20051215, 15:24  #1 
May 2003
3·7·11 Posts 
Mortarless brick constructions
How many bricks are required to build a structure such that one of the bricks overhangs at least one bricklength beyond the base of the structure, assuming nothing apart from normal gravity holds the structure together. (By an overhang of at least one bricklength, I mean that the nearest part of that brick to the base is horizontal distance >epsilon from the base, and the furthest part is horizontal distance >1+epsilon, where 1 is the brick's length, it's longest dimension.)
What about an overhang of 2? 
20051215, 15:47  #2  
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
2B79_{16} Posts 
Quote:
It's the minimum i such that the harmonic series sum exceeds 2 (resp. 3). The sum reaches 2.083333333333333333333333333 at i=4 and 3.019877344877344877344877344 at i=11. To get an overhang of ten, you need 33617 bricks, I believe. Paul 

20051215, 19:07  #3 
Jun 2003
3^{4}·5·13 Posts 
Five, counting the base brick

20051216, 08:16  #4 
May 2003
3×7×11 Posts 
The base brick indeed is counted, and with that you have the right answer.
It's pretty difficult to achieve with imperfect bricks. Here's me missing by 1, but able to get plenty of extra overhang from the extra "brick": http://fatphil.org/maths/bricks.jpg 
20051216, 13:36  #5  
"Mark"
Apr 2003
Between here and the
2^{7}·3·17 Posts 
Quote:


20051216, 13:41  #6 
Jul 2005
2·193 Posts 
Things without uniform weight distribution.
HTH. 
20051216, 14:46  #7 
May 2003
3·7·11 Posts 
The thicker plastic casing is heavier than the thin aluminium, and so I was at a _disadvantage_ that way round.
I just tried with some fairly uniform 'plastic': http://fatphil.org/maths/cards.jpg and it was much easier; I managed to hit the goal of 5. 
20051218, 11:14  #8 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
Mortarless brick constructions
This problem is better explained by using a pack of standard cards which is more practical. The problem is beautifully treated in the very first chapter by John Derbyshire in his excellent book โPrime obsessionโ The total overhang is with just 4 bricks (not counting the base card). The total over hang is 1/2 + 1/4 +1/6 + 1/8 = 25/24. Yes your are right Paul its half the harmonic series. Some interesting facts he gives are: For 51 cards the total overhang is a shade less than 2.2594065907334. For 100 cards itโs a tad less than 2.58868875887. For a trillion cards we have a bit more than 14.10411839041479 overhang. The total over hang increases very slowly as the number of cards goes up but never comes to an end as the harmonic series is divergent See proof by Nicole dโOresme. mathworld.wolfram.com/HarmonicSeries.html  32k  Cached  Similar pages Mally. P.S. 'Prime Obsession' is a must read for anyone interested in the RH. 