Olympiad Maths Prep

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Geometry Difficulty 5.8 AIME, harder Prove it Belarus

Is it possible to mark 66 distinct points (11 red, 22 blue, and 33 green points) on a plane so that the sum of the distances between the red point and the blue points is 88, the sum of the distances between the red point and the green points is 66, and the sum of the distances between the blue points and the green points is 99?

(S. Mazanik, I. Voronovich)

Solution

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 RR, B1B_1, B2B_2, and G1G_1, G2G_2, G3G_3 respectively. By triangle inequality,
B1G1+RG1≥RB1,B2G1+RG1≥RB2,B1G2+RG2≥RB1, B_1G_1 + RG_1 \ge RB_1, \quad B_2G_1 + RG_1 \ge RB_2, \quad B_1G_2 + RG_2 \ge RB_1,
B2G2+RG2≥RB2,B1G3+RG3≥RB1,B2G3+RG3≥RB2. B_2G_2 + RG_2 \ge RB_2, \quad B_1G_3 + RG_3 \ge RB_1, \quad B_2G_3 + RG_3 \ge RB_2.
Summing these inequalities, we obtain
(B1G1+B1G2+B1G3+B2G1+B2G2+B2G3)+2(RG1+RG2+RG3)≥3(RB1+RB2), (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) \ge 3(RB_1 + RB_2),
i.e. 9+2⋅6≥3⋅89 + 2 \cdot 6 \ge 3 \cdot 8, which is impossible.

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