Pe un cerc se considera punctele \( A_{1}, A_{2},\ldots A_{n} \) si \( B_{1}, B_{2}, \ldots, B_{n} \). Sa se demonstreze ca exista un punct \( P \) pe cerc astfel incat \( \sum_{i=1}^{n} PA_{i}=\sum_{i=1}^{n} PB_{i} \).
Marius Cavachi, etapa judeteana 1991, Constanta
Exista P pe cerc astfel incat \sum PA_i = \sum PB_i
Moderators: Bogdan Posa, Beniamin Bogosel, Marius Dragoi
-
Cezar Lupu
- Site Admin
- Posts: 612
- Joined: Wed Sep 26, 2007 2:04 pm
- Location: Bucuresti sau Constanta
- Contact:
Exista P pe cerc astfel incat \sum PA_i = \sum PB_i
Last edited by Cezar Lupu on Tue Nov 06, 2007 2:33 am, edited 1 time in total.
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
Fie \( A_i(cos a_i, sin a_i), B_i(cos b_i, sin b_i) \) si \( P(sin x, cos x). \)
\( PA_i =2 \left| \sin \frac {x-a_i}{2} \right| \) si \( PB_i =2 \left| \sin \frac {x-b_i}{2} \right| \). Problema revine la a arata ca exista \( c \in [0,2\pi] \) astfel incat \( \sum_{i=1}^{n} \left( \left| \sin \frac {c-a_i}{2} \right| - \left| \sin \frac {c-b_i}{2} \right| \right) = 0 \).
Calculam \( \int_{0}^{2\pi} \left( \left| \sin \frac {x-a_i}{2} \right| - \left| \sin \frac {x-b_i}{2} \right| \right) dx =\int_{-\frac{a_i}{2}}^{\pi-\frac{a_i}{2}} |\sin t| dt - \int_{-\frac{b_i}{2}}^{\pi-\frac{b_i}{2}} |\sin t|dt =\int_{0}^{\pi}|\sin t |dt - \int _{0}^{\pi} |\sin t|dt =0 \) \( \Rightarrow \) \( \int_{0}^{2\pi} \sum_{i=1}^{n} \left( \left| \sin \frac {x-a_i}{2} \right| - \left| \sin \frac {x-b_i}{2} \right| \right) =0 \Rightarrow \exists c\in [0,2\pi] \) astfel incat \( \sum_{i=1}^{n} \left( \left| \sin \frac {c-a_i}{2} \right| - \left| \sin \frac {c-b_i}{2} \right| \right) = 0 \) si de aici rezulta concluzia.
\( PA_i =2 \left| \sin \frac {x-a_i}{2} \right| \) si \( PB_i =2 \left| \sin \frac {x-b_i}{2} \right| \). Problema revine la a arata ca exista \( c \in [0,2\pi] \) astfel incat \( \sum_{i=1}^{n} \left( \left| \sin \frac {c-a_i}{2} \right| - \left| \sin \frac {c-b_i}{2} \right| \right) = 0 \).
Calculam \( \int_{0}^{2\pi} \left( \left| \sin \frac {x-a_i}{2} \right| - \left| \sin \frac {x-b_i}{2} \right| \right) dx =\int_{-\frac{a_i}{2}}^{\pi-\frac{a_i}{2}} |\sin t| dt - \int_{-\frac{b_i}{2}}^{\pi-\frac{b_i}{2}} |\sin t|dt =\int_{0}^{\pi}|\sin t |dt - \int _{0}^{\pi} |\sin t|dt =0 \) \( \Rightarrow \) \( \int_{0}^{2\pi} \sum_{i=1}^{n} \left( \left| \sin \frac {x-a_i}{2} \right| - \left| \sin \frac {x-b_i}{2} \right| \right) =0 \Rightarrow \exists c\in [0,2\pi] \) astfel incat \( \sum_{i=1}^{n} \left( \left| \sin \frac {c-a_i}{2} \right| - \left| \sin \frac {c-b_i}{2} \right| \right) = 0 \) si de aici rezulta concluzia.