Star
Posted: March 15th, 2023, 10:33 am
Shares, Investment and Personal Finance Discussion Forums
https://www.lemonfool.co.uk/
cinelli wrote:
I believe there are six essentially different solutions.
Cinelli
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9873210 wrote:
I believe you can shorten your solution significantly by observing that the sums of the vertices of both big triangles must be 13. This is a stronger theorem than the one you use that the sum must be greater than 11. Also the proof is simpler.
Divide the star into three parts,
U - the three corners of the up pointing triangle ;
D - the corners of the down pointing triangle ; and
I - the six remaining interior points.
Then, with loose notation, sum (2*U+I) = 3*26 = sum( 2*D+I) so sum(U) = sum(D) and it's given that sum(D + U) = 26
You could use this to proceed as you did. But shorten many of your cases.
. A
/ \
D---I---K---E
\ / \ /
G H
/ \ / \
C---L---J---B
\ /
F
cinelli wrote:The six rows of four must add to 26 but the sum of the points of the
start, A E B F C D, can be any number from 12 to 27, except 14 and 25,
which are impossible. A+B+C must be the same as D+E+F
Cinelli
9873210 wrote:
I may have misunderstood, but
A+B+C = D+E+F would suggest that A+B+C+D+E+F must be even. How can it be say 13 or 27?