Question 1
In the plane, let $M_0M_1\cdots M_{2n}$, $n\ge1$, be a convex polygon whose vertices occur counterclockwise on a circle $(C)$ of radius $R$. A point $A$ inside the polygon satisfies
$$\angle M_0AM_1=\angle M_1AM_2=\cdots=\angle M_{2n}AM_0=\frac{360^\circ}{2n+1}.$$
Assume $A$ is not the center. Let $B$ lie on the circle so that $AB$ is perpendicular to the diameter through $A$. Prove
$$\frac{2n+1}{\frac1{AM_0}+\frac1{AM_1}+\cdots+\frac1{AM_{2n}}}\lt AB\lt\frac{AM_0+AM_1+\cdots+AM_{2n}}{2n+1}\lt R.$$
The source calls the same circle $(O)$ in the sentence defining $B$; we use $(C)$ consistently and denote its center by $O$. Either of the two possible points $B$ gives the same length.
The supplied source has no question point values. Each problem uses 1 editorial point for import compatibility; this is not an official scoring scale. The printed dates are 16–17 May 1990; grouping Q1–3 and Q4–6 into sessions is an editorial inference. Import scoring multipliers: correct 1, incorrect 0, blank 0.
$$\angle M_0AM_1=\angle M_1AM_2=\cdots=\angle M_{2n}AM_0=\frac{360^\circ}{2n+1}.$$
Assume $A$ is not the center. Let $B$ lie on the circle so that $AB$ is perpendicular to the diameter through $A$. Prove
$$\frac{2n+1}{\frac1{AM_0}+\frac1{AM_1}+\cdots+\frac1{AM_{2n}}}\lt AB\lt\frac{AM_0+AM_1+\cdots+AM_{2n}}{2n+1}\lt R.$$
The source calls the same circle $(O)$ in the sentence defining $B$; we use $(C)$ consistently and denote its center by $O$. Either of the two possible points $B$ gives the same length.
The supplied source has no question point values. Each problem uses 1 editorial point for import compatibility; this is not an official scoring scale. The printed dates are 16–17 May 1990; grouping Q1–3 and Q4–6 into sessions is an editorial inference. Import scoring multipliers: correct 1, incorrect 0, blank 0.