(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.)![size(4cm); pair A[]; int i; draw(unitcircle,black+0.5bp); for (i=0;i<10;++i) { A[i] = rotate(40*i)*(1,0); if (i>0) { draw(A[i-1]--A[i],black+1bp); } dot(A[i]); } dot((0,0)); path p = scale(0.1)*unitcircle; draw(shift(A[0])*p); draw(shift(A[2])*p); draw(shift(A[6])*p); draw(A[0]--A[2]--A[6]--cycle,blue+1bp);](/api/ssl-img-proxy?src=http%3A%2F%2Flatex.artofproblemsolving.com%2F9%2Fb%2Fd%2F9bd111b022ba8a3ad3885d9996f1b0c4d9fa0e21.png)
(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?
+118609 PART A
Consider the nonogon vertices being labeled 1 to 9 in a clockwise direction
Let one of the triangle vertices be at 1
These are the triangles that will include the centre
126
136
137
146
147
148
156
157
158
159
That is 10 triangles that will include the centre
P(centre is included) = $$\frac{10}{8C2}=\frac{10}{8C2}=\frac{10}{28}=\frac{5}{14}$$
+118609 PART A
Consider the nonogon vertices being labeled 1 to 9 in a clockwise direction
Let one of the triangle vertices be at 1
These are the triangles that will include the centre
126
136
137
146
147
148
156
157
158
159
That is 10 triangles that will include the centre
P(centre is included) = $$\frac{10}{8C2}=\frac{10}{8C2}=\frac{10}{28}=\frac{5}{14}$$
Guys... Please don't cheat on AOPS homework. I was trying to look for a faster way to solve part a, and apparently casework is the best way, but don't do this. You won't learn anything.
(a)5/14
(b)(1, 2...(n-1), n) over 2n choose 2
(a) If the rotation is irrelevant, we have 8 choose 2 options for our triangle.
Let's call our origin point 0.
If we number the points clockwise:
Points 1 and 8 have 1 success each.
Points 2 and 7 have 2 successes each.
Points 3 and 6 have 3 successes each.
From here it would be logical to guess that points 4 and 5 have 4 successes each.
That is correct!
(Note: Triangle 0-4-6 is the same triangle as 0-6-4, so we need to divide the total result by 2.)
Adding up our successes, we get 1+2+3+4=10 successes.
Our answer is 10/28 or 5/14.
(b) In part (a) we noticed that there was a pattern in successes.
Therefore, we can say that a regular polygon with a number (2n+1) corners will have (1+2+3...(n-2)+(n-1)+n) over (2n) choose 2.
That is our answer.
