showing a bijection is true for the assumption that it takes 4 points to have 1 intersection in a circle

Question

Problem: 15 points are taken on the circumference of a circle, and through any two of them a chord is drawn. Suppose that no three chords intersect at the same point inside the circle. How many points of intersections are between these chords?

My attempt:

I know that you need atleast 4 distinct points on the circumference to have an intersection. If i can show that a bijection is true for n points then I can easily say how many points of intersections occur for fifteen points. I know to prove a bijection I must also prove that it is injective and surjective however I get stuck when creating the function.

surjective:

if(fx) = f(y) then f=y

so I have a function

k(x) = x(1/4). where x >= 4.

However my function fails when I input 5 since you cant have 1.25 intersections. any help?

score 0 · Accepted Answer · answered Nov 15 '13 at 02:59

0

Given any $4$ of our $15$ points, precisely one pair of lines determined by pairs of these points meet in the interior of the circle. Since intersection points are by assumption distinct, there are $\binom{15}{4}$ intersection points.

answered Nov 15 '13 at 02:59

André Nicolas

507,029

how do you evaluate $\binom {15}{4}$ – user2510809 Nov 15 '13 at 05:05
This is what is sometimes in high school called $C(15,4)$, or ${}^{15}C_4$, or something similar. It is equal to $\frac{15!}{4!11!}$, which simplifies to $\frac{(15)(14)(13)(12)}{4!}$. I think that's $1365$, but my arithmetic is not necessarily accurate. – André Nicolas Nov 15 '13 at 05:10
its accurate i checked on wolframalpha :) thanks a lot for the help! – user2510809 Nov 15 '13 at 05:16
You are welcome. I find calculating in my head faster than loading Alpha. But less accurate. – André Nicolas Nov 15 '13 at 05:24

showing a bijection is true for the assumption that it takes 4 points to have 1 intersection in a circle

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