Let O be an interior point of the triangle ABC. Prove that there exist positive integers p,q and r, such that ∣p⋅OA+q⋅OB+r⋅OC∣<20071.
Solution
It is well-known that there are positive real numbers β,γ such that OA+βOB+γOC=0. So for positive integer k, we have kOA+kβOB+kγOC=0. Let m(k)=[kβ], n(k)=[kγ], where [x] is the biggest integer which is less than or equal to x, and {x}=x−[x].
Assume T is an integer larger than max{β1,γ1}. Then the sequences {m(kT)∣k=1,2,…} and {n(kT)∣k=1,2,…} are increasing, and ∣kTOA+m(kT)OB+n(kT)OC∣=∣−{kTβ}OB−{kTγ}OC∣≤∣OB∣⋅{kTβ}+∣OC∣⋅{kTγ}≤∣OB∣+∣OC∣. This shows there exists infinitely many vectors such that kTOA+m(kT)OB+n(kT)OC, whose endpoint is in a circle O of radius ∣OB∣+∣OC∣. So there are two of the vectors, such that the distance between the endpoints of the two vectors is less than 20071, this means there exists two integers k1<k2, such that ∣(k2TOA+m(k2T)OB+n(k2T)OC)−(k1TOA+m(k1T)OB+n(k1T)OC)∣<20071. So, if we let p=(k2−k1)T, q=m(k2T)−m(k1T), r=n(k2T)−n(k1T), then p,q,r are integers, and ∣pOA+qOB+rOC∣<20071.
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Source: MathNet,
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