Suppose three points are chosen at random in a circle. A triangle is made with these three points as vertices.

What's the probability that the triangle contains the origin of the circle?

(Although I have an answer, I thought I would throw it out there and see what others make of it!)

  • 3
    $\begingroup$ To clarify, you mean points don't have to be on the edge of the circle? $\endgroup$
    – Quark
    Commented Feb 7, 2015 at 17:58
  • 2
    $\begingroup$ Right, the points can be anywhere either on the boundary or in the interior of the circle. $\endgroup$
    – Simon S
    Commented Feb 7, 2015 at 18:42
  • $\begingroup$ Related: math.stackexchange.com/questions/268635/… $\endgroup$
    – Simon S
    Commented Feb 7, 2015 at 23:56
  • $\begingroup$ Also don't forget to add the probability that one of the three points lies on origin :) $\endgroup$ Commented Sep 9, 2018 at 12:49

7 Answers 7


First, observe

That the triangle does not contain the center if and only if all three points are contained in a half-circle. So, only the polar angles of the points matter, not their distance to the center.

Next, let's

Label the points $A$, $B$, $C$. If we starting at point $A$ and go clockwise for half a circle, there's a $1/4$ chance that both point $B$ and point $C$ are in that half-circle, putting all the points on a half-circle. Likewise for starting at point $B$ and starting at point $C$.

These events are

Disjoint (well, overlap with probability 0), and comprise all the way the three points can be on a half-circle, since there's always exactly one for the other two are within half a circle clockwise. So, the probability of them being on a half-circle is $$3/4 = 1/4 + 1/4 + 1/4 $$and so the probability the triangle contains the center is the complement $$1 - 3/4 = 1/4$$ By the same argument, if we generalize to choosing $n$ points and connecting them in a polygon in cyclic order of polar angle, the probability that it contains the center is $$ 1 - \frac{n}{2^{n-1}}$$.

  • 6
    $\begingroup$ Elegant and a nice generalization. What I particularly appreciate about this answer is that it is 'easy' to get to the generalization; something I couldn't say about my original answer. $\endgroup$
    – Simon S
    Commented Feb 7, 2015 at 21:37

A generalization to arbitrary centrally symmetric regions

A planar region is called centrally symmetric with respect to the origin, if for every point $P$ in the region also its reflection with respect to the origin is in the region. Examples for centrally symmetric regions are for instance circles, squares, regular hexagons, regular polygons with an even number of sides.

Theorem: Let $R$ be a (bounded) region that is centrally symmetric with respect to the origin. If three points are randomly chosen from $R$, then the probability that the resulting triangle contains the origin is $1/4$.

As a special case, we derive that the corresponding probability for a circle is also $1/4$.


The proof

(1) My argument heavily uses the fact that exactly one of the following two statements holds true for any triangle $\Delta$:

  • The triangle $\Delta$ does contain the origin.
  • There exists a line that separates the origin from the triangle $\Delta$.

(2) Now consider the region $R$, and let $\ell$ be a horizontal line through the origin. We will assume throughout that none of the three randomly chosen points does ever land directly on line $\ell$ (as these cases have measure zero in the underlying probability space).
A situation where the triangle contains the origin is called "success", and a situation where the triangle does not contain the origin is called "failure".

(3) We distinguish three cases for three randomly chosen points $A,B,C$ in the region $R$:

Case (3a): All three points are above $\ell$.
Case (3b): All three points are below $\ell$.
Case (3c): Two points are above $\ell$, and one is below $\ell$.
Case (3d): Two points are below $\ell$, and one is above $\ell$.

The respective probabilities for cases (3a), (3b), (3c), (3d) are $1/8$, $1/8$, $3/8$ and $3/8$. Furthermore, cases (3a) and (3b) are clear failures (as line $\ell$ itself separates the origin from the resulting triangle). By symmetry, the success probabilities for cases (3c) and (3d) are identical and will be called $p^*$.

(4) Now let us compute the success probability $p^*$ for case (3c). Assume that two points $A$ and $B$ are above $\ell$, whereas the third point $C$ is below $\ell$. The triangle $\Delta ABC$ is a success, if and only if the line through $C$ and the origin has $A$ on one side and $B$ on the other side. Let point $C'$ be the mirror image of $C$ with respect to the origin. The triangle $\Delta ABC$ is a success,

  • if and only if the line through $C'$ and the origin has $A$ on one side and $B$ on the other side,
  • which happens if and only if in the clockwise ordering of $A$, $B$, $C'$ around the origin we have point $C'$ in the middle between $A$ and $B$.

Now note that the probability distribution for choosing $C$ is identical to the probability distributions for choosing $C'$, for choosing $A$ and for choosing $B$, as all these points are chosen from congruent regions (that are one half of the symmetri region $R$). Consider an arbitrary choice of three points $X$, $Y$, $Z$ from the half-region. If we randomly label these three points by $A$, $B$, $C'$, then exactly two of the six labelings have $C'$ in the middle between $A$ and $B$.

Summarizing, the success probability in case (3c) is $p^*=2/6=1/3$.

(5) Now let us wrap things up. The overall success probability equals the overall success probability in the four cases

Case (3a): $\frac14\cdot0$
Case (3b): $\frac14\cdot0$
Case (3c): $\frac38\cdot p^*$
Case (3d): $\frac38\cdot p^*$

As the sum of these four success probabilities is $\frac34\cdot p^* =\frac14$, the proof is complete.

  • $\begingroup$ Great. I agree with $p = 1/4$. I'll wait to see if someone posts another solution before posting mine. $\endgroup$
    – Simon S
    Commented Feb 7, 2015 at 20:06
  • $\begingroup$ (For the benefit of people just reading the thread: This answer has been generalized and rewritten.) $\endgroup$
    – Simon S
    Commented Feb 8, 2015 at 17:58

Imagine putting the three points within the circle one-by-one. We are only interested in the angular coordinates of these points. We use the first point to define the zero point for this angular coordinate. The second and third point have angular coordinates running from $-\pi$ to $+\pi$. The combinations of both angular coordinates for which the triangle contains the origin is indicated in below figure as the two blue regions. These blue regions cover $1/4$ of the total area.

enter image description here


Put the center of the circle as the origin of a polar coordinate system. Then each point is determined by and angle $\theta$ and a distance $r$ from the origin (we will assume none of the points lands on the origin). Then the triangle joining the three points will contain the origin if and only if the three $\theta$ values are not contained in any half-circle.

Given angles $\theta_1$ and $\theta_2$, let $|\theta_2-\theta_1|$ denote their angular separation, chosen so that $0^\circ\leq|\theta_2-\theta_1|\leq 180^\circ$. Then the set of andlges $\theta_3$ such that $\theta_1,\theta_2,\theta_3$ aren't contained in a half-circle is an arc, of length $|\theta_2-\theta_1|$ (this arc is antipodal to the arc connecting $\theta_1$ and $\theta_2$; see the image, where this arc is colored red). Thus the probability that a randomly chosen $\theta_3$ satisfying this condition is $$ \frac{|\theta_2-\theta_1|}{360^\circ}. $$

If $\theta_1$ and $\theta_2$ are chosen independently at random, then the expected value of $|\theta_2-\theta_1$ is $90^\circ$ (the quantity $|\theta_2-\theta_1|$ will be uniform in the interval $[0,180^\circ]$). This means the probability the center of the circle lies in the triangle with three random points as vertices is $$ \frac{90^\circ}{360^\circ}=\frac{1}{4}. $$

enter image description here


Thanks everyone for their answers. Here's mine which is close in spirit to a couple of the existing ones.

Consider a circle centered at the origin. Without loss of generality, let one of the points lie on the positive $x$-axis, call it $p_1$. We can do this just by rotating the coordinate system and the fact that the probability the origin is one of the points is zero.

Label the other two points $p_2, p_3$ such the angle they make is in ascending order, $0 = \theta_1 \leq \theta_2 \leq \theta_3 < 2\pi$.

Now observe that if a triangle formed by the three points containing the origin can be formed we necessarily have $\theta_2 < \pi$, as if $\theta_2 \geq \pi$ this implies $\theta_3$ is also greater than $\pi$ and both $p_2$ and $p_3$ lie in the 3rd and 4th quadrants. Similarly, we require $\theta_3 > \pi$, as if $\theta \leq \pi$ then $p_2$ and $p_3$ lie in the 1st and 2nd quadrants.

Those two conditions being satisfied, we also require that the difference $\theta_3 - \theta_2 < \pi$ since if the difference is greater than $\pi$ it is possible to draw a line through the origin which doesn't intersect the triangle.

The desired probability, $p$, is then the proportion of the parameter space $$S = \{ (\theta_2,\theta_3) \ : \ 0 \leq \theta_2 \leq \theta_3 < 2\pi\}$$ that meets all three conditions, $S_{Yes} = \{(\theta_2,\theta_3) \in S : \theta_2 < \pi \ \wedge \ \theta_3 > \pi \ \wedge \ \theta_3 - \theta_2 < \pi \}$:

$$p = \frac{\mu(S_{Yes})}{\mu(S)} = \frac{\frac{1}{2}\pi^2}{\frac{1}{2}(2\pi)^2} = \frac{1}{4}$$

enter image description here

The three conditions ($\theta_2 < \pi \ \wedge \ \theta_3 > \pi \ \wedge \ \theta_3 - \theta_2 < \pi$) are equivalent to saying the three points don't lie in a semi-circle. What's elegant about xnor's answer above is how it abstracts away the messiness of angular coordinates altogether and just uses that key condition.


(Note: this solution was inspired by Gamow's opening remark regarding central symmetry. It shares some ideas with Gamow's solution).

Consider a random triangle ABC.

You can build 7 other triangles by mirroring some of the points wiht respect to the origin. This creates a set of 8 triangles arranged in what looks like an octahedron.

Every triangle in the circle is part of such a set. All the triangles in a set have the same probability to be chosen in the first place. And in a set, exactly 2 of the triangles cover the origin.

This means the probability that a random triangle covers the origin is 2/8 = 1/4.

enter image description here

And this is true not only for a circle but for any figure that has central symmetry. For example the cross on the Swiss flag. Or a showflake.

Swiss flag showflake


$\bigtriangleup$ABC contains the circle's centre O if A, B, C are all in a semicircle (sc). So consider only their arguments. Write these as between 0 and $2\pi$.

Assume WLOG that A has the smallest argument. Rotate the circle so that OA points straight up. There are now 3 cases for B and C. (Cases where a point is on a boundary have 0 probability and can be ignored.) Case 1 (p=0.25): both are in the righthand sc (RSC); Case 2 (p=0.25): neither is in RSC; Case 3 (p=0.5): exactly one is in RSC. To get $\bigtriangleup$s that contain O, we need consider only Case 3. Assume WLOG that B is in RSC. For $\bigtriangleup$ to contain O, we need C's argument to be less than $\pi$ greater than B's. As B's argument varies from 0 to $\pi$, the probability of this occurring varies from 0 to 1. It can be shown that the mean probability of it is 0.5. So the probability of $\bigtriangleup$ containing O is 0.5 × 0.5 = 0.25.


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