Tim Chow (tchow@alum.mit.edu) asked me to post this to math-fun. I should also mention that this was the subject of a very nice talk by Peter Winkler at the Joint Math Meetings a few weeks ago ("Stacking Bricks and Stoning Crows") and the paper "Maximum Overhang" http://arxiv.org/abs/0707.0093 Victor Here's Tim's post: Mike Paterson and Uri Zwick have a very nice article in the current issue of the American Mathematical Monthly entitled "Overhang." Earlier versions of this work may be found, for example, on Zwick's homepage. http://www.cs.tau.ac.il/~zwick/ The topic is the old problem of stacking bricks on a table so that they hang over the edge by an arbitrarily large amount. The standard solution involves a harmonic series, but the authors show that one can do much better if one allows more than one brick at each level. One thing that I mentioned to Paterson and Zwick (though apparently too late for them to include it in their Monthly article) is that there is, in fact, at least one place in the literature where the possibility of more than one brick on each level is explicitly considered. I refer to the book "Ingenious Mathematical Problems and Methods" by L. A. Graham (Dover 1959), specifically the problem about the "Bridal Arch." In Graham's version of the problem, there are 86 bricks (smooth, uniform, etc.) with dimensions 2x4x8, and they are to be stacked to form a (symmetrical) 7'-high, 34"-wide arch. In the solutions section, Graham reproduces three proposed solutions to the problem sent in by his readers, all of which have more than one brick per level. See http://thelastdanishpastry.blogspot.com/2009/01/bridal-arch.html However, Graham dismisses all three solutions as incorrect "models of graceful instability" without any analysis! I asked Paterson and he said that at least some of these solutions might be correct, but he did not actually check them in detail. Presumably, Graham simply assumed that those other solutions could not be correct. Can anyone here either vindicate them, or show that they are wrong? Tim