# Probability that three randomly chosen points on a circle provides an acute triangle

In trying to solve the classic problem --

"Find the probability that three randomly chosen points on a circle provides an acute triangle"

I came across this page that seems to have a good explanation.

However, I do not understand how they came up with the probability as $\int_{0}^{\pi }{\frac{1}{\pi }\cdot \frac{\theta }{2\pi }\cdot d\theta }$

I understood that the probability is the length of sector of the circle between the two dotted lines divided by the total circumference, but do not see how/why the integral is needed.

## 2 Answers

I answered a more general form of this problem here. Think of the integral as the sum of all the angles you can have, and the division as averaging it out. Just as $\frac{1}{n}\sum_{i=1}^n f(i)$ is an average, so is $\frac{1}{\pi}\int_0^\pi f(\theta)\mathrm d\theta$. So, in this case, you want to take the average of all the $\theta$ you can have, but there are an infinite amount of these, so how do you take an average? By this integral, of course!

Once you have an average $\theta$, you need to recognize that the length of the arc the third point can occupy in order to form the triangle is exactly $\theta$, the length of the minor arc between the first two points. The probability of landing in this arc is $\theta/2\pi$, and thus your integral.

• Why is the integral from 0 to pi though? Shouldn't it be from 0 to 2pi? – 1110101001 Nov 9 '13 at 22:22
• The minor arc length only varies between 0 and $\pi$. A distance of $3\pi/2$, for example, is only a distance $\pi/2$. $2\pi$ is really 0 when it comes to circles. You're looking for the probability of event $E$ occurring. if $E$ had a $1/3$ probability occurring $1/2$ the time and $2/3$ the other $1/2$, you would say there is a $50%$ probability of $E$. This is the same, except with infinite probabilities. – Tim Ratigan Nov 9 '13 at 22:33
• I'm using arc length and angle interchangeably, since arc length and angle from the center are the same. This probably is unadvisable, sorry for any confusion. Don't worry about it. – Tim Ratigan Nov 9 '13 at 22:40
• No need to apologize :) . We use the minor arc because the "inverted" minor arc is where the third point needs to be in order to make an acute triangle. This is because a triangle is acute iff it contains the center of the circle that circumscribes it. – Tim Ratigan Apr 26 '14 at 9:30
• You might even be able to see that once you have chosen the first two points, you can't pick a pair of points that are $\pi$ rad apart such that if you form a triangle with each point in the pair, both triangles would be acute. That precludes the possibility of using a major arc. – Tim Ratigan Apr 26 '14 at 9:32

The probability of the third point producing an acute triangle is a continuous function of the choice of the second point. Hence the integral