Solution to "The Magic Hexagon" required
Garth Williams
please. It reads as follows:
The Magic Hexagon
Twelve circles have been placed at the vertices and intersections of a
hexagon. How can the numbers 1 to 12 be placed, one in each circle, so
that the sums of the numbers in every row, and also the sum of the six
vertices, are equal?
Thanks in advance.
Tom Maciukenas
A hexagon has only 6 vertices. Where do the other 6 circles go? Where
are the "intersections"? Could you supply a diagram?
1____2
/ \
/ \3
4\ /
\____/
5 6
-toMm
Dennis Yelle
>I need the solution to this very basic (not basic enough for me) puzzle
>please. It reads as follows:
>
>The Magic Hexagon
>Twelve circles have been placed at the vertices and intersections of a
>hexagon. How can the numbers 1 to 12 be placed, one in each circle, so
>that the sums of the numbers in every row, and also the sum of the six
>vertices, are equal?
>
>Thanks in advance.
Here is one way:
1 3
9
12 11
2 5
6 4
10
7 8
--
den...@netcom.com (Dennis Yelle)
"You must do the thing you think you cannot do." -- Eleanor Roosevelt
heus...@umanitoba.ca
Garth Williams <will...@engmlab.uct.ac.za> writes:
>I need the solution to this very basic (not basic enough for me) puzzle
>please. It reads as follows:
>
>The Magic Hexagon
>Twelve circles have been placed at the vertices and intersections of a
>hexagon. How can the numbers 1 to 12 be placed, one in each circle, so
>that the sums of the numbers in every row, and also the sum of the six
>vertices, are equal?
>
>Thanks in advance.
What do you mean by the *intersections* of a hexagon? My guess is that you
are considering the following diagram:
*
* * * *
* *
* * * *
*
I would call that a Star of David, not a hexagon.
Here is another "Magic Hexagon" problem that I have seen before.
Place numbers from 1 to 19 so that the sums along any "row" (in all 3 of
the directions parallel to the sides of the hexagon) are all the same.
* * *
* * * *
* * * * *
* * * *
* * *
How many solutions are there, up to rotations and reflections?
And what about hexagons of other sizes, instead of sides of length 3?
Ignore the "From:" line -- heuser20 is a terminal, not a person.
Barry Wolk <wo...@ccm.umanitoba.ca>
Dept of Mathematics
University of Manitoba
Winnipeg Manitoba Canada
Stephen H. Landrum
>
> I need the solution to this very basic (not basic enough for me) puzzle
> please. It reads as follows:
>
> The Magic Hexagon
> Twelve circles have been placed at the vertices and intersections of a
> hexagon. How can the numbers 1 to 12 be placed, one in each circle, so
> that the sums of the numbers in every row, and also the sum of the six
> vertices, are equal?
>
> Thanks in advance.
Something is wrong with the problem as stated. A hexagon has six
vertices, but they are the only thing that I would consider to be
intersections.
Perhaps you meant a six pointed star arranged as follows:
O
/ \
O---O---O---O
\ / \ /
O O
/ \ / \
O---O---O---O
\ /
O
And the sums of the 4 numbers on each of the six lines should
be equal to each other, as well as to the sum of the 6 numbers
at the points of the star.
--
Stephen H. Landrum voice: (415)261-2626 email: slan...@3do.com
System software programmer, M2 graphics division.
For general 3DO questions email customer...@3do.com
Philip Carter
>please. It reads as follows:
>
>The Magic Hexagon
>Twelve circles have been placed at the vertices and intersections of a
>hexagon. How can the numbers 1 to 12 be placed, one in each circle, so
>that the sums of the numbers in every row, and also the sum of the six
>vertices, are equal?
>
>Thanks in advance.
3 at top; 2,12,11,1 next row; 5,8 middle row; 6,9,7,4 next row; 10 at
the bottom.But I am sure there are many other solutions to find.
--
Philip Carter
Philip Carter
writes
>In article <315B76...@engmlab.uct.ac.za>,
>Garth Williams <will...@engmlab.uct.ac.za> writes:
>
>>I need the solution to this very basic (not basic enough for me) puzzle
>>please. It reads as follows:
>>
>>The Magic Hexagon
>>Twelve circles have been placed at the vertices and intersections of a
>>hexagon. How can the numbers 1 to 12 be placed, one in each circle, so
>>that the sums of the numbers in every row, and also the sum of the six
>>vertices, are equal?
>>
>>Thanks in advance.
>
>are considering the following diagram:
>
> *
> * * * *
> * *
> * * * *
> *
>
>I would call that a Star of David, not a hexagon.
>
>Here is another "Magic Hexagon" problem that I have seen before.
>
>Place numbers from 1 to 19 so that the sums along any "row" (in all 3 of
>the directions parallel to the sides of the hexagon) are all the same.
>
> * * *
> * * * *
> * * * * *
> * * * *
> * * *
>
>How many solutions are there, up to rotations and reflections?
>
>And what about hexagons of other sizes, instead of sides of length 3?
>
>Ignore the "From:" line -- heuser20 is a terminal, not a person.
>
>Barry Wolk <wo...@ccm.umanitoba.ca>
>Dept of Mathematics
>University of Manitoba
>Winnipeg Manitoba Canada
One solution to the Magic Hexagon puzzle is: 3,19,16 top row; 17,7,2,12
second row; 18,1,5,4,10 middle row; 11,6,8,13 fourth row and 9,14,15
bottom row. Unfortunately I cannot tell you how many solutions there
are.I suspect there are many. With Magic squares,for example there are
said to be 880 order-4 squares(magic 34) not counting rotations and
reflections, but does anyone have them all,have they in fact all been
worked out yet, and is there a formula for working out the number of
possible solutions for such puzzles?
--
Philip Carter
Dennis Yelle
[...]
>
>Here is another "Magic Hexagon" problem that I have seen before.
>
>Place numbers from 1 to 19 so that the sums along any "row" (in all 3 of
>the directions parallel to the sides of the hexagon) are all the same.
>
> * * *
> * * * *
> * * * * *
> * * * *
> * * *
>
>How many solutions are there, up to rotations and reflections?
Answer:
Just one:
18 11 9
17 1 6 14
3 7 5 8 15
19 2 4 13
16 12 10
>
>And what about hexagons of other sizes, instead of sides of length 3?
There is no solution to this one:
. .
. . .
. .
But this one:
.
Does have a solution :)
So does this one:
There is no solution to the 4 by hexagon.
heus...@umanitoba.ca
Philip Carter <phi...@knowl.demon.co.uk> writes:
:In article <heuser20.2...@umanitoba.ca>, heus...@umanitoba.ca writes
:>Here is another "Magic Hexagon" problem that I have seen before.
:>Place numbers from 1 to 19 so that the sums along any "row" (in all 3 of
:>the directions parallel to the sides of the hexagon) are all the same.
:>
:> * * *
:> * * * *
:> * * * * *
:> * * * *
:> * * *
:>
:>How many solutions are there, up to rotations and reflections?
:>And what about hexagons of other sizes, instead of sides of length 3?
:>
:>Ignore the "From:" line -- heuser20 is a terminal, not a person.
:>
:>Barry Wolk <wo...@ccm.umanitoba.ca>
[snip]
:
:One solution to the Magic Hexagon puzzle is: 3,19,16 top row; 17,7,2,12
:second row; 18,1,5,4,10 middle row; 11,6,8,13 fourth row and 9,14,15
:bottom row. Unfortunately I cannot tell you how many solutions there
:are.I suspect there are many. With Magic squares,for example there are
:said to be 880 order-4 squares(magic 34) not counting rotations and
:reflections, but does anyone have them all,have they in fact all been
:worked out yet, and is there a formula for working out the number of
:possible solutions for such puzzles?
:--
:Philip Carter
Well, the amazing thing about this "Magic Hexagon" problem is that there
is only one answer, up to rotations and reflections. Also, there are no
solutions for hexagons of other sizes. So the uniqueness of the solution
also refers to the size of the hexagon involved. Quite amazing, and exactly
the opposite of the corresponding results for magic squares.
I first saw this in the journal "Recreational Mathematics Magazine."
My recollection is that the proof was given by Charles W Trigg. However,
I do not have a specific reference.
Ignore the "From:" line -- heuser25 is a terminal, not a person.
heus...@umanitoba.ca
>I need the solution to this very basic (not basic enough for me) puzzle
>please. It reads as follows:
>
>The Magic Hexagon
>Twelve circles have been placed at the vertices and intersections of a
>hexagon. How can the numbers 1 to 12 be placed, one in each circle, so
>that the sums of the numbers in every row, and also the sum of the six
>vertices, are equal?
>
>Thanks in advance.
Several follow-ups have questioned the wording of the problem, and some of
them have given just one solution, based on one way of interpreting the
words "vertices" and "intersections". There appear to be two different
interpretations. Both possibilities use the same diagram:
*
* * * *
* *
* * * *
*
Problem 1: Place the numbers from 1 to 12 so that the 6 *inner* numbers,
and also the 4 numbers in any row, all have the same sum.
Problem 2: Place the numbers from 1 to 12 so that the 6 *outer* numbers,
and also the 4 numbers in any row, all have the same sum.
Here are all the solutions I found to problem 1, up to rotations and
reflections.
------------
12
11 3 4 8
6 1
5 2 10 9
7
------------
12
10 4 7 5
2 1
8 3 9 6
11
------------
12
11 1 4 10
7 2
6 3 9 8
5
------------
12
10 2 5 9
1 6
11 8 4 3
7
------------
12
10 6 1 9
5 2
3 4 8 11
7
------------
12
9 3 8 6
1 2
10 5 7 4
11
------------
Did I miss any solutions?
Now, what about problem 2? I find it interesting that all solutions
to this problem can be obtained from the solutions of problem 1, by
interchanging every pair of numbers that add up to 13. For example,
if you make these swaps in the last diagram that I gave, you get the
following solution to problem 2:
1
4 10 5 7
12 11
3 8 6 9
2
The outer 6 numbers, and all rows of 4 numbers, have the same sum.
Ignore the "From:" line -- heuser10 is a terminal, not a person.
Jim Waters
Dennis Yelle <den...@netcom.com> wrote:
>In article <heuser20.2...@umanitoba.ca> heus...@umanitoba.ca writes:
>[...]
>>
>>
>>Place numbers from 1 to 19 so that the sums along any "row" (in all 3 of
>>the directions parallel to the sides of the hexagon) are all the same.
>>
>> * * * *
>> * * * * *
>> * * * *
>> * * *
>>
>
>
>Just one:
>
> 18 11 9
> 17 1 6 14
> 3 7 5 8 15
> 19 2 4 13
> 16 12 10
>
>>
>
> . .
> . . .
> . .
>But this one:
> .
>Does have a solution :)
>
>So does this one:
>
>There is no solution to the 4 by hexagon.
True, IF the cells have to start with 1. But might it be possible for
there to be an order 4 magic hexagon with the cells numbered from 3 to
39, with the magic sum for each row being 111?
>
>--
>den...@netcom.com (Dennis Yelle)
>"You must do the thing you think you cannot do." -- Eleanor Roosevelt
It has been established that magic hexagons of order higher than 3 are
impossible IF the cell numbering has to start with 1. But by relaxing
that requirement and allowing a specially chosen first cell#, magic
hexagons of higher orders MIGHT be possible!
For a proposed order n, divide (n+1) by 4 to get quotient q and remainder r.
The first cell # for the following values of r would be:
r 1st cell #
0 q
1 3q
2 -3q (or 5q+1 if you insist on a positive value)
3 -q (or 7q+3 if you insist on a positive value)
So for n=3, 1st cell is 1
n=4, 3
n=5, -3 or 6
n=6, -1 or 10
n=7, 2
n=8, 6
n=9, -6 or 11
n=10, -2 or 17
etc. Of course, starting with the triangular number for (n-2) would also
work. (n-2)*(n-1)/2
-- Jim Waters (jwa...@az.com)
heus...@umanitoba.ca
jwa...@anna.az.com (Jim Waters) writes:
>In article <dennisDp...@netcom.com>,
>Dennis Yelle <den...@netcom.com> wrote:
>>In article <heuser20.2...@umanitoba.ca> heus...@umanitoba.ca writes:
>>[...]
>>>
>>>Here is another "Magic Hexagon" problem that I have seen before.
>>>
>>>Place numbers from 1 to 19 so that the sums along any "row" (in all 3 of
>>>the directions parallel to the sides of the hexagon) are all the same.
>>>
>>> * * *
>>> * * * *
>>> * * * * *
>>> * * * *
>>> * * *
>>>
>>>How many solutions are there, up to rotations and reflections?
>>
>>Answer:
>>
>>Just one:
>>
>> 18 11 9
>> 17 1 6 14
>> 3 7 5 8 15
>> 19 2 4 13
>> 16 12 10
>>
>>>
>>>And what about hexagons of other sizes, instead of sides of length 3?
[snip]
>>There is no solution to the 4 by hexagon.
>
>True, IF the cells have to start with 1. But might it be possible for
>there to be an order 4 magic hexagon with the cells numbered from 3 to
>39, with the magic sum for each row being 111?
[snip]
> -- Jim Waters (jwa...@az.com)
Well, if you change the problem, then you will almost certainly change the
answer. In the case of magic hexagons of order 3, we know that there is
a unique solution that uses the numbers [1, 19]. However, what about other
intervals of length 19? I have found several solutions that use the numbers
[-9, +9]. Here are some with interesting symmetries and relationships:
-6 7 -1 -3 7 -4
-2 -3 -4 9 -2 -6 -1 9
8 5 0 -5 -8 5 8 0 -8 -5
-9 4 3 2 -9 1 6 2
1 -7 6 4 -7 3
I suspect that there will be many solutions for all orders n > 2 , using
the numbers [-k, +k], k = 3n(n-1)/2. What about other possible ranges?
The [1, 19] range has midpoint 10, and the [-9, +9] range has midpoint 0.
For the order 3 case, the midpoint of the range must belong to the set
{-15, -10, -5, 0, 5, 10, 15}
I am not claiming that solutions exist for all of these midpoints. However,
solutions do not exist for all *other* values of the midpoint.
Similarly, for order 4 magic hexagons, the midpoint of the range of any
solution must belong to {-35, -28, -21, -14, -7, 0, 7, 14, 21, 28, 35}.
Jim Waters asked about the range [3, 39] with midpoint 21. What about the
other possibilities? I am pretty sure that solutions with midpoint 0 exist.
Ignore the "From:" line -- heuser15 is a terminal, not a person.