## January 2011

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In my January diary I posed this brain-teaser.

Here's a formal definition:If you now additionally drop the condition that all theArchimedean polyhedron: All faces are convex regular polygons, though not all the same polygon; all vertices are congruent; prisms and antiprisms don't count (because they're boring).verticesmust be the same, you get the Johnson polyhedra. That gets us back to the Egyptian-style square-based pyramid, which is the simplest Johnson polyhedron. The four base vertices are the same (square meets two triangles) but the summit vertex is different (four triangles).

There are 92 Johnson polyhedra, a fact not proved until 1969 — incredibly, it seems to me, when you consider that polyhedra have interested mathematicians since antiquity.

Here's something even more incredible. The number of Archimedean polyhedra as defined above was given as thirteen by Archimedes himself (according to Pappus), and this number was repeated down through the succeeding 22 centuries into my own schooldays. Cundy and Rollett's classicMathematical Models, for example (my edition 1961) repeats it on page 100. Yet it iswrong!And in all those centuries nobody (with the possible exception of Kepler) noticed!

The whole amazing story is told by Branko Grünbaum in an essay titled "An Enduring Error," which you can find in this splendid anthology.

OK, OK, a brainteaser. The true number of Archimedean polyhedra, on the definition above, is fourteen, not thirteen. The fourteenth one, though none of the other thirteen, is included among Johnson's 92. Can you spot it in the Mathworld list? What is its Johnson number?

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*Solution*

It is number 37, the elongated square gyrobicupola. Grünbaum, however, refers to it as the pseudorhombicuboctahedron: not to be confused (Heaven forfend!) with the pseudoquasirhombicuboctahedron, which shows up later in Grünbaum's paper.