I am struggling with this problem... pleh

CPhill: Please poke some holes in this argument !!.

There are 9C3 =84 possible triangles.

If you divide the nonagon in half by drawing a straight line from any vertex through the center of the nonagon, will have 5 vertices on either side(including the chosen vertex that divided it in two.) Any triangle formed from the 5 vertices on one side and 5 vertices on the other side will not have the center of the nonagon inside it !!

So, we have: 5C3 =10 triangles x 2 =20 triangles that will not contain the center of the nonagon.

84 - 20 =64 triangles that will contain the center of the nonagon. Or:

64 / 84 =76.2% probability. I did this for fun only !!!!!???.

(a) A regular nonagon (9-sided polygon) is inscribed in a circle as shown below. Three of the 9 vertices are selected at random, and the triangle formed by those three points is drawn. What is the probability that the center of the circle lies inside the triangle? (A successful selection of 3 such points is shown below.)

Hi Mr Owl and guest :)

Here is my thinking.

Any point can be chosen as the first vertex. So P(correct first vertex) = 1

Now draw a line through the chosen vertex and through the centre thereby bisecting the nonogon.

Now the centre will be included so long as one vertex is in side A and the other is in Side B

There are 4 points on side A and there are 4 suitable matches for each of those on part B so that is 4*4=16 possible triangles that include the centre (this is for each of the 9 initial vertices)

Now the number of ways that 2 points can be chosen from the 8 without restriction is 8C2 = 28

P(the centre is included in the triangle is) = \(\frac{16}{28}=\frac{4}{7}\)

Hi Melody: Is it just a coincidence that:

9C3 - 2*(5C3) (as I calculated it for fun!) - (4*4)(as you calculated it in earnest!) =

84 - 20 - 16 =48, which is the same as:

[9C3 - 9C2] / 9C3 =

[84 - 36] / 84 =

48/84 =4/7 will have the center of the nonagon within !!!!.