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Combinatorics Question

The vertices of a cube are numbered from 1 to 8. Each edge is then labeled with the sum of the numbers at its endpoints. Is it possible that every edge has a different label?

(Source: WOOT Olympiad 2)

Note by Alan Yan
2 years ago

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The sum of the edges is 3 times the sum of the corners which is 3(1+2...8)= 108. Now the sum of two adjacent corners belongs to the set {3,4,...15}. This set contains 13 elements, we need 12 elements since there are 12 edges. 3+4+... 15 -x =108, x=9. Now the labels 15,14,13 and 12 exist. The only way to write 15 and 14 is 8+7 and 8+6 respectively with the allowed numbers so 8 is adjacent to 7 and 6 where 6 is not adjacent to 7. The only way to write 13 is 7+6 or 8+5 with the allowed numbers but since 7 is not adjacent to 6, 5 must be adjacent to 8 whereby it is not adjacent to 7 (as 13 is an included label).12 is also included and the only way to write 12 is 7+5 or 8 + 4 but 7 is not adjacent to 5 so 8 is adjacent to 4 but 8 is adjacent to exactly 3 different vertices being 7,5 and 6 so it is not possible

- 1 year, 11 months ago