Problem of the Month
(March 2009)

This month we consider some problems involving the six letter forms below:
These have six different areas, so there are a lot of interesting packing, covering, and tiling questions available. We treat these as polyominoes, so they can be rotated and reflected if necessary.

1. Given a m×n rectangle, what is the largest area that can be filled with non-overlapping copies of these letters? In particular, what are the answers for n×n squares?

2. Given a m×n rectangle, what is the smallest area of letters that will cover that rectangle? In particular, what are the answers for n×n squares?

3. We will call two collections of letters X and Y equivalent if there is some collection of letters Z and a region S so that the letters X and Z tile S, and the letters Y and Z also tile S. Can you find some small regions S that can be tiled with two different collections of letters? Given X and Y, what is the smallest collection Z you can find to show X = Y? Can you prove that any two collections of letters with the same area are equivalent?


ANSWERS

1. Here are the densest known letter packings in rectangles:

Densest Known Letter Packings in Rectangles
m \ n3456789
5
13

16

22
6
16

20

26

32
7
16

20

31

37 (GS)

40
8
16

24

35

42

50 (GS)

58 (GS)
9
22

28

39

48

54 (GS)

63 (GS)

72 (GS)
10
26

32

44

53 (GS)

62

72 (GS)

80 (GS)
(GS) = George Sicherman

Here are the densest known squares:

Densest Known Squares

10×10 - 90 (GS)

11×11 - 111 (GS)

12×12 - 131 (GS)

13×13 - 156 (GS)


14×14 - 182 (GS)

15×15 - 205 (GS)

16×16 - 238 (GS)


17×17 - 263 (GS)

18×18 - 298 (GS)


19×19 - 330 (GS)

20×20 - 367 (GS)


21×21 - 400 (GS)
(GS) = George Sicherman


2. Here are the smallest known letter coverings of rectangles:

Smallest Known Letter Coverings of Rectangles
m \ n12345
28
12
381616
4816
16

22
5
8
16
18

24

29
616
16

20

30

34
716
24

26

34 (BH)

40 (GT)
8
16

28

32

39 (GT)

45
(BH) = Bryce Herdt
(BT) = Gavin Theobald


3. Here are the smallest known double letter tilings:

Smallest Known Double Letter Tilings
EquationAddPicture
A C = E2B
E3 = A2 FB2
D E = A2A B3 E
D F = E2E F2
(Bryce Herdt)
E4 = F5
A2 E2 = C2 D2E2 F4
B3 = A C2 EE
B3 = C E3A B
(Bryce Herdt)
D2 = F3E2 F2
(Bryce Herdt)
C2 = E FA2 B2 E
(Bryce Herdt)
A D = B EB C E2 F3
(Bryce Herdt)

Using these substitutions, we can prove that any two sets of letter polyominoes with equal area are equal. Use A D = B E repeatedly to remove B's. Then use D E = A2 repeatedly to remove D's. Then use A C = E2 repeatedly to remove A's. Then there will be an even number of C's (since they have odd area and E's and F's have even area) that can be removed with repeated use of C2 = E F. Then only E's and F's will remain, in a 4:5 ratio, and they can be removed with repeated use of E4 = F5!


If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 3/3/09.