+0  
 
0
121
4
avatar+202 

Two points on the rim of a circle are randomly chosen. What is the probability that the distance between the points is shorter than the side length of an equilateral triangle inscribed in the circle?

 Jun 12, 2021
 #1
avatar
0

Since 60 degrees is 1/6 of the circle = 360 degrees, the probability is 1/6.

 Jun 12, 2021
 #3
avatar+202 
0

Yes! thank you guys for helping me understand this problem. Aops gave me a hint and i managed to solve it without using pi!

The answer is 2/3 because like you picture melody, say the first point is exactly on a vertex of the triangle, then the other point could not land in the bottom third of the circle, so the probability is 2/3 that the triangle is shorter than a side of the triangle.

mathisopandcool  Jun 12, 2021
 #4
avatar+114919 
0

I can see where 2/3 comes from.  I read of the question as the arclength. (not just the distance)

My error.   

I turned a trivial question into something much more complicated.

 

 

And please do not post AoPS questions here. 

It is against their code of conduct. 

Plus they claim their questions are copyright so you may be breaking the law. 

Melody  Jun 13, 2021
edited by Melody  Jun 13, 2021
 #2
avatar+114919 
+1

 

I will start with an equilateral triangle inscribed in a unit circle (radius=1)

 

Using cosine rule it can be determined that the side length of the triangle is sqrt3

Choose any first point and call it A

The second point can be any point up to sqrt3 units left or right of A so that is a total of 2sqrt3 units

the circumference is 2pi units

 

So the prob that the 2 points are  closer together  than the sides of the triangle is     

\(\frac{2\sqrt3}{2\pi}=\frac{\sqrt3}{\pi}\)

 

 

 

 

 

 Jun 12, 2021

20 Online Users

avatar