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There are $6$ given points on a circle and each of the two points are connected by a line segment. Suppose that any three segments are not concurrent so that any three intersecting segments form a triangle inside the circle. Find the number of triangles formed inside the circle.

My Attempt

$\binom{6}{4}=15$ are the number of points of intersection inside the circle.

So the number of triangles formed inside the circle is the number of ways to choose three points out of $21$ points ($15$(inside the triangle) + the number of points on the circle($6$)).

The number of triangles=$\binom{15}{3}+\binom{15}{2}\times \binom{6}{1}+\binom{15}{1}\times \binom{6}{2}+\binom{6}{3}=\binom{21}{3}=1330$

But I can see that there are many instances of counting of triangles which do not exist. For e.g. a chord has a point of intersection on it. $\binom{21}{3}$ also counts the triangle formed by end-points of this chord and this point of intersection.

How should I avoid these cases.

Instead of choosing points if try to choose intersecting line segments how should I go about it.

The answer given is $111$.

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  • $\begingroup$ If we are allowed to assume the problem has a definite answer, then you can simplify the solution by considering just one figure as typical. For example, the regular hexagon inscribed in a circle. $\endgroup$
    – hardmath
    Commented Apr 7, 2023 at 19:03
  • $\begingroup$ Condition on the number of endpoints that the triangle's edges result in. There could be 3, 4, 5 or 6 endpoints. $\endgroup$
    – Calvin Lin
    Commented Apr 7, 2023 at 19:14

1 Answer 1

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let $N_3$ be the number of triangles that can be formed for which all 3 vertices are on the circle $$N_3 = \binom 63 = 20$$ let $N_2$ be the number of triangles that can be formed for which 2 out of 3 vertices are on the circle.

  • all 15 interior points are each connected to 4 points on the circle, being part of $\binom 42 -2=4$ triangles $$N_2 = 15 \left (\binom 42 -2 \right )=60$$

let $N_1$ be the number of triangles that can be formed for which 1 out of 3 vertices are on the circle.

  • all 6 points on the circle each have 3 interior lines emanating from them
  • let the six points on the circle be $P_1 ... P_6$, numbered consecutively
  • consider the point $P_1$...
    • the 3 internal lines will cross the line $P_2 \to P_6$ in 3 places, creating 3 triangles
    • the 3 internal lines will cross the line$P_3 \to P_6$ in 2 places, creating 1 triangle
    • the 3 internal lines will cross the line $P_2 \to P_5$ in 2 places, creating 1 triangle
    • the 3 internal lines will not cross any other lines more than once so no other triangles involving $P_1$ and 2 internal points can be formed.
  • all 6 points will contribute the same number of triangles to $N_1$ as does $P_1$ $$N_1 = 6(3+1+1) = 30$$

let $N_0$ be the number of triangles for which all 3 vertices are internal points. There is only 1 of these, formed by the lines $(P_1 \to P_4) (P_2 \to P_5) (P_3 \to P_6)$

$$N_{TOT} = N_3 + N_2 + N_1 + N_0 = 20+60+30+1=111$$

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    $\begingroup$ More generally, for $n$ points in general position on a circle, the number of triangles is $$\binom n3+4\binom n4+5\binom n5+\binom n6$$=A005732. $\endgroup$
    – user14111
    Commented Apr 8, 2023 at 4:31

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