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Topic: Star (Read 361 times) |
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ThudnBlunder
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On the star below arrange the integers 1 to 12, one on each vertex, so that the four integers in each line sum to 26.
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TenaliRaman
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Re: Star
« Reply #1 on: Jan 10th, 2004, 11:40am » |
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::
i will try to avoid making a diagram here...
Tilt the star a bit so that one of the pointy head is at the top.Then the numbering done should be,
.........1
3...11....8...4
..12........7
2....5....9...10
........6
There are many other combination that will work.This is just one of the many combinations.
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ThudnBlunder
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Re: Star
« Reply #2 on: Jan 10th, 2004, 11:49pm » |
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I take it you mean the configuration below.
How did you arrive at it?
...........1...................4
....................8
..........11...................7
3..........................................10
..........12...................9
....................5
...........2...................6
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TenaliRaman
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Re: Star
« Reply #3 on: Jan 11th, 2004, 1:02am » |
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Label the vertices a to l. We will get 6 distinct equations. 6 equations and 12 unknowns. So there is more than one solution which means we can start with some arbitrary value for a variable say we make a=1 and also assume some values for variables on a particular edge. Then a bit of numerical analysis gives values of all the remaining variables.
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| « Last Edit: Jan 11th, 2004, 1:03am by TenaliRaman » |
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Barukh
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Re: Star
« Reply #4 on: Jan 11th, 2004, 6:42am » |
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I’ve got another solution (as TemaliRaman pointed, there are many):
...........1...................7
....................8
..........9...................12
2..........................................5
..........10...................4
....................11
...........6...................3
I arrived at the solution thinking similar to TemaliRaman. In addition, I searched only for the configurations with the following property: as many lines as possible that may be broken into 2 pairs summing up to 13. It’s easy to see that at most two such lines are possible. Incidentally, TemaliRaman’s solution also has 2 such lines (1, 11, 12, 2; 4, 7, 9, 6). Is every solution of this type?
By the way, THUD&BLUNDER, your formulation of the problem contains redundant data 
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| « Last Edit: Jan 11th, 2004, 6:43am by Barukh » |
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ThudnBlunder
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Re: Star
« Reply #5 on: Jan 11th, 2004, 11:52am » |
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Quote:| By the way, THUD&BLUNDER, your formulation of the problem contains redundant data |
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Yes, 26 is the 'magic sum'.
Quote:| as many lines as possible that may be broken into 2 pairs summing up to 13. It’s easy to see that at most two such lines are possible. Incidentally, TemaliRaman’s solution also has 2 such lines (1, 11, 12, 2; 4, 7, 9, 6). Is every solution of this type? |
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There are C(12,4) = 495 ways of choosing a line.
Of these, only 33 sum to 26.
Below are three more solutions.
I don't think there are very many; probably around 20, ignoring rotations and reflections.
Do the triangles have anything in common?
Consider the two large equilateral triangles and the numbers at their vertices.
For the solutions below these sum to 17, 18, and 20.
Your vertices sum to 12 and TenaliRaman's to 13.
eg. in the first triangle below
12 + 2 + 3 = 9 + 1 + 7 = 17
Also, considering the six smaller equilateral triangles, two of them have the same sum if they are opposite each other; eg. in the first triangle below
9 + 6 + 5 = 8 + 10 + 2 = 20
7 + 4 + 8 = 5 + 3 + 11 = 19
4 + 6 + 12 = 10 + 11 + 1 = 22
These equations are no doubt derivable from the 6 equations in 12 unknowns.
...........12.................9
....................6
...........4..................5
7.....................................3
..........8.................11
...................10
..........2..................1
=========================
...........6.................3
...................11
..........7..................1
5.....................................8
..........9.................12
....................2
..........4.................10
=======================
...........1...................5
....................2
..........10.................12
9......................................11
..........7...................3
....................4
..........8...................6
========================
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| « Last Edit: Jan 12th, 2004, 12:49am by ThudnBlunder » |
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