The 6-square pieces were first enumerated by H. D. Benjamin in Problemist Fairy Chess Supplement, December 1934. In fact the very first dissection problem (¶1597) in PFCS/FCR is Benjamin's question whether the 35 pieces of 6 squares can be arranged to cover a rectangular area. The text reads as follows:
"A BIG CUTTING-OUT DESIGN AND A PRIZE OFFER. Here is the special poser, a real Christmas puzzle: 1597. H. D. Benjamin, Calcutta."
"Cut a board 15 by 14 into 35 pieces each of 6 squares attached to one another by their edges, all cuts along the lines, the 35 pieces to be all fundamentally different in shape. Rotation or reflection of a piece is NOT to be considered a difference."
"For a solution of this, which he is not sure can be done, Mr. Benjamin has placed a prize of 10 shillings at my disposal. Solvers can attack it in either of two ways, of courseby actually cutting or marking off the pieces, or by discovering the 35 possible 6-square shapes and then assembling them in a 15 by 14 rectangle. In the latter method, which seems more useful, any piece can be turned over, which amounts to reflection, at will, so long as no two pieces of the same shape occur in the 35 initially; or stated in another way, it does not matter in which aspect any piece is used for its one appearance in the design."
The solution of this problem appears in the next issue, as follows:
"Dr. F. Kadner has sent two elegant proofs that a 15 x 14 rectangle can not be dissected into the 35 different 6-square shapes. Hs first proof is geometrical and commences by showing that the 35 pieces will asemble to a 15 x 14 rectangle except for one single square overlapping an adjacent one. Thus to achieve the complete assembly, translations, rotations, and reflections of THIS array must end in a resultant displacement of 1 square 1 step. But each of the 35 pieces in any of these motions alters the array by an EVEN number of squares x steps. Hence the resultant of any number of such motions can never give the required ODD unit change."
"The second proof is even more simpleif the 35 pieces are chequered, it is found 24 of them must have 3 white and 3 black squares, but the remaining 11 (the piece abcde1, b2 is an example) MUST have 2 squares one colour and 4 the other. Now the 15 x 14 rectangle has equal numbers of each colour, so the 11 pieces just noted must be coloured half white and half blackand this is clearly impossible. Thus, again, the problem cannot be solved."
"These beautiful bits of reasoning amount to a complete solution, of course, and Dr. Kadner easily secures the prize offered."
The argument shows it is impossible to use the 35 hexominoes to cover any rectangle of 210 squares, 2×105, 3×70, 5×42, 6×35 , 7×30, 10×21 or 14×15. More generally it follows that any chequered area covered by the 35 hexominoes must show an excess of squares of one colour over squares of the other colour. This colour difference cannot be zero but must take one of the values 2, 6, 10, 14, 18 or 22.
Another general result about rectangles appeared much later:
F. Hansson (FCR 1950 ¶8555). "Two rectangles are joined along one edge to form a figure of 210 squares. Typical examples of joining are abc1 joined to bcd2, or cde2 or a234, and so on, but the rectangles need not be equal. What are the relative dimensions and joining squares for formation of the 210-square shape by the 6-pieces?"
Solution: "For double rectangles A,B to be possible with the 6-sq. pieces, A and B must both be odd and have the same corner colour. A and B cannot both be primes. Following are impracticable: 11, 199; 13, 197; 17, 193; 19, 191; 29, 181; 31, 179; 37, 173; 41, 169, 43, 167; 47, 143; 53, 157; 59, 151; 61, 149; 67, 143; 71, 139; 73, 137; 79, 131, 83, 127; 89, 121; 97, 113; 101, 109; 103, 107. Very doubtful is 9, 201."
The following diagrams show some of the shapes that have been devised to make use of all 35 of the hexominoes. They are classified broadly according to the symmetry of their outline shape. This of course is only one way they might be classified, but one I have found helpful. Within each section those with fewest holes come first.
F. Hansson (FCR 1950, ¶8558) "From a SQUARE cut out a centrally situated rectangle and and form the shape left with the 6-pieces." The solution is this 15×15 square with central 3×5 hole; it has no crossroads, CD = 2.
W. Stead (FCR 1954 ¶30). This 15×17 rectangle with a Greek cross-shaped 45-square hole appears in the illustrated article, CD = 2.
H. D. Benjamin (FCR 1945 ¶6375-6). Two rectangles 13×24 with holes in the shape of two versions of the Cross of Lorraine; both have CD = 2.
It could be argued that these should be classified under those with an axis of symmetry, since the holes form the design rather than the outer rectangle, but sticking to my (admittedly rather arbitrary) rule I only take account of the outer outline.
G. P. Jelliss (Chessics #28 1986 p.145). This 'rhombus' (or 'battleship' or 'aircraft carrier') pattern with serrated edges in a regular stepped form was composed specifically as an example of CD = 18. It has no crossroads (the diagram is rotated a quarter turn from that in Chessics). The containing rectangle is 13×37.
G. P. Jelliss (Chessics #28 1986 p.145). This 'diamond' or square with serrated edges of 11-square length within a 21×21 area and with a 1×11 hole was composed to show the CD = 22 case.
W. Stead (FCR 1952 ¶9331). A 15x15 square with three cross-shaped holes. Dedicated to H. E. Dudeney and his work on Greek Crosses.
G.P.Jelliss (Chessics #28 1986 p.144). A cross with 6×7 arms on a 7×7 centre, within a 19×19 area, and having 7 holes regularly spaced down the centre axis, CD = 6.
W. Stead (FCR 1954 ¶35). Rectangle 5×45 with 15 holes. CD = 2.
F. Hansson (FCR 1950 ¶8556). "Place two 3×35 rectangles in the longest possible shape formable by the 6-pieces." [This is the 105, 105 case of the double rectangle problem.]
A well-known arrangement with CD = 14 is this parallelogram with sloping edges serrated. It appears in W. Stead's notebooks with his initials alongside, and it was first published in his article in FCR December 1954 ¶32. He notes there that the central stepped piece is pre-set to echo the outline. The containing rectangle is a doubled square, 14×28, and there is a vertical fault line between the two squares, except where the stepped piece crosses it.
This serrated triangle with 20-unit sides was given by H. D. Benjamin (FCR Feb/Apr 1937 ¶2622) it has no crossroads, and is the first arrangement of the hexominoes in a form without holes, diagonal axis, CD = 10.
H. D. Benjamin (FCR Dec/Feb 1941/2 ¶4985) 15×15 square with 15 cells removed, diagonal axis CD = 2.
H. D. Benjamin (FCR Dec/Feb 1941/2 ¶4986) 15×15 square with 15 cells removed, with diagonal axis CD = 6.
H. D. Benjamin (FCR Dec/Feb 1941/2 ¶4987) 15×15 square with 15 cells removed, CD = 14.
H. D. Benjamin (FCR Dec/Feb 1941/2 ¶4988) 16×16 square with 46 cells removed, CD = 18.
H. D. Benjamin (FCR Dec/Apr 1941/2 ¶4989) 17×17 square with 79 cells removed, CD = 22.
H. D. Benjamin (FCR 1950 ¶8912). "Arrange the 35 6-pieces in a symmetrical figure, a rectangle plus a square. There are only 3 figures possible.": Solution: "The Sombrero Hat." This is 3×43 + 9×9 and has colour difference = 2.
"Other cases are 11×19 with 1×1 and 19 ×11 with 1×1." Solutions for these cases were not given. The following are my own solutions (G. P. Jelliss May 2006). The 'suitcase' has one crossroad,
the 'clipboard' has three.
H. D. Benjamin (FCR Apr/Aug 1952 ¶9235). 'The Spreading Chestnut Tree'.
W. Stead (FCR Dec/Apr 1953/4 ¶9783). Christmas Tree.
H. D. Benjamin (FCR Dec/Apr 1941/2 ¶4990) castle within 15×20 rectangle, vertical axis, CD = 2.
H. D. Benjamin (FCR Dec/Apr 1941/2 ¶4991) castle in 16×21, with vertical axis, CD = 6
H. D.Benjamin (FCR Dec/Apr 1941/2 ¶4992) another castle in 16×21, with vertical axis, CD = 6.
H. D. Benjamin (FCR Dec/Apr 1941/2 ¶4993) another castle in 17×21, with vertical axis, CD = 10.
W. Stead (FCR Dec 1954 ¶29). Another rook shape. CD = 2.
H. D.Benjamin (FCR Dec/Apr 1941/2 ¶4994) a 'mine' in 17×21, with vertical axis, CD = 14.
H. D. Benjamin (FCR Dec/Apr 1941/2 ¶4995) a 'mine' in 17×21, with vertical axis, CD = 18.
H. D. Benjamin (FCR Dec/Apr 1941/2 ¶4996) 'mine' 15×16 with 30 cells removed, CD = 22,
H. D. Benjamin (FCR Dec/Apr 1941/2 ¶4997) 'mine' 15×17 with 45 cells removed, vertical axis, CD = 22.
H. D. Benjamin (FCR Oct/Dec 1943 ¶5668) Victory V in 21×23, with vertical axis. No crossroads, all pieces on edge, CD = 10.
This 'pyramid', more precisely triangle with serrated right-angled edges (G. P. Jelliss Chessics #28 1986 p.145), CD = 14, has the added feature that the 2×3 'block' can fall down (when released by pyramid robbers entering the middle chamber!) so that the 3×5 hole is in an alternative position. The containing rectangle is 15×29. As befits a properly constructed pyramid it has no crossroads.
H. D. Benjamin (FCR Jun/Dec 1950 ¶8728). All the s in a double serrated triangle. CD = 14. The F-shaped joining piece is also formed of two triangles.
H. D. Benjamin (FCR 1950 ¶8559). "Modern Art: In the style of the double rectangle above, form a Triple Square such as a5-1-e1-5, b8-6-d6-8, c9 with no overlap of joined sides, no two squares on a common side. What shape must this be for formation by the 6-pieces?"
Solution: "The three sucessive squares are 11x11, 8x8 one square in from left, and 5x5 two squares in."
No actual dissection is given in FCR. The solution shown below was given by G. P. Jelliss, in The Games and Puzzles Journal (V1, #12, 1989, p.194), it has no crossroads or corner cuts.
W. Stead (FCR Feb/Jun 1953 ¶9455). The New York Skyline, (or Churchillian Profile).
W. Stead (FCR Dec 1954 ¶28). Knight shape, with 'eye' as unit hole. CD = 2.
W. Stead (FCR 1954 ¶26). 14×17 rectangle with holes spelling F.H. (dedicated to Frans Hansson)
W. Stead notebooks, before 1978, (Games and Puzzles Journal Vol.1 #1 p9 1987), dedicated to C. E. Kemp.