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The Clock-Tower School-Juniors Competion 1st problem
Posted: Mon Mar 22, 2010 10:26 am
by salazar
Fie un poligon regulat \( A_1A_2...A_{2010} \) avand centrul in punctul \( O \). Pe fiecare dintre segmentele \( OA_k \), cu \( k=1,2,...,2010 \), se considera punctul \( B_k \) astfel incat: \( \frac{OB_k}{OA_k}=\frac{1}{k}. \)
Determinati raportul dintre aria poligonului \( B_1B_2...B_{2010} \) si cea a poligonului \( A_1A_2...A_{2010} \).
Posted: Mon Mar 22, 2010 5:15 pm
by Claudiu Mindrila
Presupunem ca centrul circumscris poligonului regulat are raza 1 si notam cu \( \phi \) masura unghiului la centru. Intai
\( \left[A_{1}A_{2}\ldots A_{2010}\right]=2010\cdot\frac{1\cdot1\cdot\sin\phi}{2}=1005\sin\phi \). Apoi, cum \( OB_{k}=\frac{1}{k} \), avem ca
\( \left[B_{1}B_{2}\ldots B_{2010}\right]=\left[B_{1}OB_{2}\right]+\left[B_{2}OB_{3}\right]+\ldots+\left[B_{2009}OB_{2010}\right]+\left[B_{2010}OB_{1}\right]=\frac{1}{2}\cdot\sin\phi\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\ldots+\frac{1}{2009\cdot2010}+\frac{1}{2010}\right)=\frac{1}{2}\sin\phi \).
In fine, \( \frac{\left[B_{1}B_{2}\ldots B_{2010}\right]}{\left[A_{1}A_{2}\ldots A_{2010}\right]}=\frac{\frac{1}{2}\sin\phi}{1005\sin\phi}=\frac{1}{2010} \)