((Modern) plane geometry:)
The reptile is the trimino
+--+
| |
+--+--+
| | |
+--+--+
Yes, 3 connected
squares, shaped like an "
L". It's interesting because of a way you can
tile the
plane with it.
The easiest tesselation of the plane with copies of the reptile is to use 2 copies to form a 3×2 rectangle:
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| 1| 1|
+--+--+
|2 | 1|
+--+--+
|2 |2 |
+--+--+
Then use many of these 3×2 rectangles to tile the plane, in any of several
boring ways.
But the reptile can also be used to get an aperiodic tiling of the plane (i.e., a tiling which never repeats itself). To do this, first put 4 reptiles together to get a new reptile which is twice the size:
+--+--+
| 1| 1|
+--+--+
|2 | 1|
+--+--+--+--+
| 3|2 |2 |4 |
+--+--+--+--+
| 3| 3|4 |4 |
+--+--+--+--+
Use 4 of
these double-sized reptiles to get a
quadruple-sized reptile.
Lather, rinse, reptile.
By taking an appropriate limiting process, you get a tiling of the plane. One way to do this is to take the current reptile (of size 2k×2k) as tile #1 in the next reptile (of size 2k+1×2k+1). This lets you fill in the "hole" left; the resulting sequence of tiles grows rapidly to cover the entire plane. In fact, you can chose many different limiting processes; it would seem you can get 2ℵ0 (the cardinality of the continuum) different tilings using the reptile!
It's also fairly easy (though tedious) to see that the result is aperiodic. The idea behind the proof is to show that if the tiling has period (i,j) (that is, if moving the tiling i squares right and j up results in the same tiling) then it also has period (i/2,j/2). Since i and j must be integers, this is clearly impossible.
The reptile is a single tile which gives an aperiodic tiling. However, it also gives periodic tilings of the plane. Other aperiodic tilings are known -- the most famous is uses Penrose tiles -- but require more than one tile. It is an open question if there exists a single tile which admits only aperiodic tilings.