Sa se arate ca oricum am avea 2004 puncte in spatiu exista o sfera care sa aiba in interior 1002 puncte si sa lase in exterior 1002 puncte.
Lavinia Savu ,,Gh. Lazar'',2004
Puncte in interiorul si exteriorul unei sfere
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Fie \( O \) un punct care nu se afla pe niciun plan mediator al vreunui segment \( A_{i}A_{j} \), cu \( i,j=\overline{1,2004} \). Fie \( A_{1},A_{2},\dots,A_{2004} \) cele \( 2004 \) puncte din enunt si fie \( d_{1}=MA_{1},d_{2}=MA_{2},\dots,d_{2004}=MA_{2004} \).
Putem presupune fara a leza generalitatea ca \( d_{1}<d_{2}<\dots<d_{2004} \). Deoarece \( d_{1}<d_{2}<\dots<d_{1002}<\frac{d_{1002}+d_{1003}}{2}<d_{1003}<\dots<d_{2003}<d_{2004} \), considerand sfera \( \mathcal{S}\left(O,\frac{d_{1002}+d_{1003}}{2}\right) \) observam ca aceasta are in interiorul ei punctele \( \left\{ A_{1},A_{2},\dots,A_{1002}\right\} \) si lasa in afara ei punctele \( \left\{ A_{1003},A_{1004},\dots,A_{2004}\right\} \).
Putem presupune fara a leza generalitatea ca \( d_{1}<d_{2}<\dots<d_{2004} \). Deoarece \( d_{1}<d_{2}<\dots<d_{1002}<\frac{d_{1002}+d_{1003}}{2}<d_{1003}<\dots<d_{2003}<d_{2004} \), considerand sfera \( \mathcal{S}\left(O,\frac{d_{1002}+d_{1003}}{2}\right) \) observam ca aceasta are in interiorul ei punctele \( \left\{ A_{1},A_{2},\dots,A_{1002}\right\} \) si lasa in afara ei punctele \( \left\{ A_{1003},A_{1004},\dots,A_{2004}\right\} \).
elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste