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How many different triangles can be inscribed inside a regular decagon such that the triangle shares its vertices with the vertices of the decagon, but the triangle shares none of its sides?

Here is an example of what is allowed and not allowed:

Examples of triangles

I considered '10 choose 3', but the idea that the sides of the triangle cannot share it's sides with the sides of the decagon, this can't be correct. Or can it?

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I'd attack this problem by starting with $n$-gons for values of $n < 10$ and try to find a pattern I can generalize. – Michael Joyce Jan 17 '13 at 2:57
Pick a vertex. How many legal choices do you have for the second vertex? Now how many for the third one? – Maesumi Jan 17 '13 at 3:00
For a more general question, where this is a special case, see this. – Librecoin Apr 17 '13 at 18:56
up vote 2 down vote accepted

There are $\binom{10}{3}$ choices for triangle with no constraints. Now, each triangle in those $\binom{10}{3}$ shares $0,1$, or $2$ sides with the decagon.

Can you count triangles sharing exactly one edge with the decagon?

What about triangles sharing two edges with the decagon?


There are $\binom{10}{3} = 120$ triangles. We subtract out the triangles sharing one or two edges with the decagon (no triangle can share all three edges).

Triangles with one edge shared - Choose which edge is the shared one ($10$ choices), then we have $2$ of the vertices of the given triangle, so we need one more. You can't pick any vertex on or adjacent to the edge (as then the triangle would share two edges with the decagon), so there are $10-4 = 6$ choices per edge. That means there are $60$ triangles that share a single edge with the decagon.

Triangles with two edges shared - Choose the vertex incident on the two shared edges ($10$ choices). Now we have the three vertices of the triangle, so everything else is forced. That means there are $10$ triangles that share two edges with the decagon.

Putting it all together, there are $120 - 60 - 10 = 50$ triangles in the decagon that don't share any edges with the decagon.

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Attempted to find the constraints. But i am finding difficult.I first thought [10 choices for 1st one], [2 choices] for the 2nc one, in that you choose an adjacent vertex, and the 3rd one [8choose 1]. This giving 2 sides, and 3 sides triangles, but this is wrong since max amount of triangles is already 120. – MatthewL Jan 17 '13 at 6:57
If you're counting triangles that share exactly one edge with the decagon, you choose the shared edge (10 choices), then pick the 3rd vertex (6 choices for each edge). If you're counting triangles that share two edges with the decagon, you pick the pair of adjacent edges on the decagon (10 choices) and then the third edge is forced. – Michael Biro Jan 17 '13 at 7:08
Attempted to find constraints. I went and did it manually(got frustrated) and figured out that there is only 10 3 sided triangles. Now i can try and just figure out the 2 sided triangles in that, (10 choose 1) {1 choose 1, 2 choose 1 overcounts by factor of 2) and (6 Choose 1) This gives me 10 x 6. 60 triangles where one edge touches the decagon. That and including the 3 sided triangles, that is a total of 70 triangles. 120 - 70 = 50 total triangles. I don't feel comfortable with the answer, i Don't understand the logic behind it. Can you elaborate on your hints, or give me a detailed answer? – MatthewL Jan 17 '13 at 7:10
Hey btw, Thanks for your help. This website is incredible, first time i have been here, and i've attempted at other places to reach help in questions, but this site is quite impressive! I think i am going to be a frequent user here now, especially since you guys helped me out. You scratch my back, I scratch yours. If i can help at least. haha, Thanks. – MatthewL Jan 17 '13 at 7:14
No problem, I edited my answer. – Michael Biro Jan 17 '13 at 7:24

Revised: You want the number of ways of choosing $3$ of the $10$ vertices in such a way that no two of the chosen vertices are adjacent. Turn the problem around. Call the chosen vertices $A,B$, and $C$ in clockwise order, and look at how the other seven vertices of the decagon can be distributed between them if there must be at least one in each of the gaps (between $A$ and $B$, between $B$ and $C$, and between $C$ and $A$). There are four cases.

  1. If one gap contains $5$ vertices, each of the other gaps must contain one vertex. The triangle is isosceles; if $A$ is the vertex on its axis of symmetry, $A$ can be any of the $10$ vertices of the decagon, so there are $10$ such triangles.

  2. If one gap contains $4$ vertices, one of the other gaps must contain two vertices, and the other must contain just one. Reading clockwise, starting with the largest gap, the gaps can appear in the order $4,2,1$ or in the order $4,1,2$, and each of these orders can appear in any of $10$ positions around the decagon, so there are $20$ such triangles altogether.

  3. Two gaps contain $3$ vertices each, and one contains just one; this is like the first case.

  4. One gap contains $3$ vertices, and the others contain two vertices each; this is also like the first case.

The total is therefore $10+20+10+10=50$ triangles.

Thanks to Michael Biro for catching the overcounting in the original solution.

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I think this approach has a tricky issue with overcounting. It would probably be easier to look at the different (ordered) partitions of $7$ into $3$ pieces. There are $4$ of them $(5,1,1),(4,2,1),(3,3,1),(3,2,2)$. Then we'd look at the 10 different rotations per partition, as well as the reflections of the $(4,2,1)$ triangles. – Michael Biro Jan 17 '13 at 7:30
@Michael: You’re right: it has. – Brian M. Scott Jan 17 '13 at 7:31

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