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PrintSELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO
Belarus geometry
Problem
There are points on the plane with rational coordinates, all pairwise distances between them being integer. Prove that there exist points on the plane with integer coordinates such that for all .
Solution
Let be the least common multiple of the denominators of all coordinates of the points . After homothety with center at the origin and coefficient , these points will go to points , respectively, with integer coordinates. Let us switch to complex numbers and work with Gaussian integers. Let us introduce the following notation: Let be an arbitrary prime divisor of . Let us show that there is a rotation of the plane such that , that is, . We have and for any , where means norm. If for any , then everything is proven. Suppose that there exists such that . Then it can be shown that there exists such that and . Then for any : - either , whence , - or and , from which and again . Apply the following plane transformation, which is a rotation, since : Each number from can be represented as , which is sent by the rotation to the number . Thus, we get as required. Let us now consider a decomposition of the number into prime factors (not necessarily different). According to what has been proven, there is a rotation such that . Similarly, there is a rotation such that , and so on, there is a rotation such that Since , the composition of rotations sends the set to the set of integer points with the required properties.
Techniques
HomothetyRotationComplex numbers in geometryQuadratic fieldsUnique factorization