bonanova Posted April 2, 2009 Report Share Posted April 2, 2009 After studying brickmaking, Johnny wondered whether he could make a stack of building blocks that leaned in such a way that the top block, when viewed directly from above, would not overlap the bottom block at all. What's your guess? If you think it's possible, how many blocks would it take? Of course, glue is not permitted. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 2, 2009 Report Share Posted April 2, 2009 (edited) 1 block...altough I guess that's not a stack Edited April 2, 2009 by wakesnow61784 Quote Link to comment Share on other sites More sharing options...
0 preflop Posted April 2, 2009 Report Share Posted April 2, 2009 Are the blocks perfect cubes (H/W/D all the same length)? Quote Link to comment Share on other sites More sharing options...
0 Pickett Posted April 2, 2009 Report Share Posted April 2, 2009 (edited) After studying brickmaking, Johnny wondered whether he could make a stack of building blocks that leaned in such a way that the top block, when viewed directly from above, would not overlap the bottom block at all. What's your guess? If you think it's possible, how many blocks would it take? Of course, glue is not permitted. I would argue that this configuration would satisfy the OP (pretend that these are all equal sized bricks)...except there are 2 top bricks...not just one. As a result, with this configuration, it takes 4... OOps, didn't realize that they needed to be cubes...n/m :c) Edited April 2, 2009 by Pickett Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 2, 2009 Report Share Posted April 2, 2009 After studying brickmaking, Johnny wondered whether he could make a stack of building blocks that leaned in such a way that the top block, when viewed directly from above, would not overlap the bottom block at all. What's your guess? If you think it's possible, how many blocks would it take? Of course, glue is not permitted. Deja vu? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted April 2, 2009 Author Report Share Posted April 2, 2009 Are the blocks perfect cubes (H/W/D all the same length)? Yes, cubes. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted April 2, 2009 Author Report Share Posted April 2, 2009 Deja vu? What then is the minimum number of cubes that accomplishes the task? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 2, 2009 Report Share Posted April 2, 2009 What then is the minimum number of cubes that accomplishes the task? 4 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 2, 2009 Report Share Posted April 2, 2009 What then is the minimum number of cubes that accomplishes the task? Six. Assuming that there can be more than one block on the top level. One block as the base, two blocks set evently on top of the first, and then three blocks evenly spaced on top of the two middle ones. The top right block will be right next to the base block (when looking down) but it will not overlap. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 2, 2009 Report Share Posted April 2, 2009 This is an example of a 'harmonic cantilever'. I've put a picture below. You'd need six blocks to be able to see both the top and the bottom at the same time, assuming you discount the fact that some of the other blocks are in the way. Also, the picture is of long blocks, but squares will work too. The series this is based on is 1/2 +1/3 + 1/4 + 1/5 + 1/6... on into infinity. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 2, 2009 Report Share Posted April 2, 2009 I would guess onlyYou would only need three and the bottom layer of the stack would be two blocks long and the last block goes on top of one of the bottom ones but since the stack wouldn't be leaning persay maybe this is wrong Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 2, 2009 Report Share Posted April 2, 2009 (edited) 4 ahem, 5 (forgot you need one on the bottom...) Edited April 2, 2009 by d3k3 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 2, 2009 Report Share Posted April 2, 2009 This is an example of a 'harmonic cantilever'. I've put a picture below. You'd need six blocks to be able to see both the top and the bottom at the same time, assuming you discount the fact that some of the other blocks are in the way. Also, the picture is of long blocks, but squares will work too. The series this is based on is 1/2 +1/3 + 1/4 + 1/5 + 1/6... on into infinity. I think you are onto something with this harmonic cantilever; however, I think your answer of 6 is not correct. Based on your picture, the 6th block would overlap as well. The block that has 1/2 overhanging is always on top. I think then, the answer in the case of the harmonic cantilever is you would need an infinite amount of blocks. That is assuming you can only have one block on the top level. If there can be more than one, then six is the answer. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted April 2, 2009 Author Report Share Posted April 2, 2009 Five is the best claimed solution so far. Johnny imagines only a single block on the base [and every other] level. Is there a sketch? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 2, 2009 Report Share Posted April 2, 2009 I think you are onto something with this harmonic cantilever; however, I think your answer of 6 is not correct. Based on your picture, the 6th block would overlap as well. The block that has 1/2 overhanging is always on top. I think then, the answer in the case of the harmonic cantilever is you would need an infinite amount of blocks. That is assuming you can only have one block on the top level. If there can be more than one, then six is the answer. The picture isn't perfect. According to the math, you'd only need 5 blocks. A base block plus one over hanging that by 1/2 another by 1/3+1/4+1/5 adds up to more than one. Therefore, the last block overhanges the fifth by more than the total length of a block. I said six because I've actually tried it before and it was impossible to get it perfect, so it actually took me six. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 2, 2009 Report Share Posted April 2, 2009 My Answer is Five Forgive my Crude Drawing: TWO OO = 1 block OO OOOO OO OO Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 2, 2009 Report Share Posted April 2, 2009 My Answer is Five Forgive my Crude Drawing: TWO OO = 1 block OO OOOO OO OO It didn't show my spacing correctly. TWO OO = 1 block. X = Space OO OOOO XOO OO Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 2, 2009 Report Share Posted April 2, 2009 Five is the best claimed solution so far. Johnny imagines only a single block on the base [and every other] level. Is there any rule against having more than one block on the loewst level? If there isn't then I can do it with only 4 blocks. Place two cubes on the bottom with exactly enough space in between them to put another cube. Place another cube on top of the first, but turn it 45 degrees so that it's diagonal corners are resting on each cube below it. Finally place the fourth cube on top of the third, making sure it is oriented the same way as the bottom two cubes and so that it is in between the bottom two cubes. With this configuration, the top cube won't block either of the bottom two cubes. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted April 2, 2009 Author Report Share Posted April 2, 2009 Is there any rule against having more than one block on the loewst level? If there isn't then I can do it with only 4 blocks. Place two cubes on the bottom with exactly enough space in between them to put another cube. Place another cube on top of the first, but turn it 45 degrees so that it's diagonal corners are resting on each cube below it. Finally place the fourth cube on top of the third, making sure it is oriented the same way as the bottom two cubes and so that it is in between the bottom two cubes. With this configuration, the top cube won't block either of the bottom two cubes. Yes. See post 14. It's a leaning tower problem. Can you modify your solution to work with one block per level? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 2, 2009 Report Share Posted April 2, 2009 (edited) I proved this solution at home. Sorry... I couldn't resist /Uhre Edited April 2, 2009 by uhre Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted April 2, 2009 Author Report Share Posted April 2, 2009 I proved this solution at home. Sorry... I couldn't resist /Uhre Now you're pulling my leg[o]. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 2, 2009 Report Share Posted April 2, 2009 Yes. See post 14. It's a leaning tower problem. Can you modify your solution to work with one block per level? Hah, I didn't realize that Johnny was the character from the OP. I thought he was another user. I think the other solution with 1/2 overhang, then 1/3 overhang ,then 1/4, etc., could be modified by rotating the all the blocks except for the top and bottom ones 45 degrees. This would allow the top block to be moved slightly further away from the bottom block, perhaps just enough to reduce the number of required blocks from 5 to 4. I haven't done the math though to prove this. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 2, 2009 Report Share Posted April 2, 2009 Hah, I didn't realize that Johnny was the character from the OP. I thought he was another user. I think the other solution with 1/2 overhang, then 1/3 overhang ,then 1/4, etc., could be modified by rotating the all the blocks except for the top and bottom ones 45 degrees. This would allow the top block to be moved slightly further away from the bottom block, perhaps just enough to reduce the number of required blocks from 5 to 4. I haven't done the math though to prove this. works. The center of the top block could be placed over the corner of the 2nd block (from the top). The third block can be shifted sqrt(2)/4 and the last (bottom) block by sqrt(2)/6, leaving a space of 0.089 between the top and bottom blocks. Quote Link to comment Share on other sites More sharing options...
Question
bonanova
After studying brickmaking, Johnny wondered whether he could
make a stack of building blocks that leaned in such a way that
the top block, when viewed directly from above, would not
overlap the bottom block at all.
What's your guess?
If you think it's possible, how many blocks would it take?
Of course, glue is not permitted.
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