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Can you help me solve this problem?

A regular 13-sided polygon is inscribed in a circle with centre $O$. Triangles can be formed by choosing three vertices of this polygon to be the vertices of the triangle.

How many of the triangles that can be formed in this way have the point $O$ inside?

I have no idea how to approach this question, so I tried to start off simple by drawing the number of triangles that can be inscribed in a square and a pentagon. However, after finding the values 4 and 8, I have no idea where to go from here.

Any help is highly appreciated. Thank you!

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  • $\begingroup$ Use complex numbers and roots of unity. $\endgroup$
    – Neutral Element
    Mar 19, 2021 at 6:01
  • $\begingroup$ @NeutralElement That would be overkill. $\endgroup$
    – Parcly Taxel
    Mar 19, 2021 at 6:12

2 Answers 2

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Choose two points. These are then a certain number of edges apart, up to $6$. A brief investigation will show that there are a number of points that can form a centre-including triangle with those points, and that number is equal to the separation of the points.

There are equal numbers of points with all valid separations, and each triangle will be counted three times, so the answer is $$\binom{13}{2}\big / 6\cdot (1+2+3+4+5+6)/3 = 13\times 7 = 91$$

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  • $\begingroup$ > A brief investigation will show that there are a number of points that can form a centre-including triangle with those points, and that number is equal to the separation of the points. Can I ask how you arrived at this statement? $\endgroup$
    – hmmhmm hmm hmm
    Mar 19, 2021 at 6:52
  • $\begingroup$ Well, via a brief investigation. But to justify afterwards with some maths, we need a point in the long gap and it can't be more than 6 separated from either of the chosen points. So the available points increase as the separation of chosen points increases. $\endgroup$
    – Joffan
    Mar 19, 2021 at 6:56
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Let the paths between the three chosen vertices have lengths $a,b,c$. The number positive integer solutions to $a+b+c=13$ with $a,b,c\le6$, i.e. the number of nonnegative integer solutions to $a+b+c=10$ with $a,b,c\le5$. By stars and bars and inclusion and exclusion, this is $$\binom{10+2}2-3\binom{4+2}2=21$$ These $21$ solutions reduce to five upon counting permutations as identical: three with two duplicated lengths ($166,355,544$) and two with all distinct lengths ($256,346$). Each of the first set counts $13$ times and each of the second $26$ times, for a final answer of $3×13+2×26=91$.

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  • $\begingroup$ Can I ask where we get $a+b+c=13$ and $a,b,c \le 6$? $\endgroup$
    – hmmhmm hmm hmm
    Mar 19, 2021 at 7:02
  • $\begingroup$ @ChrisAbrea Obviously the three paths have to go all the way around the 13-gon, so $a+b+c=13$. There is the observation that if some variable is $7$ or more this means the region formed by the corresponding chord and the corresponding arc contains the centre, so the triangle cannot contain the centre. Hence $a,b,c\le6$. $\endgroup$
    – Parcly Taxel
    Mar 19, 2021 at 10:16

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