Square in Rubik's Cube
How many squares can you perceive in a Rubik's Cube and how many of them are visible ?
By perceive, I mean you can assume a square, if you can join four co-ordinates to form a square.
Labels: mathemagic, outside-the-box
posted by Rajesh Lal at
9:34 PM
16 Comments:
42
84
324 squares can be made and 84 of them are fully visible to us (asuming that a rubik's cube has a sigle layer of cubes composing it)
~shawn~
none
If I understand the question and assuming a generic 3x3x3 rubik's cube.. Then I say 216 perceived and 54 visible.
explation:
3x3x3=27 individual cubes percevied (1x1) times 6 sides each.
8 larger cubes perceived (2x2) times 6 sides each.
1 larger cube (3x3) times 6 sides.
All add up to 216.
Each side has 9 visible squares, 9*6 = 54.
Well, I might go with the 216. However, 9*6+4*6+1*6 = 84 would be the total "completely visible" squares (per maza).
I think the only completely hidden ones would be on the small middle cube - giving a total of 216-9 = 207 "visible" squares.
Well 9 squares per side and 6 sides would be 54 visible "small squares", but each group of 4 small squares make another larger square, and each 3x3 side makes a still larger square. Each of the six sides have four 4x4 squares and each of the six sides has 1 3x3 square so 54+24+6 = 84 visible.
Well 9 squares per side and 6 sides would be 54 visible "small squares", but each group of 4 small squares make another larger square, and each 3x3 side makes a still larger square. Each of the six sides have four 4x4 squares and each of the six sides has 1 3x3 square so 54+24+6 = 84 visible.
84 sqaures. Each of the six sides consists of 14 squares.
Potter-
i would agree most with shawn. euclid's brother has miscalculated, as he forgot to count ALL possible squares, assuming there is a center cube making the total number of cubes 27. each of these has 6 squares, and each visible side has 14 squares. i dont know the exact number, but i would lean to the side of something like 324, what shawn said.
yep.. i agree.. 84.. i didn't think to count the larger visible squares before..
Actually, I don't think any of us have come close yet (even ignoring my "9-sided" inner cube slip above.) For starters, given the problems explicit definition of "a square" each "side" has at least 4 squares nobody has mentioned yet - the 4 with an area of 2 square units centered around the vertices of the small inner square. That alone gives 108 "squares" on the outside of the cube (which would be one interpretation of "visible".)
Those of us considering the inside have some more work to do too. Actually, that could be where Shawn's 324 number comes from; but that does not jive with the 84. Also, I think there are even more than 324 and less than 635376.
Any more thoughts?
14 per side times 6 sides = 84
there are 84 visible squares:
(9*6)+(4*6)+6=84
there are 252 squares in the cube...
but i wonder, shawn?
how did you get 324?
Well... that is the first mention of 252. For those wondering Anon is assuming that ownensrx and others are correct in that there are 14 "squares" per side. The six outside surfaces (sides) then give 84. HOWEVER, if you consider there there are actually 18 surfaces if you include the interior of the cube (each consisting of 9 small squares and corresponding larger squares) you can get 18*14=252.
Shawns original insight was (possibly) that there were an additional 4 squares per side (as I hinted above) giving 18*18=324.
However, I now believe there are at least an additional 2 squares per surface giving a lower bound of 20*6=120 outside surface squares and 18*20=360 squares (inside and out).
Again, I believe that there are even more than that.
doing research on 9 sided hyper-hyper cube.looking at how it folds together---look at a go board and fold the 9 squares into a hyper-hyper cube-see why the center point is up-how many indices on the
cube----show me the math supporting your answer
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