$3$ points $A$, $B$, $C$ are randomly chosen on the circumference of a circle. If $A$, $B$, $C$ all lie on a semicircle, then what is the probability that all of the angles of triangle $ABC$ are less than $120^\circ$?

Okay, let's fix a semicircle and make an interval $[0, 1]$ along it. It's clear the condition that the condition that all angles are less than $120^\circ$ means that the two points of the triangle furthest apart on the semicircle are further than $2/3$ apart. So we want to calculate the percentage of the volume of the unit cube already satisfying the following:$$0 < x < y < z < 1$$subject to the additional inequality$$z - x > {2\over3}$$However, I'm not sure what to do from here. Any help would be well-appreciated.

UPDATE: With the helpful hint of Daniel Mathias in the comments, I was able to get the answer of ${5\over9}$. But I am wondering if there is a way to solve it in the more complicated way I proposed, where there's $3$ variables and maybe we can do some multivariable integration.

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    $\begingroup$ We can always choose a diameter of the circle such that one of the three points is an endpoint of the diameter, with the other two points on the same side of that chosen diameter. $\endgroup$ Aug 7, 2021 at 14:44
  • $\begingroup$ Thanks for your hint Daniel Mathias, I was able to solve it and got ${5\over9}$ as my answer. But I am still curious if someone can come up with a solution using $3$ variables. $\endgroup$ Aug 7, 2021 at 14:52
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    $\begingroup$ @VladimirFokow. Allright then. What prevents you from formulating an answer, if not for the bounty then for us who are curious how you did it? $\endgroup$ Aug 14, 2021 at 18:42
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    $\begingroup$ I've rediscovered Vladmir Fokow's answer. Here is a GeoGebra applet to show this, where $A$ should remain fixed, and the green line makes an angle of $120º$ with the $x$-axis, acting as the dividing line. The probability is indeed $\frac{1}{3} \cdot 1 + \frac{2}{3} \cdot \frac{1}{3} = \frac{5}{9}$, where it is the probability of $B$ first, then $C$. $\endgroup$
    – Toby Mak
    Aug 15, 2021 at 7:03
  • $\begingroup$ Using a well-known circle theorem, $\frac{1}{2}(360º - 120º) = 120º$, so for the condition to hold, the sum of the angles at the origin must be greater than $120º$. $\endgroup$
    – Toby Mak
    Aug 15, 2021 at 7:08

3 Answers 3


I don't know how to solve this problem using the "complicated multivariable integration" but here is how I did it by applying some simple reasoning: (+ did some Java simulations - see below)

I'll be using the inscribed angle theorem:

An angle $θ$ inscribed in a circle is half of the central angle $2θ$ that subtends the same arc on the circle. Therefore, the angle does not change as its vertex is moved to different positions on the circle.

Without the loss of generality, suppose that one of the points ($C$) is exactly on the end of our semicircle. Then amongst the other two points there will always be the "higher" ($A$) and the "lower" point ($B$).
(Except for when $A$ and $B$ collide, which is negligibly rare since the set of all points of a semicircle is infinite.)

So let us consider that $B$ is always between $A$ and $C$:

B is always between A and C.

Note that:

  • If $A$ is exactly on the opposite side of $C$ (on the other end of the semicircle), then $∠ABC = 90°$ (no matter where $B$ is – by the inscribed angle theorem),
    and $∠BAC < 90°$, and $∠BCA < 90°$.


  • For fixed $B$: if $A$ moves closer to $C$ along the semicircle, then $∠BCA$ decreases. Therefore $∠BCA$ cannot ever be greater than $90°$.

  • For fixed $A$: if $B$ moves closer to $C$, then $∠BAC$ decreases, and the maximum $∠BAC$ is when $B$ is as close to $A$ as possible (then $∠BAC$ approaches $90°$). Therefore $∠BAC$ cannot ever be greater than $90°$.

So the only angle that can be greater than $120°$ is $∠ABC$.

Let us now proceed to find out when $∠ABC < 120°$, and when it is $≥ 120°$:

  • From the inscribed angle theorem we know that the value of $∠ABC$ depends only on the positions of $A$ and $C$ (and remains unchanged if $B$ moves between them). Also,

  • If $A$ moves closer to $C$, then $∠ABC$ increases.

So let us find out the position of point A to make $∠ABC = 120°$.

By the inscribed angle theorem, the central angle that subtends the arc $AC$ is

$$∠AOC = ∠ABC \cdot 2 = 240°$$

making $∠EOA = 60°$:

∠EOA = 60°

We can conclude that:

(All the angles of triangle $𝐴𝐵𝐶$ are less than $120°$) $\iff $

$\iff ∠ABC < 120° $

$\iff ∠EOA < 60° $

$\iff $ (At least one of two points lies in that top region of the semicircle of 60° (in order to make $∠EOA < 60°$))

Since the point locations are independent events, the probability of BOTH points being NOT in that region is:

$$\overline{P} = \frac {120°}{180°} \cdot \frac {120°}{180°} = \frac {4} {9}$$

Then the probability of at least one point BEING IN that region is

$$ P = 1 - \frac 4 9 = \frac 5 9$$


I have done the simulations in Java to check my answer.

There I generate the points on the whole circumference first and then choose only the ones that happen to be on the same semicircle - just be sure.

However it would absolutely OK (and much simpler) to generate the points that are on the same semicircle already. This way we could avoid a lot of tricky if statements.

You can find my simulation here.

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    $\begingroup$ Very much appreciated (+1). Thanks. Personally, I have no trouble with common sense instead of "rigor". $\endgroup$ Aug 14, 2021 at 20:45
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    $\begingroup$ What your answer shows is that $\angle COB\lt 120$ and $\angle COA\lt 120$. Each of these events has a probability of $\frac23$ and they are independent. The probability that these occur simultaneously is $\frac23 \times \frac23=\frac49$, and this is the correct solution. $\endgroup$ Aug 14, 2021 at 20:46
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    $\begingroup$ Correction: At least one of the angles $\angle COA$ and $\angle COB$ must be greater than $120^\circ$, so the probability is $1-\frac49=\frac59$ $\endgroup$ Aug 14, 2021 at 20:56
  • $\begingroup$ Corrected the answer! Thanks. $\endgroup$ Aug 14, 2021 at 21:20

well, lets say you have a circle in polar coordinates, using the usual system of $r,\theta$. Lets say each point has the same radius (as they are on the circumference) but a different angle i.e. $\theta_a,\theta_b,\theta_c$ and it is clear that if $\theta_a<\theta_b<\theta_c$ and $\theta_c-\theta_a<\pi$ then the points lie on a semicircle. You could then make each of the thetas randomly distributed over $[0,2\pi]$ and come up with a formula for the great angle formed between these three points.


You are on the right path with your stick-breaking approach. However, you can let $z = 1-(x+y)$ because the angles in a triangle always sum to $180º$, reducing the problem to two variables $x,y$.

We want all the angles to be less than $120º$, or $0 < x < \frac{2}{3}, 0 < y < \frac{2}{3}, 0 < 1-(x+y) < \frac{2}{3}$, where theprobability space is given by $0 < x < 1, 0 < y < 1, 0 < 1-x-y<1$. Plotting all of this gives:

where the white region satisfies all three inequalities.

Dividing the probability space into $9$ congruent right triangles, the total probablity is $1 - \frac{3}{9} = \frac{2}{3}$.

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    $\begingroup$ Yep, this is surely wrong because this doesn't translate back to the original problem. $\endgroup$
    – Toby Mak
    Aug 15, 2021 at 6:54
  • $\begingroup$ Your idea is correct. Thre is just a mistake in your calculation. $\endgroup$
    – Hans
    Aug 16, 2021 at 14:24

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