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Title: Cuboid construction Post by NickH on Jan 10th, 2003, 1:55pm An a by b by c cuboid is constructed out of abc identical unit cubes -- a la Rubik's Cube. Identify all such cuboids with the following property: the number of external cubes (i.e., those that constitute the faces of the cuboid) is equal to the number of internal cubes. Nick |
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Title: Re: Cuboid construction Post by Garzahd on Jan 10th, 2003, 3:35pm Well, it seems to me there are two ways to answer this... 1) If you're saying (number of cubes touching outside)==(number of total cubes) then it's trivial... whenever a<3 or b<3 or c<3. 2) If you're saying (surface area)==(volume) then instead you're solving 2ab + 2bc + 2ac = abc, which has a bunch of less-than-trivial solutions, very few of which are integers... A 6x6x6 cube would work. |
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Title: Re: Cuboid construction Post by NickH on Jan 10th, 2003, 3:52pm Think of the original 3 x 3 x 3 Rubik's Cube. It has 26 external cubes and 1 internal cube. The 4 x 4 x 4 version has 60 external and 4 internal cubes. As an example of a cuboid with the required property, consider a 7 x 7 x 100 cuboid. Internal cubes: 5 x 5 x 98 = 2450. External cubes: 7 x 7 x 100 - 5 x 5 x 98 = 2450. The puzzle is to find a systematic way of finding all such cuboids. |
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Title: Re: Cuboid construction Post by Garzahd on Jan 10th, 2003, 4:48pm Oh, I see, that's more interesting; I hadn't caught onto the term "internal cubes" as being the non-external cubes. So... abc = 2*(a-2)(b-2)(c-2) abc = 2abc-4ab-4ac-4bc+8a+8b+8c-16 0 = abc-4(ab+ac+bc)+8(a+b+c)-16 That's a mess. I wonder if there's any completing-the-square mechanism that works with 3 variables... |
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Title: Re: Cuboid construction Post by NickH on Jan 13th, 2003, 5:51pm I got the following results -- highlight text below to view. Is this a neat puzzle? Or is the solution a little too long and messy? Let a <= b <= c be positive integer solutions of abc = 2(a - 2)(b - 2)(c - 2) Firstly, (1 - 2/a)(1 - 2/b)(1 - 2/c) = 1/2 Since 1 - 2/a <= 1 - 2/b <= 1 - 2/c, 1 - 2/a <= cubrt(1/2), and so a <= 2/(1 - cubrt(1/2)) So a <= 9 Expanding, abc = 2[abc - 2(ab + bc + ca) + 4(a + b + c) - 8] So abc - 4(ab + bc + ca) + 8(a + b + c) - 16 = 0 Geometrically, it's clear a > 3. Consider separately cases a = 4 to 9. If a = 4: -16(b + c) + 8(b + c) + 16 = 0 Therefore b + c = 2, contradicting a <= b <= c If a = 5: bc - 12(b + c) + 24 = 0 (b - 12)(c - 12) = 120 b - 12 = {1, 2, 3, 4, 5, 6, 8, 10}, c - 12 = {120, 60, 40, 30, 24, 20, 15, 12} Therefore (b,c) = {(13,132), (14,72), (15,52), (16,42), (17,36), (18,32), (20,27), (22,24)} If a = 6: 2bc - 16(b + c) + 32 = 0 bc - 8(b + c) + 16 = 0 (b - 8)(c - 8) = 48 Therefore (b,c) = {(9,56), (10,32), (11,24), (12,20), (14,16)} If a = 7: 3bc - 20(b + c) + 40 = 0 9bc - 60(b + c) + 120 = 0 (3b - 20)(3c - 20) = 280 Therefore (b,c) = {(7,100), (8,30), (9,20), (10,16)} If a = 8: 4bc - 24(b + c) + 48 = 0 bc - 6(b + c) + 12 = 0 (b - 6)(c - 6) = 24 Therefore (b,c) = {(8,18), (9,14), (10,12)} If a = 9: 5bc - 28(b + c) + 56 = 0 25bc - 140(b + c) + 280 = 0 (5b - 28)(5c - 28) = 504 This has no solutions with b >= 9 Total number of cuboids is 20. |
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Title: Re: Cuboid construction Post by Garzahd on Jan 14th, 2003, 10:52am Interesting. If I had the a <= 9 insight, it probably would have led me toward your guided-brute-force approach (which I tend to apply far too often anyway) |
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Title: Re: Cuboid construction Post by wowbagger on Jan 16th, 2003, 4:52am The problem (and solution) is interesting indeed. My unguided-brute-force approach - checking values for a, b, c up to 1000 - left me with the same results NickH posted. Unfortunately, this doesn't give one the certainty of having found all solutions, but only a strong clue. On the other hand, I could have invested some time and brain power in finding theses bounds... |
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