Author |
Topic: Cuboid construction (Read 1229 times) |
|
NickH
Senior Riddler
Gender: 
Posts: 341
|
 |
Cuboid construction
« on: Jan 10th, 2003, 1:55pm » |
 Quote  Modify
|
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
|
|
 IP Logged |
Nick's Mathematical Puzzles
|
|
|
Garzahd
Junior Member
Gender:
Posts: 130
|
|
Re: Cuboid construction
« Reply #1 on: Jan 10th, 2003, 3:35pm » |
Quote Modify
|
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.
|
|
IP Logged |
|
|
|
NickH
Senior Riddler
Gender:
Posts: 341
|
|
Re: Cuboid construction
« Reply #2 on: Jan 10th, 2003, 3:52pm » |
Quote Modify
|
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.
|
|
IP Logged |
Nick's Mathematical Puzzles
|
|
|
Garzahd
Junior Member
Gender:
Posts: 130
|
|
Re: Cuboid construction
« Reply #3 on: Jan 10th, 2003, 4:48pm » |
Quote Modify
|
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...
|
|
IP Logged |
|
|
|
NickH
Senior Riddler
Gender:
Posts: 341
|
|
Re: Cuboid construction
« Reply #4 on: Jan 13th, 2003, 5:51pm » |
Quote Modify
|
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.
|
|
IP Logged |
Nick's Mathematical Puzzles
|
|
|
Garzahd
Junior Member
Gender:
Posts: 130
|
|
Re: Cuboid construction
« Reply #5 on: Jan 14th, 2003, 10:52am » |
Quote Modify
|
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)
|
|
IP Logged |
|
|
|
wowbagger
Uberpuzzler
 
Gender:
Posts: 727
|
|
Re: Cuboid construction
« Reply #6 on: Jan 16th, 2003, 4:52am » |
Quote Modify
|
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...
|
|
IP Logged |
"You're a jerk, <your surname>!"
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print