2006 SMT/General Problems/Problem 16

Revision as of 17:35, 14 January 2020 by Dividend (talk | contribs) (Solution)

Problem

Points $A_1, A_2, \cdots$ are placed on a circle with center $O$ such that $\angle OA_nA_{n+1}=35^\circ$ and $A_n\not=A_{n+2}$ for all positive integers $n$. What is the smallest $n>1$ for which $A_n=A_1$?


Solution

The condition we are looking for is $35(n-1)\equiv 0 \mod 360$, because if any other $A_n = A_1$ than $35(n-1)$ must be a multiple of 360 as we have rotated $(n-1)$ by the time we place $A_n$.

\[35(n-1)\equiv 0\mod 360 \Rightarrow 5(n-1)\equiv 0\mod 360 \Rightarrow x\equiv 73,1\mod 360\]

We can ignore the case of $n=1$ because be are looking for an $n\neq1$. Therefore, our answer is the second lowest option, $n=73$ or $\boxed{A_{73}}$