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Problem 1694

National Olympiad, first round

Geometry Difficulty 6.3 Prove it The 26th Hong Kong (China) Mathematical Olympiad · Hong Kong

Let $A_1A_2\dots A_n$ be a convex polygon where $n \ge 4$ . For each $i = 1, 2, \dots, n$ , let $f(i)$ be the maximum of the angles $\angle A_iA_jA_{i+1}$ where $A_j$ can be any vertex different from $A_i$ and $A_{i+1}$ (with $A_{n+1} = A_1$ ). For each $n$ , find the smallest possible value of $f(1) + f(2) + \dots + f(n)$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

For each $k = 2, 3, \dots, n-1$ , define $S_k = \sum_{i=1}^{n} \angle A_i A_{i+1} A_{i+1}$ where the indices are taken modulo $n$ . Note that
$S_2 + S_3 + \dots + S_{n-1} = \angle A_1 + \angle A_2 + \dots + \angle A_n = (n-2) \cdot 180^\circ \quad (1)$
because each angle $\angle A_i A_j A_k$ with distinct $i, j, k$ is counted exactly once. By definition, the given sum $S$ is at least $S_k$ for each $k = 2, 3, \dots, n-1$ . Therefore, by equation (1), we obtain
$S \ge \frac{S_2 + S_3 + \dots + S_{n-1}}{n-2} = 180^\circ.$
On the other hand, if $A_1 A_2 \dots A_n$ is cyclic (for example, it is a regular $n$ -gon), then
$f(1) + f(2) + \dots + f(n) = A_1 A_2 + A_2 A_3 + \dots + A_{n-1} A_n = 180^\circ.$
Therefore, the smallest possible value of $S$ is $180^\circ$ .

(In the figure below, we have $n=5$ . Also, $S_2$ is the sum of all angles in white, $S_3$ is the sum of all angles in grey, and $S_4$ is the sum of all angles in black.)

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.