3. There are n different points on a circle. The number of triangles whose vertices are among the n points is positive, and equal to the number of hexagons whose vertices are among the n points. What is n?

mathisopandcool May 1, 2021

#2**+1 **

Hello and welcome to web2.0calc :)))

(n)(n-1)(n-2)/3/2 is how many points of 3 you can select from n points (or number of triangles)

(n)(n-1)(n-2)(n-3)(n-4)(n-5)/6/5/4/3/2 is how many points of 6 you can choose from n points (number of hexagons)

(n)(n-1)(n-2)/3/2 = (n)(n-1)(n-2)(n-3)(n-4)(n-5)/6/5/4/3/2

1 = (n-3)(n-4)(n-5)/6/5/4

120 = (n-3)(n-4)(n-5)

n = 9

=^._.^=

catmg May 1, 2021

#3**+1 **

Since 6 points in a circle determine a hexagon, and 3 points determine a triangle:

\({n \choose 6} = {n \choose 3}\\ \frac{n!}{6!(n-6)!}=\frac{n!}{3!(n-3)!}\\ 6!(n-6)!=3!(n-3)!\\ 4\cdot5\cdot6=(n-3)(n-4)(n-5)\\ \)

Notice that the terms on both sides form an arithmetic progression. If you set a term on the RHS to the corresponding term on the LHS, you get (I used the smallest term as an example):

\(4=n-5\\\boxed{n=9}\)

oh no I got sniped

textot May 1, 2021