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Suppose that someone randomly picks $3$ points $A, B$ and $C$ on a fixed circle. What is the probability of triangle $ABC$ to be acute?

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    $\begingroup$ Do you mean on the circle, or inside the circle? $\endgroup$ – Robert Israel Apr 24 '15 at 20:02
  • $\begingroup$ On the bounddary of the circle @RobertIsrael, not inside. $\endgroup$ – brick Apr 24 '15 at 20:05
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    $\begingroup$ Possible duplicate of math.stackexchange.com/questions/560441/… $\endgroup$ – Ojas Apr 24 '15 at 20:18
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Let one point be fixed and let circle have radius $1$. Now find posibility for other two points. First integrate all possibilities: $$I_1=\int_0^{2\pi}\int_0^{2\pi}dbda=4\pi^2$$ Then integrate required case: $$I_2=\int_0^{\pi}\int_a^{2\pi}dbda+\int_{\pi}^{2\pi}\int_0^{a}dbda=3\pi^2$$ So, your probability is $\frac {I_2} {I_1}=\frac34$.
Explanation: let second point be $A$ and third be $B$. When we choose $A$, we cannot choose point $B$ between first point and $A$. Because of that reason, we integrate from $0$ to $a$.

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  • $\begingroup$ It looks to me like the second integral (you really should use something other than $A$ and $B$ for them) is almost the space where the three points form an obtuse triangle. For instance, in the first part, if we let $C$ be at argument $0$ on the unit circle, $A$ be at $a$, and then if $B$ is anywhere between $0$ and $a+\pi$, then an obtuse triangle will be formed. Of course, then the second integral should have inner limits $-\pi$ to $a-\pi$. The area will still be $3\pi^2$, but then the ratio of the two integrals yields the probability of an obtuse triangle, not an acute one. $\endgroup$ – Brian Tung Apr 24 '15 at 22:22
  • $\begingroup$ @BrianTung. Yes, you are right. I edited integral limits. $\endgroup$ – user164524 Apr 24 '15 at 22:58
  • $\begingroup$ But in addition, the probability of an acute triangle is $1$ minus the ratio of the two integrals, or $1/4$, isn't it? The probability that the third point fits in the same semicircle as the first two (and therefore creates an obtuse triangle) is always greater than $1/2$, isn't it? $\endgroup$ – Brian Tung Apr 24 '15 at 23:24
  • $\begingroup$ @BrianTung. Why you think that triange is acute iff third point isn't in the same semicircle as the first two? $\endgroup$ – user164524 Apr 25 '15 at 10:13
  • $\begingroup$ Because if they all fit in the same semicircle, then the center point will be opposite an expanse of circle that is greater than 180 degrees, and therefore the angle it subtends will be greater than half of that, or 90 degrees. The triangle is then obtuse. If they don't all fit in the same semicircle, then no point is opposite an expanse of circle greater than 180 degrees, and each angle is less than half of 180 or 90 degrees. It is therefore if and only if. $\endgroup$ – Brian Tung Apr 25 '15 at 15:55
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Put the first point at the top of the circle. Randomly select the second point uniformly from the circle; without loss of generality, assume the point falls on the right half. (If it's on the left, the arguments that follow can just be mirror-reversed.)

Let the angle between the two points (as measured from the circle's center) be $x$ radians; clearly, $0 \leq x \leq \pi$. We claim (and will shortly demonstrate) that the triangle will be acute if and only if the third point is chosen such that no semicircle contains all three points. This happens if the third point is chosen on the left side, between $\pi-x$ and $x$ radians from the first point at the top of the circle. The probability of this, given $x$, is $x/(2\pi)$. Since $x$ ranges freely and uniformly from $0$ to $\pi$, $x/(2\pi)$ ranges freely and uniformly from $0$ to $1/2$, and the desired probability is the average of $x/(2\pi)$, or $1/4$.

Argument that a triangle inscribed in a circle is acute if and only if the three points cannot be contained by a semicircle: Let $A, B, C$ be three points on a circle. Suppose that all three points are contained in a semicircle, in the order $A, B, C$ (without loss of generality). Then $B$ is opposite a section of the circle subtending at least $\pi$ radians. Since m$\angle B$ is half of the section of circle it subtends, $m\angle B \geq \pi/2$ and the triangle is obtuse (or right).

Suppose, on the other hand, that no semicircle contains the three points. Then no point is opposite a section of circle $\geq \pi$ radians, and therefore each of the three angles is less than half that, or $\pi/2$ radians; the triangle is therefore acute.

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