Competition problem — Combinatorics

Library / /10 of 54

Combinatorics Difficulty 5.3 AIME, harder Prove it China

Put numbers $1, 2, 3, 4, 5, 6, 7$ and $8$ at the vertices of a cube, such that the sum of any three numbers on any face is not less than $10$ . Find the minimum sum of the four numbers on a face. (posed by Qiu Zonghu)

Solution

Suppose that the four numbers on a face of the cube are $a_1, a_2, a_3, a_4$ such that their sum reaches the minimum and $a_1 < a_2 < a_3 < a_4$ . Since the maximum sum of any three numbers less than $5$ is $9$ , we have $a_4 \ge 6$ , and then $a_1 + a_2 + a_3 + a_4 \ge 16$ .

As seen in the figure, we have $2+3+5+6=16$ , and that means the minimum sum of the four numbers on a face is $16$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.