60th Belarusian Mathematical Olympiad

Question

Is it possible to mark 6 distinct points (1 red, 2 blue, and 3 green points) on a plane so that the sum of the distances between the red point and the blue points is 8, the sum of the distances between the red point and the green points is 6, and the sum of the distances between the blue points and the green points is 9? (S. Mazanik, I. Voronovich)

Accepted Answer

Suppose, contrary to our claim, that there exist points satisfying the problem condition. We label the red point, two blue points, and three green points R, B_1, B_2, and G_1, G_2, G_3 respectively. By triangle inequality, B_1G_1 + RG_1 RB_1, B_2G_1 + RG_1 RB_2, B_1G_2 + RG_2 RB_1, B_2G_2 + RG_2 RB_2, B_1G_3 + RG_3 RB_1, B_2G_3 + RG_3 RB_2. Summing these inequalities, we obtain (B_1G_1 + B_1G_2 + B_1G_3 + B_2G_1 + B_2G_2 + B_2G_3) + 2(RG_1 + RG_2 + RG_3) 3(RB_1 + RB_2), i.e. 9 + 2 6 3 8, which is impossible.