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 A Tri Star Issue (Posted on 2008-06-01)
The letters, A to L, within this star represent intersections of unique pairings of its 6 lines, and α, β, γ, δ, ε and ζ are sums of intersections defined as:
```α = A + D + G + K     β = E + G + J + L    γ = K + J + I + H
δ = L + I + F + B     ε = H + F + C + A    ζ = B + C + D + E
```
```
A α
/ \
ζ  B---C---D---E  β
\ /     \ /
F       G
/ \     / \
ε  H---I---J---K  γ
\ /
L  δ

```
Assign values from 1 to 12 to each of the locations A to L such that each sum is an element of an arithmetic progression with an arithmetic difference of two (2) but not necessarily as adjacent vertex values.

Secondly, attempt the same task but with a difference of four (4) as the outcome.

And for a tease... can you offer a solution if all such vertex sums are equal, ie, 26?

Note:
Discounting rotations and reflections, more than one possibility exists for each of the first two tasks.

 See The Solution Submitted by brianjn Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): computer solution; spoiler | Comment 7 of 17 |
(In reply to re: computer solution; spoiler by Charlie)

I'm afraid the requirement that H, A and K be in ascending order, and that the 1 be in position H, C, G or I was not sufficient to prevent reflections. For example, both the following are present:

`      5                  6  7   9  4             11      2               3  1  12 10                  8     `
`and `
`     6                5  7  11  3            9      1             4  2  12  8               10     `

each being a reflection of the other along a line joining the 7 and the 12.

Off hand, I can't think of a way of better eliminating reflections, other than brute comparison of each pair of solutions.

 Posted by Charlie on 2008-06-02 00:24:25

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