$6$ given points on a circle are joined by line segments of which no three are concurrent. Find number of triangles formed inside the circle.

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 $$111$$.

• 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. Commented Apr 7, 2023 at 19:03
• Condition on the number of endpoints that the triangle's edges result in. There could be 3, 4, 5 or 6 endpoints. Commented Apr 7, 2023 at 19:14

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 $$(P_1 \to P_4) (P_2 \to P_5) (P_3 \to P_6)$$

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

• 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. Commented Apr 8, 2023 at 4:31