Let $A_1, A_2, . . . , A_{2021}$ be $2021$ points on the plane, no three collinear and $$\angle A_1A_2A_3 + \angle A_2A_3A_4 +... + \angle A_{2021}A_1A_2 = 360^o,$$in which by the angle $\angle A_{i-1}A_iA_{i+1}$ we mean the one which is less than $180^o$ (assume that $A_{2022} =A_1$ and $A_0 = A_{2021}$). Prove that some of these angles will add up to $90^o$. Proposed by Morteza Saghafian - Iran
Problem
Source: Iranian Geometry Olympiad 2021 IGO Elementary p5
Tags: geometry, angles
qwertyboyfromalotoftime
25.01.2022 20:17
parmenides51 wrote: Let $A_1, A_2, . . . , A_{2021}$ be $2021$ points on the plane, no three collinear and $$\angle A_1A_2A_3 + \angle A_2A_3A_4 +... + \angle A_{2021}A_1A_2 = 360^o,$$in which by the angle $\angle A_{i-1}A_iA_{i+1}$ we mean the one which is less than $180^o$ (assume that $A_{2022} =A_1$ and $A_0 = A_{2021}$). Prove that some of these angles will add up to $90^o$. Proposed by Morteza Saghafian - Iran
$\boxed{Solution.}$ Let $\alpha_i$ be the angle $ \angle A_{i-1} A_i A_{i+1}$ which is less than $180^{\circ}.$
Starting from $A_1$, we walk on the perimeter of the (not necessarily simple) polygon $A_1A_{2} . . . A_{2021}$.
As we reach a vertex $A_i$, we turn by angle $180^{\circ}- \alpha_i$ in clockwise or counterclockwise direction. After walking one round and returning back to the edge $A_1A_2,$ we have turned by a multiple of $360^{\circ}$ in total. Therefore, the signed sum of turning angles is a multiple of $360^{\circ}.$ More formally, if we define $C_1$ and $C_2$ for the set of clockwise and counterclockwise angles, then for some integer
number $k$ we have $$360^{\circ}k=\sum_{\alpha_i \in C_1}(180^{\circ}-\alpha_i)-\sum_{\alpha_j \in C_2}(180^{\circ}-\alpha_j)$$But the total number of angles we have is $2021$ which is an odd number, so if we cancel numbers $180$ as much as possible from the above expression, we can conclude that $360^{\circ}t+180^{\circ}=\sum_{\alpha_i \in C_1}\alpha_i-\sum_{\alpha_j \in C_2}\alpha_j$
for some integer number $t.$ On the other hand, by the assumption of the problem, the sum $\sum_{\alpha_i \in C_1}\alpha_i+\sum_{\alpha_j \in C_2}\alpha_j$ is equal to $360^{\circ}.$ This implies that $\sum_{\alpha_i \in C_1}\alpha_i-\sum_{\alpha_j \in C_2}\alpha_j=\pm 180^{\circ}.$ Therefore, the two sums should be $90^{\circ}$ and $270^{\circ}.$
Attachments:
