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Problem 1708

National Olympiad, first round

Combinatorics Difficulty 6.5 Prove it Selection Examination A · Greece

We consider on a circle $n$ different points such that there are no three chords with ends from these points passing through the same point lying in the interior of the circle.

a. Find the value of $n$ , if you know that the number of triangles with vertices three from the $n$ points is equal to $2n$ .

β. Find the value of $n$ , if you know that the number of the intersection points of the chords, lying in the interior of the circle is equal to $5n$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

(a) Any three from the $n$ points lying on the circle define a triangle. Therefore the number of triangles defined by the $n$ points is equal to the number of combinations of $n$ elements by $3$ , that is
$\binom{n}{3} = \frac{n(n-1)(n-2)}{6}$
Therefore we get the equation: $\frac{n(n-1)(n-2)}{6} = 2n \Leftrightarrow n = 5$ .

(β) Any four different points from the $n$ points lying on the circle define a convex quadrilateral whose diagonals intersect at an interior point of the circle and conversely. Therefore the number of interior points of the circle which are points of intersection of two chords is equal to
$\binom{n}{4} = \frac{n!}{4!(n-4)!} = \frac{n(n-1)(n-2)(n-3)}{24}$
Therefore we have the equation
$\frac{n(n-1)(n-2)(n-3)}{24} = 5n \Leftrightarrow n = 7$

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.