# Probability of an obtuse triangle in a circle.

Suppose we randomly pick 2 points A, B within a circle centered at point O. What is the probability that the triangle formed by ABO is an obtuse one? (Note that A and B are not exclusively on the circumference).

And what is the conclusion extended to A, B within a ball instead?

Thanks!

PS this is from a Quant interview.

The following is what I have derived during the exam: (Edited, thank you for your corrections!) consider the joint probability of x, y coordinate for any point in a unit circle, then $$f_{XY}(x, y) = \frac{1}{\pi}$$, uniformly distributed inside the circular region. The distance between the point and the center, has thus a distribution $$f_Z(z) = 2z$$ for $$z$$ in [0, 1] ($$z^2 = x^2 + y^2$$). Randomly pick an A, rotate the circle so that A is right on top of the center O. Suppose now A has a distance $$z = a$$ $$(a > 0)$$ away from the origin, then B could only be chosen in the region of

1. $$y > a$$
2. $$y < 0$$
3. within an inner circle whose diameter is $$OA$$

Therefore, given that the distance is $$z = a$$, the probability of ABO being an obtuse triangle is given corresponds to area $$\arccos(a) - a\sqrt{1-a^2} + \frac{a^2\pi}{4} + \frac{\pi}{2}$$ (Upper, inner, and lower). By this conditional probability, we could derive the total probability, which is $$\int_0^1 \frac{1}{\pi}\left(\arccos(a) - a\sqrt{1-a^2} + \frac{a^2\pi}{4} + \frac{\pi}{2}\right) \cdot 2a \; da = \frac{3}{4}$$

But is there an easier way to solve this? This looks like a math competition style of question and I expect some tricks to be at play. Thanks!

• If the obtuse angle was at point O, the possibility depends on the distance between point $A$ and point $B$, to the radius $r$ of the circle – Aderinsola Joshua Mar 29 at 7:57

The probability that point $$A$$ is at a distance from the center in $$[a,a+da]$$ is $$2\pi a\,da/\pi=2a\,da$$. The probability that, fixed $$A$$ as above, $$\angle ABO>90°$$, is the same as the probability that $$B$$ lies inside a circle of diameter $$OA$$, that is $$\pi(a/2)^2/\pi=a^2/4$$. Hence the overall probability that $$\angle ABO>90°$$ is: $$p(\angle ABO>90°)=\int_0^1 {a^2\over4}\cdot 2a\,da={1\over8}.$$

The probability that triangle $$ABO$$ is obtuse is then: $$p(\angle ABO>90°)+p(\angle BAO>90°)+p(\angle AOB>90°) ={1\over8}+{1\over8}+{1\over2}={3\over4}.$$

EDIT.

For a 3D sphere one can repeat the same argument, obtaining: $$p(\angle ABO>90°)=\int_0^1 {a^3\over8}\cdot 3a^2\,da={1\over16}.$$

• Thank you! You made me realize that we should not use the marginal w.r.t. $y$ coordinates (feasible but convoluted) but the distance from a given point to the center. And I believe that is $f_Z(z) = 2z$ for $z$ in [0, 1]? This is what you are suggesting right? – Sheng Yang Mar 29 at 11:49
• Yes, if $z$ is the radial distance, then its probability density is $2z$. – Intelligenti pauca Mar 29 at 12:43

Did you forget about the possibility of $$\angle ABO > 90^\circ$$? If you know some geometry, this happens when point B is located inside the circle with diameter $$AO$$, which easily has the area of $$\frac{\pi a^2}{4}$$. Now, we can easily calculate the value of $$P(\angle ABO > 90^\circ)$$. Now by symmetry, the value is the same for $$P(\angle BAO > 90^\circ)$$. The value of $$P(\angle ABO > 90^\circ)$$ is trivially $$0.5$$ as you said, and all those probabilities are exclusive and add up to the answer you need.

This answer extends naturally to the 3D case.

• Thank you! I miss that situation when $P(\angle ABO > 90^\circ)$. But it looks like two quantities are not symmetric if we were to measure by $a$? One corresponds to the area of the little inner circle whose diameter is $AO$ and the other corresponds to the region inside the unit circle above $y > a$. – Sheng Yang Mar 29 at 9:36
• Well I think it's fine. First you set out to calculate the probability $P(\angle ABO > 90^\circ)$ in the general condition, then you may argue to fix the position of $A$. Similarly, you may do identical thing for $B$, but now you fix the position of $B$. Of course fixing $A$ when calculating $P(\angle BAO > 90^\circ)$ will meet the integral you posted in the question. – Bimo Adityarahman Mar 29 at 9:38

Using $$f(a)=\left(2\pi \arccos(a) - a\sqrt{1-a^2} + \frac{1}{2}\right) \, \frac{4}{\pi}\sqrt{1 - a^2}$$ $$I=\int_0^1 f(a)\, da=\frac{5}{2}-\frac{1}{\pi }+\frac{\pi ^2}{2}$$ is not too difficult provided that we have some time.

For an approximation, write for example $$I_n=\int_0^{\frac 12} f(a)\, da+\int^1_{\frac 12} f(a)\, da$$ and use Taylor series of $$f(a)$$ around $$a=0$$ to $$O(a^{n+1})$$ and $$a=1$$ to $$O((1-a)^{n+1})$$.

$$I_0=\frac{1}{\pi }+2 \pi\qquad \qquad\text{relative error = 7.23 %}$$

$$I_1=1-\frac{7}{6 \pi }+2 \pi\qquad \qquad\text{relative error = 2.88 %}$$

I first compute the probability $$p$$ that the triangle $$\triangle:=OAB$$ is acute.

When$$|OA|=r$$ we may assume $$A=(r,0)$$. The point $$B=(u,v)$$ then has to satisfy $$0, and $$B$$ must not lie in the small disc with diameter $$OA$$. The feasible area for $$B$$ therefore is $$b(r):=2\int_0^r\sqrt{1-x^2}\>dx-{\pi r^2\over4}\ .$$ Since $$|OA|$$ is distributed with density $$2r$$ with respect to $$dr$$ we therefore obtain $$p=\int_0^1 {b(r)\over\pi}\>2r\>dr=\ldots={1\over4}\ .$$ It follows that $$\triangle$$ is obtuse with probability $${3\over4}$$.

• This is in agreement with my result: I edited my answer to show that. – Intelligenti pauca Mar 29 at 13:38

Wherever A and B are, one of them is going to be at least as far away from O as the other. Which point has this property is even odds and independent of the obtusity of the triangle, so assume WLOG that A is at least as far away as B.

Once you've "rotated" the point A to be directly above O, you know B is somewhere in (or on) the concentric circle passing through A, with a uniform distribution.

You know where A is relative to this configuration, but you aren't sure about where B is. Regardless, scale the configuration so that this new A-circle has a radius of $$\frac{1}{\sqrt{\pi}}$$ (and thereby an area of $$1$$). This will preserve the obtusity of ABO and avoids the need to integrate wrt the radius $$a$$.

Since the interior angle sum of ABO is $$180^\circ$$, we know there's at most one obtuse angle ($$\ge 90^\circ$$) in the triangle.

As Bimo points out:

• Our first case, $$\angle BOA\geq 90^\circ$$, corresponds to B in the lower half of the A-circle.
• Our second case, $$\angle ABO\geq 90^\circ$$, means B is in a smaller circle, half as tall as the A-circle and fitting neatly in its upper half (by Thales' theorem).
• The third case one might consider, $$\angle OAB\geq 90^\circ$$, is impossible, since this would imply that B were above A (not directly above).
• If none of these are true, B must be elsewhere in the circle.

Adding the (almost) disjoint first and second cases gives a probability of $$\frac{1}{2}+\frac{1}{4}$$ quite simply.

Integration is certainly an appropriate tool here but, as your instincts suggest, should not be used where easier, less mistake-prone solutions can be found.