PP. Let $ABC$ be an acute triangle, where $A\left(z_1\right)$ , $B\left(z_2\right)$ , $C\left(z_3\right)$ and $\{z_1,z_2,z_3\}$ are pairwise distinct complex numbers satisfying $|z_1| = |z_2| = |z_3| = 1$ and $\sum\frac{1}{2 + |z_2 + z_3|} =1\ (*)\ .$ Prove that $\triangle ABC$ is equilateral. Proof. Suppose w.l.o.g. that the circumcircle of $\triangle ABC$ is $w=C(O,1)$ and $O(0)$ is the origin of the complex plane. Thus, the midpoints of the sides $[BC]$ , $[CA]$ , $[AB]$ are $M\left(\frac {z_2+z_3}{2}\right)$ , $N\left(\frac {z_3+z_1}{2}\right)$ , $P\left(\frac {z_1+z_2}{2}\right)$ respectively. Therefore, the relation $(*)\iff$ $\sum\frac {1}{2+2\cdot OM}=1\iff$ $\boxed{\sum\frac {1}{1+\cos A}=2}\ (1)$ . Since $\sum (1+\cos A)=$ $3+\sum \cos A=$ $4+\frac {r}{R}$ and $\sum\frac {1}{1+\cos A}\cdot\sum(1+\cos A)\ge 9$ obtain that $2\cdot \left(4+\frac rR\right)\ge 9\iff$ $2r\ge R\ \stackrel{(R\ge 2r)}{\iff}R=2r\iff\ A=B=C$ . Remark 1. Prove easily that remarkable identity $\boxed{\ \sum bc(s-b)(s-c)=r^2\left[s^2+(4R+r)^2\right]\ }\ (2)$ . Observe that the relation $(1)$ $\sum\frac {1}{1+\cos A}=2\iff $ $\sum\frac {1}{\cos^2\frac A2}=4\iff$ $\sum\frac {bc}{s-a}=4s\iff$ $\sum bc(s-b)(s-c)=4s^2r^2\stackrel{(2)}{\iff}$ $\underline{4R+r=s\sqrt 3}$ . I"ll show that $4R+r=s\sqrt 3\iff A=B=C$ . Indeed, $\left\{\begin{array}{c} r_a+r_b+r_c=4R+r\\\\ r_ar_b+r_br_c+r_cr_a=s^2\end{array}\right\|$ and $\left(\sum r_a\right)^2\ge 3\left(\sum r_br_c\right)\iff$ $(4R+r)^2\ge 3s^2\iff 4R+r\ge s\sqrt 3$ . In conclusion, $4R+r=s\sqrt 3\iff \left(\sum r_a\right)^2=3\left(\sum r_br_c\right)\iff$ $A=B=C$ . Remark 2. I"ll prove the identity $(2)$ . Thus, $\sum bc(s-b)(s-c)=\sum \left[s(s-a)+(s-b)(s-c)\right](s-b)(s-c)=$ $3S^2+\sum(s-b)^2(s-c)^2=$ $3S^2+\left[\sum (s-b)(s-c)\right]^2-2(s-a)(s-b)(s-c)\sum (s-a)=$ $3S^2+[r(4R+r)]^2-2S^2=$ $s^2r^2+[r(4R+r)]^2\implies$ $\sum bc(s-b)(s-c)=r^2\left[s^2+(4R+r)^2\right]$ . I used the well-known identity $\sum(s-b)(s-c)=r(4R+r)$ . See PP20 from here.