My favorite scissors dissection problem is this one.
From a square S=ABCD with center M and vertices labelled clockwise from its northwest corner, cut out the triangle T = CDM.
Next, reattach T to what was the original square by rotating it ninety degrees clockwise about C until DC coincides with BC. Here's picture of the resulting 6 sided figure. Call it H https://www.flickr.com/photos/thane/26339632922/in/dateposted-public/ Let's say we want to cut H into two pieces that can be reassembled to form a new square congruent to S. Of course, one way to do that is simply return the triangle we cut out back to where it came from. The puzzle is to find a different solution. On Thu, Apr 14, 2016 at 9:21 AM, Dan Asimov <asimov@msri.org> wrote:
This is part of a letter by Morris W. Hirsch (who was my thesis advisor) to the Notices of the AMS (2014):
----- The term "scissor congruent," introduced in [1], refers to a pair of planar convex bodies
A = A_1 u ··· u A_n, B = B_1 u ··· u B_n,
where A_i and B_i are congruent compact 2-cells, andfor i,j the interiors of A_i and A_j (respectively, B_i and B_j) are disjoint. The main result is that A and B are scissor congruent iff they have the same area and their boundaries have partitions
∂A = I_1 u ··· u I_m, ∂B = J_1 u ··· u J_m,
where I_k and J_k are congruent 1-cells. It follows that a square and a circle are not scissor congruent, andscissor congruent ellipses are congruent. No analog in higher dimensions is known.
The term "scissor congruent" was introduced by Dubins et al. in [1] to describe a pair of convex planar bodies A, B having cell decompositions such that there is a bijection between the two-cells of A and those of B with corresponding two-cells congruent through rigid motions.
The main result in [1] is that A and B are scissor congruent iff they have the same area and their boundaries have partitions
∂A = I_1 u ... u I_m, ∂B = J_1 u ... u J_m
with I_k and J_k congruent arcs. It follows that a square is not scissor congruent to any strictly convex body, and if two ellipses are scissor congruent, then they are congruent.
References [1] L. Dubins, M. Hirsch, and J. Karush, Scissor congruence, Israel J. Math. 1 (1963), 239–247.
[2] S. Gold, The sets that are scissor congruent to an unbounded con- vex subset of the plane Trans. Amer. Math. Soc. 215 (1976), 99–117.
[3] R. Gardner, A problem of Sallee on equidecomposable convex bod- ies, Proc. Amer. Math. Soc. 94 (1985), 329–332.
[4] C. Richter, Affine congruence by dissection of discs—appropriate groups and optimal dissections, J. Geom. 84 (2005), 117–132.
[5] ________, Squaring the circle via af- fine congruence by dissection with smooth pieces, Beiträge Algebra Geom. 48 (2007), no. 2, 423–434. -----
—Dan
On Apr 14, 2016, at 5:18 AM, Bill Gosper <billgosper@gmail.com> wrote:
http://home.btconnect.com/GavinTheobald/HTML/Octagon.html#Hexadecagon I don't recall curved cuts in a polygonal dissection. Can't the circular segments be replaced by segments of hexadecagons?
Are curves essential to any known minimal dissection?
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-- Thane Plambeck tplambeck@gmail.com http://counterwave.com/