Some Bridges among Contest,
Research and Unsolved Problems
Andrejs Cibulis, University of Latvia
WFNMC-6, July 27, 2010, Riga
1. Magic Constants of Polyominoes
The notions: magic square, magic number, magic constant or magic sum
are well-known from recreational mathematics.
Let us assume that different integer numbers 1, 2, …k, are written in
the vertices of any unit square of some polyomino P.
The magic constant S(P) of a polyomino P is the one and the same
sum of the four different integer numbers written in the vertices of
any unit square of P.
Magic Constants of Pentominoes
1 2 3 4 5 6 7 2 3 5 2 10
8 6 12 4
12 11 10 9 8 7 6 11 10 8
5 7 1 9
S(I) = 26 S(X) = 26 S(V) = S(T) = 26
Pentominoes with 6 magic constants: F, L, N, Y
23; 24; 25; 27; 28; 29
S (F) 26, S (L) 26, S ( N) 26, S (Y) 26
1 2 3 4 5 6 7 8 9 10 11 12 78
3S 78 x1 x9 S 26
78 12 1 3S 78 1 12
67 3S 89 23 S 29
Proof for the other cases: S (L) 26, S ( N) 26, S (Y) 26
can be carried out in a similar way.
All the magic constants for pentominoes have been determined in .
 Cibulis, A., Pentomino Magic Constants and Pentomino Twins,
Riga, LU, 2009, 68 pp. (In Latvian)
Table of Magic Constants of Pentominoes
Pentomino Number of S Values of S
I, T, V, X 1 26
F, L, N, Y 6 23 -- 25, 27 -- 29
P 9 20 --28
U, Z 9 22 -- 30
W 11 21 -- 31
Pentomino W is that one having a maximum number of magic constants.
Vadims Novaks and Kristaps Freibergs, pupils of Riga Secondary School
No. 64 were able to elaborate the computer programme in Pascal for
solving this problem. They checked 56 first polyominoes (all up to
hexominoes) and determined all their magic constants.
Some conclusions from their contest paper
Problem of Magic Constants for Polyominoes, Riga, 2010.
▪ There is no exceptional values for the magic constants of hexominoes.
▪ A number of magic constants of hexomino is always an odd number.
▪ The minimum value of a magic constant is 22.
O-hexomino is unique one with S = 22.
▪ A record-holder is the step type hexomino
having 15 magic constants from 23 up to 37.
Hexomino Having a Maximum Number of Magic Constants
Contest or Olympiad Problems
Several Problems for mathematical Olympiad is proposed in , e. g.
▪ Show that pentomino X has only one magic constant
▪ Let M(P) be a set of all magic configurations of pentomino P.
Show that both sets M(X) and M(T) consist of equal number of elements.
Problems for research:
▪ Has any polyomino at least one magic constant?
▪ What is the maximum nymber of magic constants for n-mino?
▪ Has W-type n-mino a maximum number of magic constants?
▪ Hypothesis: a magic constant is unique for those polyominoes
having odd number of unit squares in each row and in each column.
Pentomino twins are two equal polyominoes which can be assembled
Let us consider only few slides about this very broad subject.
H=4 H=6 H = 11
Problems for research:
▪ Find symmetric twins with H = 3 or prove that they do not exist
▪ Find twins with H = 4 and with two axis of symmetry
or prove that they do not exist?
▪ What is the maximum height for symmetric twins?
What are the Highest Pentomino Twins?
A systematic investigation of small twins (10-mino) leads to H =16.
D. Hromakovs (Form 10, Riga) has found 227 little twins.
Is H = 16 the maximum height?
What is the Maximum Number of Corners?
36-gones without holes 38-gones
with 2 unit holes
Twins as Two Equal Parts of Rectangles
A systematic analysis of pentominoe twins fitted in the rectangle 6×10 has been considered in .
There are 2,339 solutions for the rectangle 610, 527 of them are twins.
There are 64 different twins for the rectangle 610.
Exercise. From twins 5 6 one can assemble the rectangles both of
5 12 and 6 10. Are these twins the only ones having this property?
The main results of the contest paper Pentominoes’ twins by M. Virza and I. Saknīte
have been exposed in  as well. They investigated Pentomino twins problem
for chess-board, namely: find all pentomino twins fitted on a chess-board 88.
Twins as Two Equal Parts of Rectangles
M. Virza was able to create computer programmes
by means of which one can find all p-twins.
The teacher G. Lāce has reported about her pupils’ achievements on pentomino
twins at the Congress on Mathematical Education, Mexico.
Cibulis A. Lāce G., The Experience of Development of Pupils’ Creativity in Latvia,
pp. 217-223, Proceedings of the Discussion Group 9: Promoting Creativity for All
Students in Mathematical Education, The 11th International Congress on Mathematical
Education, Monterrey, Mexico, 2008. http://dg.icme11.org/document/get/279
Twins on the Chess-board
Number of boundaries
Number of p-boundaries 440
between p-boundaries and
p-twins is not bijective.
Number of p-twins 3,182 Two different p-twins
may correspond to one
Number of p-assembles 13,427 and the same boundary.
It is clear because we
have 4 uncovered squares.
28th International Puzzle Party, Prague, 2008
Three colour problem Semi-pentacubes
by A. Cibulis by A. Liy
A lot of problems, from easy to unsolved; possibilities of
using them in extracurricular mathematical activities.
A. Liu proposed two basic challenges:
1. Construct a symmetric shape with six pieces that fits inside the 2×2×4 box.
2. Construct a symmetric shape with the six pieces that fits inside the 3×3×2 box.
Some other challenges:
▪ Find all solutions for these two challenges.
▪ Find twins that fit inside the 2×2×4 box.
▪ Find all twins that assemblable from semi-pentacubes.
An idea to elaborate the Three colour puzzle or the puzzle Vee-27 is arose from
the Puzzle Vee-21 which contains 21 V trominoes 7 in each of three colors.
Norton Star, Oriel Maxime, Kate Jones Vee-21
Vee-21TM, A suite of tilings with 21 trominoes,
A product of Kadon Enterprises, Inc., 40 pp.,
An unsolved challenge: Can the pieces of the
same color be separated so that they don't meet
even at corners? Doing this with one color is easy.
Total separation of two colors is very hard. All
three colors? Tell us if you find such a solution.
Solution of the Puzzle Vee-21
The minimum number of contact points is 1.
The puzzle Vee-21 has 8 optimal solutions
Atis Blumbergs (Latvia, 2003)
Marina Kļimova, Contest Paper (Riga, 2004).
Contact point – a point where trominoes of equal colours
touch each other.
The main problem is to fill the tray (square) by V-trominoes
so as to minimize the number of contact points.
The main task for V-27. Fill the tray 9 × 9 so that
each colour meets exactly in one point.
The solution is unique
with a precision to rotations
▪ Prove that any solution of 9 × 9 contains at least 4 rectangles 2 × 3.
▪ Cover 42 squares of the tray 9 × 9 so that it is impossible
to put the fifteenth V-tromino in the remaining region.
▪ What is the range of edges or vertices of three n-gons like?
In what limits can vertices of polygons change?
8+8+8 10 + 10 + 10 30 + 30 + 30
From Bachelor’s Paper by E. Pūre, 2009
Is n = 30 a maximum number of vertices?
▪ What is a maximum of r + b + g?
6+8+6 34 + 40 + 34 = 108 30 + 36 + 42 = 108
min = 20 2010, Alfa
M:= max (r + b + g) = ?
M > 110
The solution with record-holder number 110 has been published in
[Cibulis A., Three Colour Problem and its Usage in Work with Pupils and Students,
Proceedings of the 10th International Conference Teaching Mathematics:
Retrospective and Perspectives, Tallinn, 2009, pp. 137-144.]
Cut a minimum number of corners of the square 6×6
so that none corner can be cut from the remaining part.
45th & 46th MO (Regional Level) Form 9, (1994-1995;
At least 2 • 9 = 18 squares
must be covered;
18 : 3 = 6
trominoes must be used.
What is a minimum V-covering for the square 8×8?
Form 9, Open MO of Latvia, Form 9, XXI USSR
(1986-1987) Математика в школе, Москва,
1987, N6, 56-57 c.
What is the minimum number of squares that can be covered
by V-trominoes so that the remaining part does not contain
Problem for small n (n = 6, 8) is known in mathematical Olympiads and have
the elegant solution.
The known Olympiad method does not work for n > 8.
The On-Line Encyclopedia of Integer Sequences
does not contain the sequence of minimum numbers V(n).
Problem 1. Prove that 9×9 is not tileable with
13 rectangles 2×3 and one V-tromino.
Problem 2. Prove that 9×9 is not tileable with
12 rectangles 2×3 and three V-tromino.
Hint. Use a solution of Problem 1.
Problem 1 has an elegant solution and was proposed at 60 MO of Latvia, 2010,
Can the square 9 × 9 be divided in
2 × 3: 2, 2, 2
13 rectangles 2 × 3 and one corner?
Is there a rectangle which can be divided in the rectangles
2 × 3 and three corners? Answer is not known.
Or in other words, is it possible to complete
three corners with 2 × 3 to have some rectangle?
What is the minimum number of contact
points for the square 3n×3n ?
Min(6) = 1 Min(9) = 3
Min(12) = ?
4. Pentomino Bridges
Area = 262
The problem on the maximum bridge has been proposed by E. Pegg
in the magazine “World Game Review” (January 1994).
With 12 pentominoes make the bridge with the biggest area.
The best solution is of Pieter Torbijn with 275 units.
The next contributions are done by many authors.
Puzzle Fun, N2, December, 1994; N4, April 1995.
(by R. M. Kurchan, Argentina)
Hector San Segundo, Argentina, 1975, 278.
Torbijn P., Pentomino Bridges, Cubism for Fun, 59, November 2002, pp. 20 – 21.
R. Baranovskis, S. Krasņikovs, Bridges of Pentomino and Tetromino,
Contest Paper, Krāslava, 2010. (Form 11)
E. Liepiņa, Maximum Pentomino Bridges, Bachelor’s Thesis, Riga, 2010.
Maximum areas A(h)
depending on the height h of a bridge
A(1) = 36 A(11) = 272
371589120 A(2) = 72 A(12) = 276
A(3) = 107 A(13) = 278 4608
A(4) = 136 A(14) = 275
A(5) = 165 A(15) = 271
A(6) = 193 A(16) = 263
A(7) = 215 A(17) = 250
A(8) = 234 A(18) = 232 576
A(9) = 250 A(19) = 207
A(10) = 263 A(20) = 172
Intel Core2 Duo CPU 2.40GHz
Now a few slides form a pupils’ presentation.
A(14) = 275.
A biggest bridge of tetrominoes, Area = 28.
A bridge with the maximum height
L(19) = 207.
L(20) = 172.
Maximum Pentomino Ring
Area = 128.
Pentomino X always touches the base of the biggest bridge
(A = 278, length of base = 25), X--I or X--Y.
Does this fact have an olympiad type proof ?
Thank you for your attention!