Re: Brick wall: find the distribution

From: Randy Poe (poespam-trap_at_yahoo.com)
Date: 06/28/04 

Date: 28 Jun 2004 07:53:56 -0700

pfreedenberg@hotmail.com (phil) wrote in message news:<200406280023.i5S0Nq926063@proapp.mathforum.org>...
> Imagine a 2-D brick wall of arbitrary extent in which each brick
> rests on the 2 bricks immediately beneath. A weight W
> rests on the center brick of the top course. If each brick transmits
> forces equally to the 2 bricks immediately beneath, how is the weight
> distributed at the bottom of the wall, which is n courses high?
> Ignore the bricks' own weight.
>
> phil

Cute.

The answer may at first surprise you, but then the reason
becomes obvious. Hint: Each brick is also the sum of
the two bricks above it (divided by 2).

How far have you gotten?

          - Randy

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