Here is a little riddle:
Take a checkerboard, and remove two squares at opposite corners. Is it then possible to find an exact and complete cover of the remaining board using dominoes (two are shown in orange), without overlap and overhang?
There is a surprisingly simple and elegant proof for the negative answer to this question. I've just learned it this afternoon, in a great public talk by mathematician Günter Ziegler, coauthor of "Proofs from the book", current president of the German Association of Mathematicians, and main organiser of the "Year of Mathematics 2008" in Germany.
Starting with the "Year of Physics" in 2000, the German Federal Ministry of Science and Education has dedicated each year since to one particular discipline, and following the humanities in 2007, this year is all about math. As Ziegler writes in the March 2008 issue of the Notices of the American Mathematical Society, "The entire year 2008 has been officially declared Mathematics Year in Germany. This has created an unprecedented opportunity to work on the public's view of the subject."
And he used the opportunity, in a talk this afternoon on the occasion of the opening of the Mathematics Year for Frankfurt. He discussed the role of proof in mathematics, and then gave examples of actual elegant proofs of geometrical problems using colorations of the plane. The checkerboard riddle was just the first of them - he ended explaining the steps of a quite surprising proof that a square can not be decomposed into an odd number of triangles of equal area. I was amazed to see how I was guided by him through the steps and the idea of the proof - it's exciting to follow a talk like this! And my impression was that the 300 or so people in the audience have felt the same. Most of them, however, were faces I knew from the math department, or students and teachers. It would be great if such an event will attract even more people from the interested public.
Have a nice weekend - and if you want to solve the checkerboard riddle by yourself, don't read the comments - I'm convinced the answer will be there pretty soon!
Unfortunately, the slides of the talk are not online. Here are, roughly, the steps of the proof of the impossibility to decompose the square into an odd number of triangles of equal area: Start by colouring the rational points of the unit square in three different colours, using a scheme depending on the enumerator and denominator of the coordinates of the point. Then, convince yourself that each decomposition of the square into triangles (with corners in the rational points) contains at least one triangle with three different colours for the three corners. It comes out that this triangle, because of the rules chosen for colouring, has an area with an even denominator. Hence, in a decomposition into triangles with equal area, it cannot be part of a decomposition into an odd number of triangles. Use some heavy machinery to promote this proof from rational corner points to any corner points of the triangles, and you're done. If you find an error in this description, it's probably my fault - you may consult the original papers, "A Dissection Problem" by John Thomas, Mathematics Magazine 41 No. 4 (Sep. 1968), 187-190 (via JSTOR; subscription required), and "On Dividing a Square Into Triangles" by Paul Monsky, The American Mathematical Monthly 77 No. 2 (Feb. 1970) 161-164 (via JSTOR, subscription required).
Tag: Mathematics Year