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)

#Combinatorics

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Siddharth Iyer - 5 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$