
to this page of 3dimensional
magic star patterns. New material will be added here as it becomes
available and time permits. 

An 8point 3D magic star

This 3dimensional 8pointed magic star contains 12
lines with 3 numbers per line. It consists of two tetrahedrons
(pyramids) each with six lines of three cells.The pyramids intersect
at the midpoints of each of the 12 lines.
Pyramid ABCD has D pointing away from the viewer.
Pyramid EFGH has H pointing toward the viewer.
ABCD and EFGH each appear in three lines.
IJKLMN each appear in two lines. 
The smallest possible series of numbers that can be arranged in this pattern
to make each of the 12 lines sum to the same value is numbers 1 to 15 but
without the 8. Because of parity, all the points must be odd and the midline
values even or all the points even numbers and the midline values odd.
The equation for this pattern is 3(A+B+C+D+E+F+G+H) + 2(I+J+K+L+M+N) = 12S
However, the series from 1 to 15 produces only 7 correct lines and the series
from 2 to 16 produces only 8 correct lines.
I sent Aale de Winkel the results of my search and a day or two later he
emailed me the following solution which requires that 3 numbers must be left
out of the series of consecutive numbers.

This solution, using the numbers from 1 to 17, but without
the 3, 9 and 15, forms a magic 3D star with 12 lines of 3 numbers all
summing to 27. This is the only solution possible using this number set
(not counting rotations or reflections). With different number sets,
smaller possible magic constants are possible . See addendum at end.
The three unused numbers 3, 9 and 15 (shown in green) may be
incorporated in the pattern as follows:
Place the number 9 in the center of the star. It forms a magic line with
it's two 'satellites', the 3 and the 15.
The 9 also forms magic lines in conjunction with each of the 4 star point
pairs, and with each of the 3 star midline pairs.
Thus we have a pattern using the consecutive numbers from 1 to 17,
forming 22 lines of 3 numbers, all summing to 27. 
Like all magic stars, there is a companion solution to this one.
Subtract each number in the above solution from 18 to obtain the
complementary solution.
A wooden block model of the 3_d magic star (12 lines
of 3 = 27) and 3 unused numbers which also sum to the constant.
This and other models are shown on my
ms_models.htm 
Here all 17 numbers are used, producing a figure
with an additional 9 lines summing to 27.
(The line (3, 9,15) is held in place with rubber bands) 
My special thanks to
Aale de Winkel for helping me to visualize a 3dimensional star as two
intersecting tetrahedrons.
He also came up with the solution using the numbers 1 to 17 when my
attempts using numbers 1 to 15 and 2 to 16 failed.
And, while I was busy putting this all together in a drawing, he realized
that the 3 unused numbers (3, 9 and 15) could be incorporated as the star
center and two satellites. [1]
[1] Emails from Aale de Winkel April 5, 1999 and May16,
1999 ( and several inbetween).
A 10point 3D star  magic
?

This is the cell name diagram for a 10pointed magic
star consisting of two intersecting squarebase pyramids.
The blue pyramid has the tip pointing toward the viewer. The red
pyramid has the tip pointing away.
It is the same basic design as the triangle based star above and
contains 18 numbers in 16 lines of 3 numbers each.The 10 point
star would use 18 numbers from 1 to ? arranged in 16 lines of 3
numbers.
It consists of 2 pentahedrons with the side midpoint cells of one
being the base midpoint cells of the other one.
In attempting to construct a 10 point star by copying features of
the 8 point one, I finally realized there is a problem with parity,
and all even point 3D stars are probably impossible to construct
using these features.

Peak A and J are complements of each other.
Base corner B, C, D, E are complements of base corners F, G, H, I.
Midpoint cells of the base are complements of midpoint cells of the vertical
sides.
All point cells are even, with the midpoint cells odd, or vice versa.
Magic Sum (S) equation is 4(A+J)+3(B+C+D+E+F+G+H+I)+2(K+L+M+N+O+P+Q+R)= 16S
BT using complement pair totals (and assuming we are using the series from 1 to
27) this simplifies to
4(1x28)+3(4x28)+2(4x28) = 16S, so in this case S=42
However, each line (in this case) has two even and one odd number, requiring an
odd magic sum.
Sets of 5 even numbers with their complements, 5 different numbers, are only
found in number ranges from 1 to 27, 1 to 31, 1 to 35, etc. In each case, the
resulting magic sum is even., when it is required to be odd.
Sets of 5 odd numbers with their complements, 5 different numbers, are only
found in number ranges from 1 to 29, 1 to 33, 1 to 37, etc. In each case, the
resulting magic sum is odd., when it is required to be even.
I was starting to look for solutions using all even or all odd numbers only,
when I received the following impossibility proof from
Hermann Mierendorff. [2]
Hermann Mierendorff's impossibility proof
Let the values of the 10point star be A,B,...,R.
We consider the following sums of values which lay on a line:
s1:=A+K+B, s2:=B+P+C, s3:=C+Q+D, s4:=D+M+A,
s5:=F+K+G, s6:=G+P+J, s7:=J+Q+H, s8:=H+M+I.
Let all the values on a line sum up to the magic number m.
This means s1=s2=s3=s4=s5=s6=s7=s8=m.
Hence s1+s2s3+s4+s5s6+s7s8=0.
On the other hand, we see easily
s1+s2s3+s4+s5s6+s7s8=FI.
Hence FI=0, i.e. F=I.
Therefore, at least F and I are identical, i.e. there is no magic star of this
kind.
The same proof also works for all 3D stars which are build up using two
nsided pyramids as soon as n>3 (hh).
[2]
Hermann Mierendorff email of
Dec. 4, 2000 (His first email on this subject was Aug. 9, 2000)
The de Winkel 10 point
star
This magic object may not look much like a star,
but it is magic! It consists of two square based triangles with the
midpoints of the vertical sides sharing common numbers.
It uses 22 of the numbers from 1 to 33 to form 16 lines of 3
numbers. Each line sums to 36.
Aale de Winkel has discovered (about March, 2001) 3
other similar magic stars with sums of 42, 66 and 78.
See them and other magic objects at his
Magic
Encyclopedia site.


Hermann Mierendorff's
3D magic stars
Addendum  2009
In an above section on this page, Hermann Mierendorff gives an
impossibility proof for a 10 pointed magic star of the style of the 3D star
shown.
In a paper published in 2003
[1], Mierendorff used pyramids and
prismoids (or truncated pyramids) for building the polyhedra, the intersection
of which forms the stars he was investigating. He found there are many magic
stars consisting of two interconnected trumcated pyramids (he calls these twin
body stars), if one restriction is applied.
Consider the number of sides in the base of the pyramid to be
the width (w), and the number of layers when moving from 1 point to the
diametrically opposite point the height (h). For example, the 3D 8
point magic star above has w = 3 and h = 4; the 10 point nonmagic
star has w = 4 and h = 4. De Winkel’s 10 point star has w =
4 and h = 5 and is magic.
The paper shows that magic star labelling exists for any star (w,h)
if and only if h = w + 1. The following information is condensed
from the cited paper with the author’s consent.
Some solutions for magic stars (w,h) of 3,4.

A 
B 
C 
D 
E 
F 
G 
H 
I 
J 
K 
L 
M 
N 


p, q 
1 
2 
8 
12 
14 
6 
16 
10 
4 
11 
5 
1 
13 
17 
7 
117, no 3,9,15. 
S = 27 
18, 18 
2 
1 
4 
6 
7 
3 
8 
5 
2 
14 
11 
9 
15 
17 
12 
117, no 10,13,16. 
S = 22 
9, 26 
3 
10 
13 
15 
16 
12 
17 
14 
11 
6 
3 
1 
7 
9 
4 
117, no 2,5,8. 
S = 32 
27, 10 
4 
1 
9 
10 
12 
6 
15 
7 
4 
11 
3 
2 
13 
14 
5 
115, no 8. 
S = 24 
16, 16 
Solution 1 is the one shown on this web page. I consider it the prettiest
because all point values are even numbers and all crossings (valleys) are odd
numbers. Mierendorff calls this feature perfectly shuffled. As previously
mentioned on this page, all solutions can generate a complementary solution
simply by subtracting each number from the maximum number in the range plus 1.
The other 3 solutions are other possibilities for the 8 point star
Another feature that is important in the analysis of this type of magic
figure is the sum of opposite point values. Similarly the sum of opposite cross
values. He designates them respectively as p and q. 2p+q=S,
so q is obviously always even. This result has been proven and it was one
of the reasons to introduce p and q.
In the above table, the arbitrarily labelled points are A to H, with the
valleys labeled I to N. Comparing these with the first diagram on the page we
see that the opposite points are AF, BG, CE, and DH; opposite valleys are IN, JL,
and KM.
Mierendorff's paper incorporates what he calls a flat
representation to illustrate the magic lines of these stars without
resorting to complicated confusing diagrams of the stars.
The diagram above represents the star at the top of this page.
The diagram below represents the 22 point star to the right.
These two charts have been changed from those in his paper, to
correspond to the two diagrams on this page.

A 22 point star of form 5,6. Each ring has 5 points, and there are 6
layers. 3 in each of the two interlocking pyramids (polyhedra).
S = 69 for each of the 40 lines of 3 numbers. It uses the numbers from 1
to 45 with no 18, 23, or 28.
The paper [1] shows solutions for
magic stars of this type for stars of 3,4 to 6,7. 
[1] Hermann Mierendorff, Magic
twin body stars generated from chains of prismoids, Discrete Mathematics,
268 (2003) 315323.
(Although I was advised of this paper, I did not actually see it until
2009.)
