Define the point $ C'$ as the simetrical of $ C$ with respect to $ O$ then construct the paralelogram. $ OA'C'B'$ with $ A' \in \vec{OA},B' \in \vec{OB}$ then exists $ p_r,q_r \in \mathbb R^ +$ such $ p_r\cdot\vec{OA} + q_r\cdot\vec{OB} + \vec{OC} = \vec{0}$ pick $ \alpha_p, \alpha_q$ such $ |\alpha_p \cdot\vec{OA} + \alpha_q \cdot\vec{OB}| < \epsilon$ for any $ \epsilon >0$ and $ p_r = \frac {a}{b} + \alpha_p,q_r = \frac {c}{d} + \alpha_q$ then $ p = ad, q = cb,q = bd$ works.

Exist $ (p,q,r)$ for this vector equalities 0 .

mszew wrote: Define the point $ C'$ as the simetrical of $ C$ with respect to $ O$ then construct the paralelogram. $ OA'C'B'$ with $ A' \in \vec{OA},B' \in \vec{OB}$ then exists $ p_r,q_r \in \mathbb R^ +$ such $ p_r\cdot\vec{OA} + q_r\cdot\vec{OB} + \vec{OC} = \vec{0}$ pick $ \alpha_p, \alpha_q$ such $ |\alpha_p \cdot\vec{OA} + \alpha_q \cdot\vec{OB}| < \epsilon$ for any $ \epsilon > 0$ and $ p_r = \frac {a}{b} + \alpha_p,q_r = \frac {c}{d} + \alpha_q$ then $ p = ad, q = cb,q = bd$ works. Your solution will be complete if you wrote that the main inequality follows from Dirichlet theorem.

Erken wrote: Let $ O$ be an interior point of the triangle $ ABC$.Prove that there exist positive integers $ p,q$ and $ r$,such that $ |p\cdot\vec{OA} + q\cdot\vec{OB} + r\cdot\vec{OC}| < \frac {1}{2007}$. Settings $ (p,q,r)\to (0,0,0)$.

N.T.TUAN wrote: Erken wrote: Let $ O$ be an interior point of the triangle $ ABC$.Prove that there exist positive integers $ p,q$ and $ r$,such that $ |p\cdot\vec{OA} + q\cdot\vec{OB} + r\cdot\vec{OC}| < \frac {1}{2007}$. Settings $ (p,q,r)\to (0,0,0)$. But $ p,q,r$ are positive integers,i.e natural numbers.

Any elementary solutions?

jbmorgan wrote: Any elementary solutions? What is non-elementary in the solution above?

I'm not sure what you classify as elementary, but I would think Dirichlet's Theorem is not

Quote: I'm not sure what you classify as elementary, but I would think Dirichlet's Theorem is not Perhaps i misunderstood something but Dirichlet theorem says: For arbitrary $ \alpha\in\mathbb{R}$ and $ m\in\mathbb{N}$ exist such $ p\in\mathbb{Z}$ and $ q\in\mathbb{N}$,$ m\ge q$. We have $ \mid\alpha-\dfrac{p}{q}\mid<\dfrac{1}{qm}$.In our country we classify this theorem as quite elementary...