I have to get $p$,$q$ and $r$.

$p$ = the probability of triangle $ABC$ is an acute-angled triangle

$q$ = the probability of triangle $ABC$ is a right-angled triangle

$r$ = the probability of triangle $ABC$ is an obtuse-angled triangle.

I have an idea that the diameter of O is important in determining the angle. But I cannot get exact $p$,$q$ and $r$.

Please help.

  • $\begingroup$ Using polar coordinates may help you. $\endgroup$ – Kenny Lau May 15 '16 at 14:15
  • $\begingroup$ But I cannot use polar coordinate since this course is not handlinh that concepts... $\endgroup$ – user331899 May 15 '16 at 14:16
  • $\begingroup$ Hint: try to fix $A$ and $B$ and see how $C$ determines what the triangle is. $\endgroup$ – Kenny Lau May 15 '16 at 14:18
  • $\begingroup$ If AB is the diameter of O then no matter c's position, It is a right-angled triangle. $\endgroup$ – user331899 May 15 '16 at 14:21
  • $\begingroup$ What if AB is not the diameter of O? You're getting there. $\endgroup$ – Kenny Lau May 15 '16 at 14:21

enter image description here

Hint: Given $A$ and $B$ as in the picture above,

  1. $\triangle ABC$ is acute-angled only if $C$ lies in the red arc, which has probability $\frac{\alpha}{2\pi}$.
  2. $\triangle ABC$ is right-angled only if $C$ is either $A^\prime$ or $B^\prime$, which has probability $0$.
  3. $\triangle ABC$ is obtuse-angled otherwise, which has probability $1-\frac{\alpha}{2\pi}$.

Now, of course $p$, $q$, and $r$ should not depend on $\alpha$, so how do we compute those?

  • $\begingroup$ I don't know why but I think probability of $\triangle{ABC}$ being right-angled $\rightarrow 0$ instead of being $0$. $\endgroup$ – Dragonemperor42 May 15 '16 at 15:49

The probability $q$ of getting a right angled triangle $\to 0,$
since $2$ of the $3$ points need to be exactly on the diameter.

For obtuse and acute angled triangles:

Choose $3$ random points on the circumference, draw diametral lines, and randomly choose a pole of each. Any 3 consecutive points necessarily lie in the same semicircle, and will form an obtuse angled triangle. There are $6$ such combos (starting from $1$ through $6$) against a total of $2^3 = 8$, thus $r = \dfrac68 = \dfrac34$,
and $p = 1-r = \dfrac14$


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