23. Twenty points are equally spaced around the circumference of a circle. Kevin draws all the possible chords that connect pairs of these points. How many of these chords are longer than the radius of the circle but shorter than its diameter?
A 90
B 100
C 120
D 140
E 160
Solution
Each of the 20 points is connected with the 19 other points on the circumference, forming 19 chords. One of these 19 chords is exactly equal to the diameter, leaving 18 chords to consider.

For a chord to be larger than the radius, the angle it subtends at the centre has to be larger than $60^{\circ}$. This is because the chord and two radii form an equilateral triangle as shown in the diagram.

Since the 20 points form a regular 20-gon, the central angle of the chord connecting two adjacent points is $360^{\circ} \div 20$ which is $18^{\circ}$. A chord $P Q$ has a length greater than the radius but less than the diameter provided that it subtends an angle strictly between $60^{\circ}$ and $180^{\circ}$. Since $3 \times 18^{\circ}=54^{\circ}<60^{\circ}$. this means that there must be at least 3 of the 20 points between $P$ and $Q$, the shortest way round, and that $P Q$ must not be a diameter.

That is, three pairs of chords from a point to its three nearest points along the circumference (on either side) are going to be shorter than the radius. This leaves us with six pairs of chords, or twelve chords per point. Since each chord is counted twice (once per endpoint) the total number of chords satisfying the condition is $(20 \times 12) \div 2=120$.