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The vertices of a triangle are three independent uniformly random points on (the circumference of) a circle.

Let $X_k=\text{number of angles larger than $k\pi$}$.

Is the following conjecture true: $\mathbb{E}(X_k)=3(1-k)^2$.

If it is true, then, given its simplicity, I wonder if there is an intuitive explanation.

Evidence for my conjecture

I worked out the value of $\mathbb{E}(X_k)$ for four values of $k$:

  • $\mathbb{E}(X_0)=3\space$ (obvious)
  • $\mathbb{E}(X_{1/3})=\frac43\space$ (explanation below)
  • $\mathbb{E}(X_{1/2})=\frac34\space$ (explanation below)
  • $\mathbb{E}(X_1)=0\space$ (obvious)

enter image description here

These points suggest the neat relationship $\mathbb{E}(X_k)=3(1-k)^2$.

Explanations

The only possible value of $X_{1/2}$ are $0$ and $1$. We know that $P(X_{1/2}=0)=\frac14$. Therefore $P(X_{1/2}=1)=\frac34$, and so $\mathbb{E}(X_{1/2})=\frac34$.

The only possible values of $X_{1/3}$ are $1$ and $2$. Let $\theta$ be the angle subtended at the circle's centre by the first two chosen points. For each value of $\theta$, we can calculate $P(X_{1/3}=1)$ by considering the total length of the arcs along which the third point must lie in order to get $X_{1/3}=1$. We get the following graph.

enter image description here

Averaging over $\theta$, we get $P(X_{1/3}=1)=\frac23$. Therefore $P(X_{1/3}=2)=\frac13$, and so $\mathbb{E}(X_{1/3})=\frac43$.

Using this approach with a general $k$ seems difficult. I wonder if there is an easier approach.

Dan
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  • Isn't this just geometric probability? Cut up the $[0, 1]^2$ square into how many angles are bigger than $k \pi$, and sum it up. – Calvin Lin Oct 11 '24 at 08:56
  • @CalvinLin I'm not sure I follow. In the $[0,1]^2$ square, what do the two axes represent? Please could you elaborate? – Dan Oct 11 '24 at 09:14
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    Similar to what Henry did. Fix $A$ (at the top), then $x \in [0, 1]$ represents the placement of $B$ and $y \in [0, 1]$ represents the placement of $C$. Then figure out $\angle ACB, \angle BAC, \angle CBA > k \pi $ and the implications on $x, y$. (EG $ y > x > k/2 \Rightarrow \angle ACB > k \pi$. – Calvin Lin Oct 11 '24 at 12:47

1 Answers1

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If you are picking points at random from the circumference of a circle, you get something like this, with $A,B,C$ being labelled clockwise round the circle:

triangle in circle

You can then say

  • The event $\angle ACB > k\pi$ corresponds to $\angle AOB > 2k\pi$ with the inscribed angle theorem

  • $\mathbb P(\angle AOB > 2k\pi) = (1-k)^2$ for $0 < k < 1$ since you need both $B$ and $C$ further clockwise round the circumference from $A$

  • The expected number of central angles greater than $2k\pi$ is $3 (1-k)^2$ by linearity of expectation as there are $3$ central angles

  • The expected number of triangle angles greater than $k\pi$ is $3 (1-k)^2$ as equivalent events

Henry
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  • It seems like you're treating the events (of each angle being larger than $2k\pi$) as independent events, when you multiply by $3$. How can that be justified? – Dan Oct 11 '24 at 09:53
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    @Dan: He’s not, it’s “just” the linearity of expected values! Let’s call the points $A,B,C$, and let $E_A$ be the event “the angle $A$ is greater than $k\pi$” (same for $E_B$, $E_C$). Then $X_k$ is the sum of the indicator functions of $E_A,E_B,E_C$. Then the expected value of $X_k$ is the sum of the expected values of the indicator functions of $E_A,E_B,E_C$, ie the probabilities of $E_A,E_B,E_C$. All of these probabilities are equal to $(1-k)^2$, whence the conclusion. – Aphelli Oct 11 '24 at 10:51