Sum of Two Arcs

Date: 01/30/2003 at 20:46:53
From: Abishek
Subject: Probability

Hi Dr. Math,

This is the question:

Three points are taken at random on the circumference of a circle. 
What is the chance that the sum of any two arcs so determined is 
greater than the third?

Date: 01/31/2003 at 01:56:00
From: Doctor Greenie
Subject: Re: Probability

Hi, Abishek -

I enjoyed the mental exercise I got thinking this one through!

If the sum of two of the three arcs is greater than the third, then 
the largest arc is less than half the circle - i.e., less than 180 
degrees. So to solve your problem, let's look at the conditions under 
which the largest of the three arcs is GREATER than 180 degrees.

Suppose we call the points A, B, and C, and suppose points A and B 
have been chosen, and let x be the degree measure of minor arc AB 
(i.e., of the smaller of the two arcs determined by points A and B).

Now consider diameters AD and BE (draw a picture...).  If the third 
point C is on the same side of diameter AD as point B, then the 
largest of the three arcs will be greater than 180 degrees. And if the 
third point C is on the same side of diameter BE as point A, then the 
largest of the three arcs will be greater than 180 degrees. So the 
range where point C can be chosen so that the largest of the three 
arcs is LESS than 180 degrees is the minor arc DE, whose measure is 
the same as the measure of minor arc AB, which is x degrees.

So if the measure of minor arc AB is x degrees, then the probability P
(x) that the largest of the three arcs is less than 180 degrees is

  P(x) = (x/360)

If we graph this probability function over its domain from 0 to 180 
degrees, we get a straight line with P(0) = 0 and P(180) = 1/2.  The 
overall probability that the largest of the three arcs is less than 
180 degrees is the average value of the function P(x) over its 
domain, which is 1/4.

So...

Given three points randomly chosen on a circle, the probability that 
the largest of the three arcs is less than 180 degrees - that is, the 
probability that the sum of any two arcs is greater than the third - 
is 1/4.

I looked in the Dr. Math archives after I wrote this response, 
performing a search using the keywords "probability point 
circumference arc".  That search turned up this link which discusses 
this same problem but never - as far as I could tell - comes to a 
final answer:

   A Triangle in a Circle
   http://mathforum.org/library/drmath/view/51787.html 

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 02/01/2003 at 19:54:24
From: Abishek
Subject: Probability

I could follow till the part where you said p(x)= x/360, but after 
that I don't understand why we have to take the average value for the 
required probability.

Date: 02/02/2003 at 00:22:27
From: Doctor Greenie
Subject: Re: Probability

Hello, Abishek -

The statement

  P(x) = x/360

says that the probability is x/360 that the largest of the three arcs 
is less than 180 degrees if the minor arc between the first two of 
the three points measures x degrees.

So, for example, if the minor arc between the first two points is 45 
degrees, then the probability of choosing the third point so that the 
largest arc is less than 180 degrees is 45/360 = 1/8.  Or, if the 
minor arc between the first two points is 120 degrees, then the 
probability of choosing the third point so that the largest arc is 
less than 180 degrees is 120/360 = 1/3.

To find the overall probability that three points randomly chosen 
will result in the largest of the three arcs measuring less than 180 
degrees, we need to consider the probabilities P(x) over the whole 
range of possible value of x, which is from 0 to 180 degrees.  The 
probability function in this case is a linear function, with value 0 
at x=0 and value 180/360 = 1/2 at x=180.  The average value of a 
linear function over its range is the average of its values at the 
two endpoints of its range -- in this case, the average of 0 and 1/2, 
which is 1/4.

I hope this helps.  Please write back if you still have questions 
about this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/