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Difference between revisions of "2006 SMT/General Problems/Problem 16"

(Created page with "==Solution== The condition we are looking for is <math>35(n-1)\equiv 0 \mod 360</math>, because if any other <math>A_n = A_1</math> than <math>35(n-1)</math> must be a multip...")
 
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==Problem==
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Points <math> A_1, A_2, \cdots </math> are placed on a circle with center <math> O </math> such that <math> \angle OA_nA_{n+1}=35^\circ </math> and <math> A_n\not=A_{n+2} </math> for all positive integers <math> n </math>. What is the smallest <math> n>1 </math> for which <math> A_n=A_1 </math>?
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==Solution==
 
==Solution==
  
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The condition we are looking for is <math>35(n-1)\equiv 0 \mod 360</math>, because if any other <math>A_n = A_1</math> than <math>35(n-1)</math> must be a multiple of 360 as we have rotated <math>(n-1)</math> by the time we place <math>A_n</math>.  
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The condition we are looking for is <math>110(n-1)\equiv 0 \mod 360</math>, because if any other <math>A_n = A_1</math>, than <math>110(n-1)</math> must be a multiple of 360. This is because the central angle formed between two consecutive points is 110 degrees, and because we have rotated <math>(n-1)</math> times by the time we place <math>A_n</math>.  
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<cmath>35(n-1)\equiv 0\mod 360 \Rightarrow 5(n-1)\equiv 0\mod 360 \Rightarrow x\equiv 73,1\mod 360</cmath>
 
  
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We can ignore the case of <math>n=1</math> because be are looking for an <math>n\neq1</math>. Therefore, our answer is the second lowest option, <math>n=73</math> or <math>\boxed{A_{73}}</math>
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<cmath>110(n-1)\equiv 0\mod 360 \Rightarrow 10(n-1)\equiv 0\mod 360 \Rightarrow x\equiv 37\mod 360</cmath>
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We want the first <math>A_n</math> that satisfies the conditions from the problem, therefore, our answer is <math>n=37</math> or <math>\boxed{A_{37}}</math>

Latest revision as of 11:43, 15 January 2020

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 $110(n-1)\equiv 0 \mod 360$, because if any other $A_n = A_1$, than $110(n-1)$ must be a multiple of 360. This is because the central angle formed between two consecutive points is 110 degrees, and because we have rotated $(n-1)$ times by the time we place $A_n$.

\[110(n-1)\equiv 0\mod 360 \Rightarrow 10(n-1)\equiv 0\mod 360 \Rightarrow x\equiv 37\mod 360\] We want the first $A_n$ that satisfies the conditions from the problem, therefore, our answer is $n=37$ or $\boxed{A_{37}}$

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