Se considera un poligon regulat \( A_1A_2...A_n \) inscris in cercul de centru \( O \) si multimile de vectori:
\( X=\{\vec{OA_1},\ \vec{OA_2},\ ...,\ \vec{OA_n}\}\ , \)
\( Y=\{\vec{OA_1},\ \vec{OA_1}+\vec{OA_2},\ ...,\ \vec{OA_1}+\vec{OA_2}+...+\vec{OA_n}\}\ . \)
Sa se determine multimea \( X\cap Y \).
Concursul Gheorhge Mihoc, 2008
Doua multimi de vectori
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Mateescu Constantin
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Mateescu Constantin
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Observam ca \( \vec{OA_1}\in X\cap Y \). Sa consideram punctele din plan \( B_1,\ B_2,\ .\ .\ .,B_n \), astfel incat \( \vec{OB_k}=\sum_{p=1}^k \vec{OA_p},\ k=\overline{1,n} \)
Constatam ca \( B_1=A_1 \). De asemenea, avem:
\( \vec{B_kB_{k+1}}=\vec{OB_{k+1}}-\vec{OB_k}=\vec{OA_{k+1}},\ k=\overline{1,n-1} \).
\( \Longrightarrow m(\angle OB_1B_2)=m(\angle B_1B_2B_3)=\ _{\dots} =m(\angle B_{k-1}B_kB_{k+1})=180^{\circ}-\frac{360^{\circ}}{n}=\frac{(n-2)\cdot 180^{\circ}}{n} \).
Daca exista \( k,\ l\in\{1,2,\ .\ .\ .,n\} \) astfel incat \( \vec{OA_l}=\vec{OA_1}+\vec{OA_2}+\ .\ .\ .+\vec{OA_k} \),
rezulta ca poligonul \( OB_1B_2\ .\ .\ .B_{k-1}A_l \) este regulat, care are unghiurile de masura \( \frac{(n-2)\cdot 180^{\circ}}{n} \).
Prin urmare \( k-1+2=n \Longrightarrow k=n-1 \), si \( m(\angle A_1OA_l)=(l-1)\frac{360^{\circ}}{n}=\frac{n-2}{n}\cdot 180^{\circ} \)
\( \Longrightarrow l=\frac{n}{2} \), ceea ce este posibil cand \( n \) este numar par.
\( \Longrightarrow X\cap Y=\left\{ \begin \left\{\vec{OA_1},\ \vec{OA_{\frac{n}{2}}\right\},\ \mbox{daca n este par} \\ \{\vec{OA_1}\},\ \mbox{daca n este impar \)
Constatam ca \( B_1=A_1 \). De asemenea, avem:
\( \vec{B_kB_{k+1}}=\vec{OB_{k+1}}-\vec{OB_k}=\vec{OA_{k+1}},\ k=\overline{1,n-1} \).
\( \Longrightarrow m(\angle OB_1B_2)=m(\angle B_1B_2B_3)=\ _{\dots} =m(\angle B_{k-1}B_kB_{k+1})=180^{\circ}-\frac{360^{\circ}}{n}=\frac{(n-2)\cdot 180^{\circ}}{n} \).
Daca exista \( k,\ l\in\{1,2,\ .\ .\ .,n\} \) astfel incat \( \vec{OA_l}=\vec{OA_1}+\vec{OA_2}+\ .\ .\ .+\vec{OA_k} \),
rezulta ca poligonul \( OB_1B_2\ .\ .\ .B_{k-1}A_l \) este regulat, care are unghiurile de masura \( \frac{(n-2)\cdot 180^{\circ}}{n} \).
Prin urmare \( k-1+2=n \Longrightarrow k=n-1 \), si \( m(\angle A_1OA_l)=(l-1)\frac{360^{\circ}}{n}=\frac{n-2}{n}\cdot 180^{\circ} \)
\( \Longrightarrow l=\frac{n}{2} \), ceea ce este posibil cand \( n \) este numar par.
\( \Longrightarrow X\cap Y=\left\{ \begin \left\{\vec{OA_1},\ \vec{OA_{\frac{n}{2}}\right\},\ \mbox{daca n este par} \\ \{\vec{OA_1}\},\ \mbox{daca n este impar \)