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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.)

 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);

(b) Generalize part (a): A regular $(2n+1)$-sided polygon is inscribed in a circle. Three of the $2n+1$ vertices are selected at random, and the triangle formed by those three points is drawn. What is the probability (in terms of $n$) that the center of the circle lies inside the triangle?

Guest Apr 18, 2015

Best Answer 

 #1
avatar+90988 
+6

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)   =    $$\frac{10}{8C2}=\frac{10}{8C2}=\frac{10}{28}=\frac{5}{14}$$

Melody  Apr 19, 2015
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8+0 Answers

 #1
avatar+90988 
+6
Best Answer

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)   =    $$\frac{10}{8C2}=\frac{10}{8C2}=\frac{10}{28}=\frac{5}{14}$$

Melody  Apr 19, 2015
 #2
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0

can someone post part b? im having trouble

Guest Aug 23, 2016
 #3
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0

how did u get 8 choose 2 as the denominator 

Guest Aug 23, 2016
 #4
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0

I still don't understand it. And i'm having a lot of trouble with b.

Guest Aug 23, 2016
 #5
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0

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.

Guest Aug 23, 2016
 #6
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0

how you get 8 C 2

Guest Aug 23, 2016
 #7
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0

I dont know gow to do part b.

Guest Sep 25, 2016
 #8
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+1

(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.

Guest Sep 29, 2016

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