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Two points $A,B$ fixed on a plane(distance = 2).

C - random choosen point inside circle with radius $R$ with center at the center of $AB$

Find probability of triangle $ABC$ to be obtuse

My thoughts:

  1. If $C$ lies in the circle - $ABC$ will be rectangular
  2. Opposite the larger angle is large side, so using cosine theorem: $$ b^2+c^2<a^2=4R^2, $$ where $a$ - the diagonal, $b=AC, c=BC$

And then I don't know...

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Any such triangle $ABC$ with $AB$ diameter and $C$ any point within the circle will be obtuse.

Proof:-

enter image description here

(Note:-$AO=OB=r$)

Let,C be any point in the circle.$AC$ is extended to meet the circumference at $X$.So,$\angle AXB=\alpha=90^0$ and $\angle \beta>0^0$.So,$\angle ACB=\gamma=\alpha+\beta=90$+something $>0.$So,$\angle ACB$ will always be obtuse and thus $\triangle ACB$ is always obtuse.

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  • $\begingroup$ $AO=OB=1\neq R$ (The solution should be given in terms of R?) $\endgroup$ – Logophobic Apr 6 '16 at 15:56
  • $\begingroup$ @Logophobic, we were thinking along the same lines (see my answer). $\endgroup$ – Barry Cipra Apr 6 '16 at 16:47
  • $\begingroup$ @Logophobic If center of circle is at center of line,then line must be diameter...it's actually $AO=OB=1=R$... $\endgroup$ – tatan Apr 6 '16 at 17:00
  • $\begingroup$ @tatan See images in my post, $R$ is variable, not fixed at $1$. Point $C$ can occur outside the circle with diameter $AB$ when $R>1$. $\endgroup$ – Logophobic Apr 6 '16 at 17:03
  • $\begingroup$ @Logophobic Point $C$ is within the circle as mentioned in the question. $\endgroup$ – tatan Apr 6 '16 at 17:04
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Note that $A$ and $B$ are diametrically opposite points on a circle of radius $r=1$. As tatan's answer shows, if $R\le r$, then all points in the (concentric) circle of radius $R$ produce obtuse triangles. If $R\gt r$, however, the probability of producing an obtuse triangle turns out to be

$${\pi r^2+2(R^2\arccos(r/R)-r\sqrt{R^2-r^2})\over\pi R^2}$$

The idea here, taking the diameter $AB$ as horizontal, is to draw vertical chords through $A$ and $B$ and note that obtuse triangles correspond to points inside the smaller circle and in the two "ears" to the smaller circle's left and right (see the pictures in Logophobic's answer, posted simultaneously with mine); a little trig shows that each ear has area $R^2\arccos(r/R)-r\sqrt{R^2-r^2}$.

This can be nicely rewritten with $\rho=r/R$:

$$P(\rho)=\rho^2+{2\over\pi}\left(\arccos\rho-\rho\sqrt{1-\rho^2}\right)$$

Note that $P(1)=1$, in agreement with tatan's answer. Note also that $P(0)=1$, which can be confirmed by drawing a picture with a very tiny unit circle (see, again, the pictures in Logophobic's answer).

If you know a little calculus, you can derive

$$P'(\rho)=2\rho-{4\over\pi}\sqrt{1-\rho^2}$$

and conclude that the probability of producing an obtuse triangle is minimized when $\rho=2/\sqrt{4+\pi^2}$. In the OP's setting, where $r=1$, this corresponds to $R=\sqrt{4+\pi^2}/2\approx1.862095889$. (Geometrically, this is when the vertical chords through $A$ and $B$ are of length $\pi$.) The minimal probability computes to

$$P\left(2\over\sqrt{4+\pi^2}\right)={2\over\pi}\arccos\left(2\over\sqrt{4+\pi^2}\right)={2\over\pi}\arctan\left(\pi\over2\right)\approx0.639092927$$

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Any point in the small circle built on $AB$ as the diameter will give an obtuse triangle, to the probability is the ratio of the area of the big circle to the area of the small circle, namely $R^2$.

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  • $\begingroup$ (+1) - I drew a picture, but you have already answered $\endgroup$ – Roman83 Apr 6 '16 at 14:39
  • $\begingroup$ What will be big circle? $\endgroup$ – Evgeny Semyonov Apr 6 '16 at 14:39
  • $\begingroup$ Roman83, can you post your picture, please? $\endgroup$ – Evgeny Semyonov Apr 6 '16 at 14:40
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Not a solution, this image should make it clear that the solution is a function of $R$

Any point $C$ within the purple region will form an obtuse triangle $ABC$, while points within the blue region will form an acute triangle, and points on the border of the two regions will form a right triangle.

enter image description here

On the left, $R=2$, and on the right, $R=14$

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  • $\begingroup$ Very nice pictures! $\endgroup$ – Barry Cipra Apr 6 '16 at 16:52

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