$n$ points ($n>1$) are taken on the circumference of a circle. Through any two of them a chord is drawn. No three chords intersect at one point inside the circle. i) Find how many points of intersections of these chords are inside the circle? ii) Find how many parts do these chords divide the circle?
I know one solution is to make a graph and use Euler's formula $v-e+f=2$. But that idea I would have never come up with. Is there any other way to approach this?
If $U$ is the unit disc centered at the origin consider $n$ chords drawn through the interior of $U$ such that no two chords are parallel and no three chords intersect at the same point. The arrangement graph $G$ induced by the discs and the chords has a vertex for each intersection point in the interior of $U$ and $2$ vertices for each chord incident to the boundary of $U.$ Naturally $G$ has an edge for each arc directly connecting two intersection points.
A point $p$ of $G$ is said to be incident to the interior of $G$ if $p$ is in the interior of $U.$ Similarly $p$ is said be incident to the boundary of $G$ if $p$ is on the boundary of $U.$ In this sense and without loss of generality I speak of the boundary of $U$ as the boundary of $G$ and the interior of $U$ as that of $G.$ I denote the number of arcs incident to $p$ by $v(p),$ called the valency of $p.$ It is straightforward to see that if $p$ is incident to the interior of $G$ then $p$ is $4-$valent. Likewise if $p$ is incident to the interior of $G$ then $p$ is $3-$ valent. From this we see that the maximum valency of $G$ is $4$ and the minimum degree of $G$ is $3.$ By construction $G$ is planar and $3-$ connected.
lemma 1: $G$ has $N(N+3)/2, N(N+2)$ and $(N^2+N+4)/2$ points, arcs and faces respectively.
Each chord has two distinct points incident to the boundary of $G.$ Being there are $N$ such chords it is seen there are $2N$ points along the boundary of $G.$ Observe there are $N(N-1)/2$ points in the interior of $G.$ Indeed the number of points in $G$ is equal to sum of those points in the interior and along the boundary which is $$2N+N(N-1)/2=N(N+3)/2.$$ The number of arcs in $G$ follows from the handshaking lemma. In particular twice the number of arcs is equal to the sum of all valances. The later is equal to $$[3N+4N(N-1)/2]/2=N(N+2)$$ The number of faces in $G$ follows from Euler's characteristic formula for planar graphs.$$2-N(N+3)/2+N(N+2)=(N^2+N+4)/2 $$ This establishes the lemma.