VN-TST 1990 Problem 1

Let be given a convex polygon $M_0M_1\ldots M_{2n}$ ( $n\ge 1$ ), where $2n + 1$ points $M_0$ , $M_1$ , $\ldots$ , $M_{2n}$ lie on a circle $(C)$ with diameter $R$ in an anticlockwise direction. Suppose that there is a point $A$ inside this convex polygon such that $\angle M_0AM_1$ , $\angle M_1AM_2$ , $\ldots$ , $\angle M_{2n - 1}AM_{2n}$ , $\angle M_{2n}AM_0$ are equal. Assume that $A$ is not coincide with the center of the circle $(C)$ and $B$ be a point lies on $(C)$ such that $AB$ is perpendicular to the diameter of $(C)$ passes through $A$ . Prove that $$\frac {2n + 1}{\frac {1}{AM_0} + \frac {1}{AM_1} + \cdots + \frac {1}{AM_{2n}}} < AB < \frac {AM_0 + AM_1 + \cdots + AM_{2n}}{2n + 1} < R$$