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# I am struggling with this problem... pleh

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(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.)

(b) Generalize part (a): A regular -sided polygon is inscribed in a circle. Three of the  vertices are selected at random, and the triangle formed by those three points is drawn. What is the probability (in terms of ) that the center of the circle lies inside the triangle?

I am terrible with Geometry. I hope I can get some help.

Mr.Owl  Nov 19, 2017
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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 !!!!!???.

Guest Nov 20, 2017
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(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) = $$\frac{16}{28}=\frac{4}{7}$$

Melody  Nov 20, 2017
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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 !!!!.

Guest Nov 20, 2017
edited by Guest  Nov 20, 2017
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I had not worked through it before, sorry :)

p.s. I know it doesn't show (because of someone else) but I did give you a point for your answer :/

Melody  Nov 20, 2017
edited by Melody  Nov 20, 2017
edited by Melody  Nov 20, 2017

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