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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!

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  • $\begingroup$ 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 $\endgroup$ – Aderinsola Joshua Mar 29 at 7:57
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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: $$ p(\angle ABO>90°)=\int_0^1 {a^3\over8}\cdot 3a^2\,da={1\over16}. $$

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  • $\begingroup$ 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? $\endgroup$ – Sheng Yang Mar 29 at 11:49
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    $\begingroup$ Yes, if $z$ is the radial distance, then its probability density is $2z$. $\endgroup$ – Intelligenti pauca Mar 29 at 12:43
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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.

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  • $\begingroup$ 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 $y > a$. $\endgroup$ – Sheng Yang Mar 29 at 9:36
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    $\begingroup$ 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. $\endgroup$ – Bimo Adityarahman Mar 29 at 9:38
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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 $O((1-a)^{n+1})$.

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

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

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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 $0<u<r$, 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 ${3\over4}$.

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

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