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Choose three points randomly $A,B,C$ on a circle with center $O$ and radius $R$. Denote the distances from $O$ to $BC,CA,AB$ as $d_a,d_b,d_c$ respectively. Find the probability that $d_a+d_b+d_c > \frac{3}{2}R$.

First, we can claim that, such triangle $ABC$ must be obtuse. Indeed, if it is any acute or right triangle, then by Jensen's inequality, it holds that $$d_a+d_b+d_c=R(\cos A+\cos B+\cos C)\leq R\cdot 3\cos \left(\frac{A+B+C}{3}\right)=\frac{3}{2}R,$$ which contradicts the demand.

Now, we assume $A$ is just the obtuse angle, and $\angle B=x,\angle C=y$, where $x,y>0$ and $x+y<\frac{\pi}{2}$. Then we obtain $$d_a+d_b+d_c=R[\cos x+\cos y-\cos(\pi-(x+y))]=R[\cos x+\cos y+\cos(x+y)]>\frac{3}{2}R,$$ which implies $$ \cos x+\cos y+\cos(x+y)>\frac{3}{2}.$$ Thus, we consider a square with vertices $(0,0),(0,\pi/2),(\pi/2,\pi/2),(\pi/2,0)$ and the contour line $\cos x+\cos y+\cos(x+y)=\frac{3}{2}.$ Denote the area of the square above as $S$, and the one of the region enclosed by $x-$axis, $y-$axis and the contourline as $S'$. Thus we can claim that $$ P=\frac{S'}{S}$$ is the probability we want.

Am I right? How to go on with this?

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  • $\begingroup$ Minor clarification: You mean the distance from the centre $O$ and the midpoint of the line segments $AB$, $BC$, $AC$? $\endgroup$ – Satwik Pasani yesterday

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