The four 'T' Puzzle: Torsten Sillke, FFM, 1993-12 and 1999-12 Last Update 2002-07-20 Given four T-hexominoes there are three ways to fit them into a 6-square. . a a a . . square 1 . b a c c c . b a d c . . b a d c . b b b d c . . . d d d . a . . b b b square 2 a a a a b . a d . . b . . d . . b c . d.c c c c d d d . . c . . . a b . square 3 a a a a b . d d d a b . . d c b b b . d c c c c . d c . . . . a a a c c c fitting into a 5x7 rectangle . b a . d c . . b a . d c . . b a . d c . b b b d d d . . b a a a d c c c fitting into a 4x9 rectangle . b . a . d . c . . b . a . d . c . b b b a d d d c . . b a . . . c c c fitting into a 4x9 rectangle . b a a a a d c . . b a d d d d c . b b b . . . d c . But there is an even smaller square into which they fit. >>> Solutions and extensions <<< Smallest enclosing rectangles: Calculate the side-lengths of the smallest enclosing rectangle in each case. rectangle x_1, y_1 and area A_1 for square 1, rectangle x'1=y'1 and area A'1 for square 1 (shifting the two parts delta), rectangle x"1, y"1 and area A"1 for square 1 (axis parallel), rectangle x_2=y_2 and area A_2 for square 2, rectangle x_3=y_3 and area A_3 for square 3, rectangle x'3=y'3 and area A'3 for square 3 (axis parallel). More general we consider symmetric T pieces consisting of - two a*b rectangles or - one 3b*b and one a*b rectangle. Type: two equal rectangles -------------------------- case: ratio 3:1 (Hexomino) (a,b) = (3, 1) y_1 = 26/sqrt(17) x_1 = 22/sqrt(17) A_1 = 572/17 = 33.647059 = 6.305926251 = 5.335783751 x'1 = 6.111897751 with delta = 4/5 the delta-shift forms a square x_2 = 6 A_2 = 36 x_3 = 9*sqrt(10)/5 = 5.692099788303 A_3 = 162/5 = 32.4 x'3 = 6 (axis parallel) A'3 = 36 x_4 = 7 y_4 = 5 A_4 = 35 unit = 1.75 cm x_2 = 10.5 cm x_3 = 9.96 cm case: rectangle (a,b) = (4,1) a a a a a a a a square 1' a a a a a a a a b b a a c c c c c c c c b b a a c c c c c c c c b b a a d d c c b b a a d d c c b b a a d d c c b b a a d d c c b b a a d d c c b b a a d d c c b b b b b b b b d d c c b b b b b b b b d d c c d d d d d d d d d d d d d d d d a a b b square 3 a a b b a a b b a a a a a a a a a a b b a a a a a a a a a a b b d d d d d d d d a a b b d d d d d d d d a a b b d d c c a a b b d d c c b b b b b b b b d d c c b b b b b b b b d d c c c c c c c c c c d d c c c c c c c c c c d d c c d d c c d d c c c_1 = sqrt(26) c_2 = sqrt(170)/2 c_3 = sqrt(73)/2 y_1 = 75/2/sqrt(26) x_1 = 69/2/sqrt(26) A_1 = 5175/104 = 49.759615 = 7.354355067 = 6.766006662 x'1 = 7.256297000 with delta = 1/2 the delta-shift forms a square x_2 = 7.5 A_2 = = 56.25 x_3 = 6.846907111 A_3 = = 46.880137 x_4 = 9 y_4 = 6 A_4 = 54 = 54 case: rectangle (a,b) = (9,2) a a a a b b b b a a a a b b b b a a a a b b b b square 3 a a a a b b b b a a a a b b b b a a a a b b b b a a a a b b b b a a a a a a a a a a a a a a a a a a a a a a b b b b a a a a a a a a a a a a a a a a a a a a a a b b b b a a a a a a a a a a a a a a a a a a a a a a b b b b a a a a a a a a a a a a a a a a a a a a a a b b b b d d d d d d d d d d d d d d d d d d a a a a b b b b d d d d d d d d d d d d d d d d d d a a a a b b b b d d d d d d d d d d d d d d d d d d a a a a b b b b d d d d d d d d d d d d d d d d d d a a a a b b b b d d d d c c c c a a a a b b b b d d d d c c c c a a a a b b b b d d d d c c c c a a a a b b b b d d d d c c c c b b b b b b b b b b b b b b b b b b d d d d c c c c b b b b b b b b b b b b b b b b b b d d d d c c c c b b b b b b b b b b b b b b b b b b d d d d c c c c b b b b b b b b b b b b b b b b b b d d d d c c c c c c c c c c c c c c c c c c c c c c d d d d c c c c c c c c c c c c c c c c c c c c c c d d d d c c c c c c c c c c c c c c c c c c c c c c d d d d c c c c c c c c c c c c c c c c c c c c c c d d d d c c c c d d d d c c c c d d d d c c c c d d d d c c c c d d d d c c c c d d d d c c c c d d d d c c c c c_1 = 5 sqrt(5) c_2 = 5 sqrt(34)/2 c_3 = sqrt(373)/2 y_1 = 176/5/sqrt(5) x_1 = 67/2/sqrt(5) A_1 = 5896/25 =235.84 = 15.7419185616 = 14.9816554492 x'1 = 15.62495501 with delta = 17/26 = 0.653846 x"1 = 16.5 y"1 = a + 3b = 15 x_2 = 16.5 A_2 = 272.25 x_3 = 573/2/sqrt(373) = 14.8344076887 A_3 = 573^2/1492 =220.059651 x_4 = 20 y_4 = 13 A_4 = 260 =260 unit = 2/3 cm [Manfred Unnerstall] y_1 = 352/15/sqrt(5) x_1 = 67/3/sqrt(5) A_1 = 23584/225 =104.817777 = 10.494612374 = 9.9877702995 x"1 = 11 y"1 = 10 x_2 = 11 A_2 = 121 x_3 = 191/sqrt(373) = 9.8896051257993 A_3 = 191^2/373 = 97.804290 Note: a + 3b = 10 is nearly x_3 and y_1. So a frame with sidelength 10 seems optimal for puzzling. case: rectangle (a,b) = (6,1) c_1 = 5*sqrt(2) c_2 = sqrt(370)/2 c_3 = 13/2 y_1 = 133/10/sqrt(2) x_1 = 137/10/sqrt(2) A_1 = 18221/200 = 91.105 = 9.404520190 = 9.687362902 x_2 = 10.5 A_2 = 110.25 x_3 = 237/26 = 9.1153846153846 A_3 = 237^2/26^2 = 83.090236 x_4 = 13 y_4 = 8 A_4 = 104 =104 case: x_1 = y_1 c_1 = sqrt((a+b)^2 + b^2) x_1 = (3a^2 + 4ab + 5b^2)/2/c_1 y_1 = (a+b)(2a+7b)/2/c_1 let b = 1 then x_1 = y_1 means 3a^2 + 4a + 5 = (a+1)(2a+7) a^2 - 5a - 2 = 0 a = (5 + sqrt(33))/2 (a,b) = (5.372281323269, 1) x_1 = 8.765010139729 A_1 = 76.825402749559 x_3 = 8.406381725332 A_3 = 70.667253711994 case: rectangle (a,b) = (2,0) x_1 = 3 y_1 = 2 A_1 = 6 x_2 = 3 A_2 = 9 x_3 = sqrt(5) = 2.236 A_3 = 5 x_4 = 4 y_4 = 2 A_4 = 8 case: rectangle (a,b) with a >= 3b c_1 = sqrt((a+b)^2 + b^2) c_3 = sqrt( a^2 + ((a-b)/2)^2 ) P_x = (2b^2(a+b)+(a+b)^3/2)/c_1^2 + (a+b)/4 P_y = (3b(a+b)^2/2)/c_1^2 + (a-b)/2 Q_x = 3(a+b)/4 Q_y = (a-b)/2 R_x = (a+b)/4 R_y = (a+3b)/2 cosp = (a+b)/c_1 sinp = b/c_1 delta = (-a^2 + 5ab + 2b^2)/(2a + 4b) for 3 <= a/b <= (5 + sqrt(33))/2. x_1 = cosp P_x - sinp P_y = cosp Q_x - sinp Q_y = (3a^2 + 4ab + 5b^2)/2/c_1 y_1 = sinp P_x + cosp P_y = sinp R_x + cosp R_y = (a+b)(2a+7b)/2/c_1 x'1 = (a+b)(2a+7b)/2/c_1 - delta b/c_1 x"1 = 3(a+b)/2 y"1 = a + 3b x_2 = 3(a+b)/2 x_3 = c_3 + b(3a-b)/c_3 x'3 = 3(a+b)/2 (axis parallel) x_4 = 2a + b y_4 = a + 2b case: rectangle (a,b) = (2,1) a a a a square 1 a a a a can be reduced to rectangle 4 if a<2b. b b a a c c c c b b a a c c c c b b a a d d c c b b a a d d c c b b b b d d c c b b b b d d c c d d d d d d d d a a b b square 3 a a a a a a b b a a a a a a b b d d d d a a b b d d d d . b b b b d d c c b b b b d d c c c c c c d d c c c c c c d d c c x_3 = 4.486909063 minimal slanted square x'3 = 4.5 axis parallel square c c c c a a a a rectangle 4 c c c c a a a a d d c c b b a a d d c c b b a a d d c c b b a a d d c c b b a a d d d d b b b b d d d d b b b b x_4 = 5 y_4 = 4 case: x_3 = x'3 (minimal slanted square = axis parallel square) If c_3 = 2 b then the eight triangles at the border of the slanted square are congruent. And two of these triangles form the empty rectangles at the border of the axis parallel square. a = (1 + 2 sqrt(19))/5 = 1.943559577 b = 1 c_3 = 2 x_3 = 3 (3 + sqrt(19))/5 = 4.415339366 Type: enlarged hexomino ----------------------- +----+ +----+ | | | | | | | +-----------------+ | | | | | | | | ++--------------+-+ | | | | | | | | | | | | +----+ D----+ +----+ | | C----+ +----+ +----+ | | | | | | | | | | | | +-+--------------++ | | | | | | | | F-----------------G | | | | | | | B----E +----A Where the points A, E, and G are colinear. a = FG = AD b = BE = EF (3b = BC) then as the triangles ABE and EFG are similar we have AB : BE = EF : FG that is AB = b*b/a. Let d=DC be the width of the empty cross in the middle we know that the gap d shall be positive a - 3b - b^2/a = d >= 0 This gives the condition on (a,b) a^2 - 3ab - b^2 >= 0 (nonoverlapping condition) case: a^2 - 3ab - b^2 = 0 (this is d = 0) Let b = 1 then a = (3 + sqrt(13))/2 = 3.302775637732 = [3,3,3,3,3,3,...] (continued fraction) c = sqrt((13 + 3sqrt(13))/2) = sqrt(a^2 + b^2) x_2 = a + 3 = 6.3027756377320 A_2 = 39.724981 x_3 = 6(3+sqrt(13)) = 5.7425521589343 A_3 = 32.976905 ------------------ sqrt(26+6sqrt(13)) x_4 = 7 y_4 = a + 2 = 5.3027756377320 A_5 = 37.119429 case: a^2 - 3ab - b^2 >= 0 c_1 = sqrt((a+b)^2 + b^2) c_3 = sqrt(a^2 + b^2) d = a - 3b - b^2/a x_1 = (5a + 7b)b/c_1 y_1 = (a^2 + 4ab + 5b^2)/c_1 x_2 = a + 3b x_3 = (a+b)(b/c_3 + c_3/a) = (a+b)(a+b+b^2/a)/c_3 x_4 = 7b y_4 = a + 2b case: rectangle (a,b) = (7,2) c_1 = sqrt(85) c_3 = sqrt(53) d = 3/7 = 0.428571 x_1 = 98/sqrt(85) y_1 = 125/sqrt(85) A_1 = 144.117647 = 10.629592433 = 13.55815361 x_2 = 13 A_2 = 169 x_3 = 603/7/sqrt(53) = 11.832631437 A_3 = 140.011167 x_4 = 14 y_4 = 11 A_4 = 154 case: rectangle (a,b) = (10,3) continued fraction a/b = [3,3] (Second order approximation of d=0) c_1 = sqrt(178) c_3 = sqrt(109) d = 1/b = 1/10 = 0.1 x_1 = 213/sqrt(178) y_1 = 265/sqrt(178) A_1 = 317.106742 = 15.965025 = 19.862590 x_2 = 19 A_2 = 361 x_3 = 1807/10/sqrt(109) = 17.307920974 A_3 = 299.564128 x_4 = 21 y_4 = 16 A_4 = 336 unit = 2.6 mm [Peter Hajek] (a, b) = (26, 7.8) mm and d = 0.26 mm x_2 = 49.4 mm x_3 = 45.0006 mm unit = 2.9 mm [Binary Arts] (a, b) = (29.0, 8.7) mm and d = 0.29 mm x_2 = 55.10 mm x_3 = 50.193 mm case: rectangle (a,b) = (109,33) continued fraction a/b = [3,3,3,3] (Forth order approximation of d=0) c_1 = sqrt(21253) c_3 = sqrt(12970) d = 1/b = 1/109 = 0.00917 x_2 = 208 A_2 = 43264 x_3 = 2352514/109/c_3 = 189.511570490 A_3 = 35914.635350 case: rectangle (a,b) = (5,1) now the T is too tall, such that the smallest square is . a a a c c c . b a . d c . . b a . d c . . b a . d c . . b a . d c . . b a . d c . b b b d d d . x_4 = y_4 = 7 case: rectangle (a,b) = (8,1) . b a a a d c c c . b . a . d . c . . b . a . d . c . . b . a . d . c . . b . a . d . c . . b . a . d . c . . b . a . d . c . . b . a . d . c . b b b a d d d c . x_5 = y_5 = 9 Type: reduced hexomino ---------------------- case: (3 + sqrt(13))/2 = 3.302775637732 >= a/b >= 1.5 = 3/2 c_1 = sqrt((a+b)^2 + b^2) c_3 = sqrt(a^2 + b^2) x_1 = (5a + 7b)b/c_1 y_1 = (a^2 + 4ab + 5b^2)/c_1 x_2 = a + 3b x_3 = 6 a b / c_3 case: rectangle (a,b) = (3,1) continued fraction a/b = [3] See (Hexominoe), negative gap = 1/a = 1/3 x_3 = 18/sqrt(10) = 5.6921 case: rectangle (a,b) = (33,10) continued fraction a/b = [3,3,3] See [Heinrich Hemme], negative gap = 1/a = 1/33 x_3 = 1980/sqrt(1189) = 57.421 case: x_2 = x_3 let b = 1 then a + 3 = 6 a / sqrt(a^2 + 1). This means a^4 + 6a^3 - 26a^2 + 6a = 0 a = 2.600101927 x_2 = x_3 = a+3 = 5.600101927 case: rectangle (a,b) = (sqrt(6),1) = (2.44949,1) x_2 = sqrt(6) + 3 = 5.44949 x_3 = 6 sqrt(6) / sqrt(7) = 5.55492 The four Ts fit into a circle of radius sqrt(10). This is done by arangement 3, where 8 points touch the circumference. [Bartl] case: rectangle (a,b) = (2,1) a . b b b x_2 = 5 a a a b . a d . b c . d.c c c d d d . c . . . a . . x_3 = 12/sqrt(5) = 5.366563146 . a a a b . x_3' = 4*sqrt(2) = 5.656854249 d d d a b . . d c b b b . d c c c . . . c . . . case: rectangle (a,b) = (13,5) c = sqrt(194) x_2 = 28 x_3 = 390/c = 28.000368186 case: rectangle (a,b) = (3,2) a a . b b b b b b a a . b b b b b b a a a a a b b . . a a a a a b b c c a a d d . b b c c a a d d c c c c c . . d d c c c c c d d d d d d . c c d d d d d d . c c . . . . . . a a . . . . . . . . . . a a . . . . . . . a a a a a . . . . . . . a a a a a b b . . d d d d d d a a b b . . d d d d d d a a b b . . . . d d c c b b b b b b . . d d c c b b b b b b . . d d c c c c c . . . . . . . c c c c c . . . . . . . c c . . . . . . . . . . c c . . . . . . x_2 = 9 x_3 = 36/sqrt(13) = 9.9846 case: 1.5 >= a/b >= 0 x_2 = 3b + max(a,b) x_3 = 18b/sqrt(13) Type: general T --------------- +---------------------+ | | b b +---w---+ +---w---+ | | | | | | | | h h | | | | | | | | +--b--+ case: (h, b, w) = (15, 3, 5) b b b a a a b b b a a a b b b a a a b b b a a a b b b a a a b b b a a a a a a a a a a a a a a a a a a b b b a a a a a a a a a a a a a a a a a a b b b a a a a a a a a a a a a a a a a a a b b b d d d d d d d d d d d d d a a a b b b d d d d d d d d d d d d d a a a b b b d d d d d d d d d d d d d a a a b b b d d d a a a b b b d d d c c c a a a b b b d d d c c c b b b d d d c c c b b b b b b b b b b b b b d d d c c c b b b b b b b b b b b b b d d d c c c b b b b b b b b b b b b b d d d c c c c c c c c c c c c c c c c c c d d d c c c c c c c c c c c c c c c c c c d d d c c c c c c c c c c c c c c c c c c d d d c c c d d d c c c d d d c c c d d d c c c d d d c c c d d d d = 1 x_3 = 72/sqrt(10) = 22.768399 case: positive gap +-----R +-----+ | | | | | | | | | | | | | +-----------------------------+ | | | | | | | | +-+-------------------------+-+ | | | | | | | | | | | | +---------+ D---------+ | | | | C-----+ . | | | | | | +-----K | | | | +---------+ +---------Q | | | | | | | | | | | | +-+-------------------------+-P | | | | | | | | F-----------------------------G | | | | | | | | | | | | | B-----E H-----A I Where the points P, Q, and R are on a line (as well as A, E, and G). Let 'I' be a corner of the surrounding square. h = FG = AD b = BE = GR w = EF Let c = EG = sqrt(h^2 + w^2) by the Pythagorean theorem. As the triangles ABE and EFG are similar we have AB : BE = EF : FG that is AB = b*w/h. Let d=DC be the width of gap in the middle. As the gap can not be negative we know d = h - 2w - b - b*w/h >= 0 How long is the enclosing square matching the points A, E, G, and I? The side lenght x is equal x = EG + AE + AI + HI Note that the triangles EFG, ABE, and HIA are similar. So we get x = c + b*c/h + h*b/c + w*b/c Coordinates (as complex number) of P, Q, R. The origin is in the middle of the figure. 2P = 2h + 2b - f - f*i 2Q = 2w + 2b + f + (2b-f)*i 2R = 2b + f + (2h + 2b - f)*i with f = b + d + w = h - w - w*b/h the horizontal difference between C and K. Configuration with integral h, b, w, and d with d <= 3 and h <= 21: h b w d Area x --------------------------------------- 10 5 2 0 108^2/26 21.180543 12 3 4 0 62^2/10 19.606121 12 4 3 1 88^2/17 21.343135 12 6 2 1 153^2/37 25.153045 14 7 2 2 206^2/50 29.132799 15 3 5 1 72^2/10 22.768399 15 5 3 3 134^2/26 26.279562 16 8 2 3 16 4 4 3 18 3 6 2 21 7 6 0 21 3 7 3 case: negative gap +-----+ | +-----Q | | | | | | +-------------------+ | | | | | | | | +-+-------------------+ | | | | | | | | | | +-------+ H-----F-+ | | | | G-+-----K +-------P | | | | | | | | | | +-------------------E-S | | | | | | | | C-------------------D | | | | | | +-----+ | A-----B Where the points P and Q are on the surrounding square (as well as B and D). (We assume that point S is inside this square.) d = h - 2w - b - b*w/h < 0 Coordinates (as complex number) of P and Q. The origin is in the middle of the figure. 2P = 4w + 2b - g - g*i = 3w + 3b - (w-b)*i 2Q = 2w + 2b - g + (2h - g)*i = w + 3b + (2h - w + b)*i with g = w - b the signed size of the central square and f = b + w the horizontal difference between H and K. Now rotate the square around the center such that P and Q are one a vertical line. Let c = PQ = sqrt(h^2 + w^2) then exp(i*alpha) = cos(alpha) + i*sin(alpha) = (h - w*i)/c Re( 2P exp(i*alpha) ) = ( 3(w + b)h - (w - b)w )/c Re( 2Q exp(i*alpha) ) = ( 3(w + b)h - (w - b)w )/c As it should be both values are equal to the side lenght of the square square length = ( 3(w + b)h - (w - b)w )/c case: zero gap approximation The gap equation was d = h - 2w - b - wb/h >= 0 For d = 0 we get h = 2w+b + wb/h (*) Determine h in equation (*) gives h = ( 2w+b + sqrt((2w+h)^2 + 4wb )/2 Approximation for d=0 by iteration of equation (*). This gives a series of continued fractions h[1] = 2w+b h[2] = 2w+b + wb ---- 2w+b h[3] = 2w+b + wb ---- 2w+b + wb ---- 2w+b For w=b=1 we get the standard continued fractions. Rewriting the sequence as the quotiont of a 2nd order linear recursion gives x[-1] = 0 x[0] = 1 x[n] = (2w+b) x[n-1] + wb x[n-2] h[n] = x[n] / x[n-1] d[n] = h[n] - h[n+1] = (-wb)^n / ( x[n] x[n-1] ) the gap Example: w=b=1 n : 0 1 2 3 4 x[n] : 1 3 10 33 109 1/d[n] : - -3 30 -330 3597 Let's make w=b=10mm then h[2] = 33.333mm and d[2] = 0.333mm. So you can barely see the gap. But in the next approximation h[3] = 33mm the gap d[3] = -0.030303mm is invisible. case: h >= b + 2w (axis parallel square) . . . . . . +-----+-----+ . . | | | . | | | | | | . | | | +-----------------------------+ | | | | | . | | | +-------------------------+---+ | | | | | | | . | | | | | +---------+ D---------+ | | | | | | | | . | C-----+ +-----+ | . | | | | | | | | +---------+ +---------Q | | | | | . | | | | | | | +---+-------------------------R | | | . | | | | | F-----------------------------G | | | . | | | | | | . | | | . . H-----B-----E . . . . . square length = h + 2b + w case: h >= w >= b (axis parallel square) +-----+ . . . +-------------------------R | | | | | | | | | | +---------+ +---------+ | | | | | +-----------------------------+ | . | | | | | | . | +---+-----+-------------------+ | | | | | | | . | | | | | | | | | | | | . | | | | | | +-----+ | | | | +-----+ | | | | | | . | | | | | | | | | | | | . | | | | | | | +-------------------+-----+---+ | . | | | | | | . | +-----------------------------+ | | | | | +---------+ +---------+ | | | | | | | | | | +-------------------------+ . . . +-----+ square length = h + 2b + w rectangles: + . +-----+-----+ . . . +-------------------------+ | | | | | . | | | | | | | | +---------+ +---------+ . | | | | | | | +-----------------------+-----+ | . . | | | | | | | | | | . . | | +-----------------------+ | | | | | | | | . . | | |-----------------------+ | | | | | | | . . | | | | | | +-----+-----------------------+ | | . | | | | | +---------+ +---------+ | | | . | | | | | | | | | | +-------------------------+ . . . +-----+-----+ . + References: - Anatoly Kalinin; The annual puzzle party, Series: toys corner, Quantum 4:6 (July August 1994) 56-57, 55 (describes the t-puzzle of Peter Hajek) Figure 21 shows an enlarged hexomino with (a, b) = (26, 7.33) mm and gap d should be 1.93 mm. But the figure shows no gap. So not the smallest square is shown. - Jean-Claude Constantin, How the Three-L Puzzle was Developed, CFF (= Cubism For Fun) 50 Part 3 (Oct. 1999) 3-5 Figure 1 shows the 4 'T' puzzle made from hexominoes Then a 4 'Z' puzzle made from pentominoes is shown which is a minor variation of the T-puzzle as the solution is the same. Then he develops a 5 'L' puzzle which is made of different L pieces. - Rik van Grol; Puzzle News (2), CFF (= Cubism For Fun) 58 (July 2002) 49-51 Section: T-time - 4T puzzles and variants (describes the T-Party puzzle of Perry McDaniel.) Puzzles: enlarged hexomino - Peter Hajek, 13th International Puzzle Party, 1993, Breukelen, NL mailto:P.Hajek@mds.qmw.ac.uk unit = 2.6 mm, (a, b) = (10, 3), wood unit = 2.6 mm, (h, w, b) = (10, 3, 3) Two squares of different size can be filled. - 2 tantalizing Puzzles: 4-T Binary Arts, 5601 Vine Street, Alexandria, VA 22310, USA (the description wrongly says 14th IPP) unit = 2.9 mm, (a, b) = (10, 3), plastic (Hajek's version) unit = 2.9 mm, (h, w, b) = (10, 3, 3) Two squares of different size can be filled. - Jean-Claude Constantin, P\"uscheldorf 13, D-91238 Offenhausen, Germany unit = 5 mm, (a, b) = (10, 3), wood in 4 types, 1998 unit = 5 mm, (h, w, b) = (10, 3, 3) Two squares of different size can be filled. - Jean-Claude Constantin, P\"uscheldorf 13, D-91238 Offenhausen, Germany unit = 10 mm, (a, b) = (7, 2), wood, 1998 unit = 10 mm, (h, w, b) = (7, 2, 2) - Heinrich Hemme, Cogito Preisr\"atsel Juni 1999 Bild der Wissenschaft, 6/1999, p110 (solution: 9/1999) unit = 1 mm, (a, b) = (33, 10) unit = 1 mm, (h, w, b) = (33, 10, 10) - Heinrich Hemme; Mensch, \"argere dich nicht, 72 Kopfn\"usse und Knobeleien f\"ur jede Gelegenheit, ro ro ro science, Hamburg, 2003, ISBN 3-499-61575-4 - p61-62, 167 Taxi mit vier "T" unit = 1 mm, (a, b) = (33, 10) unit = 1 mm, (h, w, b) = (33, 10, 10) - Der kleine T-Tisch (the small T-Table) Bartl 2053 (Mini Games) Bartl GmbH, D-84518 Garching a. d. Alz, Germany fax: +49-8634-6411 unit = 5 mm, (a, b) = (2.5, 1), wood, 1998 The 4 T's must fit into a circle of radius 3.2 This is max(sqrt(a^2 + 4b^2), sqrt(10)*b). (Vier Tanten trinken Tee am T(ee)-Tisch.) - T-Games made in Taiwan, unit = 45/43 mm, (a, b) = (33, 10), plastic in 4 colors, 2001 unit = 45/43 mm, (h, w, b) = (33, 10, 10), x_3 = 60.09 mm. Two squares of different size can be filled. (Hajek's version) Puzzles: two rectangles - Manfred Unnerstall, Astrein und kniffelig, Denk- und Geduldspiele aus Holz, Ostenfelder Str. 25, D-49326 Melle/Neuenkirchen, Germany unit = 2/3 cm, (a, b) = (9, 2), wood, 1995? unit = 2/3 cm, (h, w, b) = (9, 3.5, 2) - The four T-tables, Explora Exhibition, Glauburg am Glauburgplatz, D-60318 Frankfurt/M, Germany tel: +49-69-788888 fax: +49-69-787777 http://www.EXPLORAMuseum.de mailto:GoStief@aol.com massures ??, tables, 1994 - Ruth Pfeffer, Meisenweg 12, D-65779 Kelkheim, Germany wood, 1995 (construction follows the Explora T-puzzle) - Das T-Pausen Puzzle Bartl 2807 (Midi Games) Bartl GmbH, D-84518 Garching a. d. Alz, Germany http://www.bartlgmbh.de/ fax: +49-8634-6411 unit = 1.1 cm, (a, b) = (3, 1), wood, 1998 unit = 1.1 cm, (h, w, b) = (3, 1, 1) - unknown, found on a flea-marked, unit = 1.75 cm, (a, b) = (3, 1), wood unit = 1.75 cm, (h, w, b) = (3, 1, 1) Special: the T-pieces can be separated into the two parts: x x x x x x y y y y y y y x x x x y y y y y y y This looks simmilar to the Bartl puzzle, but is of high quality. Puzzles: General T - Jean-Claude Constantin, Puzzle P 211 P\"uscheldorf 13, D-91238 Offenhausen, Germany unit = 1 mm, (h, w, b) = (18, 9, 2), T - steal, 1998? Each T exactly fits into a cube of size 20 mm. - Perry McDaniel; T-Party, similar to Constantin's T-steal puzzle, but the tray is filled from the side. Appendix: The five 'L' Puzzle of Jean-Claude Constantin: ---------------------------------------------- . . . . d . 4 'Z' puzzle a a d d d . . a d c c . . a a b c . . b b b c c . b . . . . smallest square = 22/sqrt(17) = 5.33578 a a a b b b 5 'L' pieces in a 5x6 rectangle c c a b . . d c a b . e d c a b . e d d d e e e The smallest square into which these piece fit is smaller than 6 units wide. -- mailto:Torsten.Sillke@uni-bielefeld.de http://www.mathematik.uni-bielefeld.de/~sillke/