286 (which is double my 'alternative' answer because it "re-counts" in an anticlockwise direction?) includes swathes of both similar and congruent triangles!
It seems to me that any sensible answer to this problem first requires a precise definition of "different"; "distinct" (although it 'fits' with C$\displaystyle {13 \choose 3}$ as an 'answer') seems far too "free" a definition (IMHO).

We should write either $\displaystyle {13 \choose 3}$ or C(13,3), not C$\displaystyle {13 \choose 3}$, which is an odd hybrid!

As I see it, "different" is too vague to be used in formal writing in math; "distinct" carries one of its meanings (namely, not being the exact same object), and "non-congruent" and "non-similar" carry others (not being the same shape in one sense or another). But to me it is clear from what the OP did that they had the former in mind: triangles that are not the very same triangle, because they are formed from different sets of vertices of the n-gon, but are allowed to be the same shape.

So, yes, "different" needs to be clarified, or replaced with a clearer word. But no, "distinct" is not "too free" to be used at all, and it is not unlikely to be what was intended. That sort of question is not uncommon (and is much easier than the number of non-congruent triangles that can be formed, which can't really even be asked without saying whether the polygon is regular). It might be helpful, though, to ask this:

@AvgStudent, can you tell us where the problem came from, and whether you have quoted it exactly? And if it is your own question, what did you mean by "different"? And are you supposing the polygon is regular?

 

[AvgStudent]

@AvgStudent, can you tell us where the problem came from, and whether you have quoted it exactly? And if it is your own question, what did you mean by "different"? And are you supposing the polygon is regular?

Thank you for the responses. This question was presented by my lecturer as a "brain teaser". My professor physically showed us a tridecagon with nails hammered at each vertex. Then he gave us a rubberband and asked if we can "count how many triangles". The question wasn't written down but verbally asked and demonstrated so if it was badly worded it's my fault.

He showed us a regular tridecagon, or it did look like one to me, but I think it should be convex polygons, not necessarily regular. I don't see it makes difference. Perhaps I'm wrong here.

Triangles that are not the very same triangle, because they are formed from different sets of vertices of the n-gon, but are allowed to be the same shape.

This is what I have in mind and the intent of the question.

Why do you not think you've proved it? Perhaps because you haven't clearly defined the problem in your own mind

I came up with the answer with my "intuition" and wasn't entirely confident in my reasoning.

Somewhat related to the topic. As the sides approach infinity, do we get a circle? If so, can we still count the triangles?
It's a bit too abstract for me to think about with what I know.

PS: This is mostly for my curiosity.