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Let there be nine fixed points on a circumference of a circle. Each of these points is joined to every one of the remaining eight points by drawing a line and the points are so positioned on the circumference of the circle that atmost $2$ straight lines can intersect at a point. The number of such interior intersection points is ?

This is how I approached the problem : Total number of lines drawn = $\binom{9}{2} = 36$ Now the number of intersection points is the number of unique pairs of lines out of thirty six lines , i.e. $\binom{36}{2}$ . And the answer comes out to be $630$. However the correct answer is $126$. Where am I wrong, and how do I solve it ?

Aditi
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  • There will be $630$ intersection points, but you are asked about the interior ones, i.e., inside the circle. – cronos2 Oct 30 '17 at 12:31
  • @cronos2 thanks, but how do I know which of these lines intersect inside the circle ? – Aditi Oct 30 '17 at 12:35

1 Answers1

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Each intersection point is determined by two line segments. Those two line segments are determined by choosing their four endpoints. Since there are nine available points, each intersection is determined by choosing four of those nine points, which can be done in $$\binom{9}{4}$$ ways.

N. F. Taussig
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  • Thank you. Could you please tell me where I was missing out ? – Aditi Oct 30 '17 at 12:33
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    As @cronos2 pointed out in the comments, you counted all the intersection points between the lines, including those lying outside the circle. We want the points where the chords intersect inside the circle, which is why I focused on the line segments. – N. F. Taussig Oct 30 '17 at 12:35
  • Ahh ! Line segment ! Now I realise why is that so! Thank you very much . – Aditi Oct 30 '17 at 12:39