# Probability of $d_a+d_b+d_c > \frac{3}{2}R$

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?

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