Answer to the Question 03/00
LEANING TOWER OF BRICKSThe question was:
By placing brick on top of another brick, slightly shifted as shown in the figure, we can create a situation that the top brick in such a pile is displaced horizontally by h relative to the bottom brick. Can this be done for an arbitrary h? What is the optimal way to do it?
(3/2000) The problem has been solved by Avi Nagar (e-mail anagar@bigyellow.com), by Lluis Batet from Technical University of Catalonia, Barcelona, Spain (e-mail lluis.batet@upc.es), by Andrew Wiggin (e-mail awiggin@hotmail.com), and by Robert White (e-mail fiziks@acsworld.net).
The solution: If the topmost brick is shifted by 1/2 brick relatively to the brick below it, and the second-from-the-top brick is shifted by 1/4 brick relatively to the brick below it, ... , n-from-the-top brick is shifted by 1/(2n) brick relatively to the brick below it, - then there is no limit to the displacement of the bricks, because the series Sum{n=1 to n=infinity}(1/n) diverges(*). At the same time one can directly verify the the center of mass of any k top-most bricks is exactly above the edge of the brick just below them: Let us calculate the horizontal position of the center-of-mass of k top-most bricks relative to the brick just below them. The displacement of the top-most brick in such a situation is (1/2)[1+1/2+1/3+...+1/k], the location of the second-from-the top is (1/2)[1/2+1/3+...+1/k], etc., and consequently position of the center of mass
(1/k)*(1/2)*[1+2*(1/2)+3*(1/3)+k*(1/k)]=1/2
i.e. it is exactly above the edge of the brick below them.
Since at every level the bricks above a given brick are shifted to their extreme positions, the solution presented here is optimal for any number of bricks.
* Note: For finite sum (up to a large integer N) Sum{n=1 to n=N}(1/n)=ln(N)+Gamma, where Gamma is Euler's constant. Thus, for a particular displacement h expressed in the units of single brick length we will need exp(2*h-Gamma) (rounded to the next integer) bricks.
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