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Print60th Belarusian Mathematical Olympiad
Belarus geometry
Problem
4 blue, 10 green and several (at least one) red points are marked on a plane. All points are distinct. It is known that the sum of the distances between the red points and the blue points is , the sum of the distances between the red points and the green points is . Can the sum of the distances between the blue points and the green points be equal to a) ? b) ?
Solution
a) Let denote the number of red points. Then the following inequality holds (see (*) in the solution of Problem 2, Category C) It is easy to see that there are no integers satisfying this condition.
b) We show that it is possible to mark blue, green, and red points on a plane so that the problem conditions hold. For the convenience we multiply all distances by . Let denote respectively the sums of the distances between red and green points, between blue and red points, green and blue points. We have to construct the example with , , . All points we place on the number axes.
First, we place blue points at the point , green points at the point , red points at the point , and the other red points at the point . Then , , and . It remains to move slightly the points so that all points become distinct but keep their values. We can replace blue points with the common coordinate by the points , , , ; green points with the common coordinate by the points red points with the common coordinate by the points and, finally, red points with the common coordinate by the points We can choose sufficiently small to keep the order of the points on the number axes.
b) We show that it is possible to mark blue, green, and red points on a plane so that the problem conditions hold. For the convenience we multiply all distances by . Let denote respectively the sums of the distances between red and green points, between blue and red points, green and blue points. We have to construct the example with , , . All points we place on the number axes.
First, we place blue points at the point , green points at the point , red points at the point , and the other red points at the point . Then , , and . It remains to move slightly the points so that all points become distinct but keep their values. We can replace blue points with the common coordinate by the points , , , ; green points with the common coordinate by the points red points with the common coordinate by the points and, finally, red points with the common coordinate by the points We can choose sufficiently small to keep the order of the points on the number axes.
Final answer
a) no; b) yes
Techniques
Triangle inequalitiesDistance chasingCartesian coordinatesConstructions and loci