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Geometry Difficulty 5.8 AIME, harder Prove it China

Let OO be an interior point of the triangle ABCABC. Prove that there exist positive integers p,qp, q and rr, such that
pOA+qOB+rOC<12007. |p \cdot \vec{OA} + q \cdot \vec{OB} + r \cdot \vec{OC}| < \frac{1}{2007}.

Solution

It is well-known that there are positive real numbers β,γ\beta, \gamma such that
OA+βOB+γOC=0. \vec{OA} + \beta \vec{OB} + \gamma \vec{OC} = \vec{0}.
So for positive integer kk, we have
kOA+kβOB+kγOC=0. k \vec{OA} + k\beta \vec{OB} + k\gamma \vec{OC} = \vec{0}.
Let m(k)=[kβ]m(k) = [k\beta], n(k)=[kγ]n(k) = [k\gamma], where [x][x] is the biggest integer which is less than or equal to xx, and {x}=x[x]\{x\} = x - [x].

Assume TT is an integer larger than max{1β,1γ}\max\left\{\frac{1}{\beta}, \frac{1}{\gamma}\right\}. Then the sequences {m(kT)k=1,2,}\{m(kT) \mid k = 1, 2, \dots\} and {n(kT)k=1,2,}\{n(kT) \mid k = 1, 2, \dots\} are increasing, and
kTOA+m(kT)OB+n(kT)OC={kTβ}OB{kTγ}OCOB{kTβ}+OC{kTγ}OB+OC. \begin{aligned} & |kT \vec{OA} + m(kT) \vec{OB} + n(kT) \vec{OC}| \\ & = |-\{kT\beta\} \vec{OB} - \{kT\gamma\} \vec{OC}| \\ & \leq |\vec{OB}| \cdot \{kT\beta\} + |\vec{OC}| \cdot \{kT\gamma\} \\ & \leq |\vec{OB}| + |\vec{OC}|. \end{aligned}
This shows there exists infinitely many vectors such that
kTOA+m(kT)OB+n(kT)OC, kT \vec{OA} + m(kT) \vec{OB} + n(kT) \vec{OC},
whose endpoint is in a circle OO of radius OB+OC|\vec{OB}| + |\vec{OC}|. So there are two of the vectors, such that the distance between the endpoints of the two vectors is less than 12007\frac{1}{2007}, this means there exists two integers k1<k2k_1 < k_2, such that
(k2TOA+m(k2T)OB+n(k2T)OC)(k1TOA+m(k1T)OB+n(k1T)OC)<12007. \begin{aligned} & | (k_2 T \vec{OA} + m(k_2 T) \vec{OB} + n(k_2 T) \vec{OC}) \\ & \quad - (k_1 T \vec{OA} + m(k_1 T) \vec{OB} + n(k_1 T) \vec{OC}) | \\ & < \frac{1}{2007}. \end{aligned}
So, if we let p=(k2k1)Tp = (k_2 - k_1)T, q=m(k2T)m(k1T)q = m(k_2T) - m(k_1T), r=n(k2T)n(k1T)r = n(k_2T) - n(k_1T), then p,q,rp, q, r are integers, and
pOA+qOB+rOC<12007. |p \vec{OA} + q \vec{OB} + r \vec{OC}| < \frac{1}{2007}.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.