Among the 54 pieces of Polyspidrons, there are 2 equilateral triangles (blue and red below) :
If we take the length of the side of the little triangle as unit, the length of the side of the big one is ; for convenience, afterwards, I shall note r this number.
If we take the area of the little triangle as unit, the area of the big one is r², that is 3 : it's understandable, because the big triangle is a trispidron.
Below we can see the 3 monospidrons, with area 1, the 10 bispidrons, with area 2, and the 41 trispidrons, with area 3 :
If we make, in the tray, a grid from the little triangle, we obtain :
This grid allows us to make exercises ; we colour triangles yellow to have a yellow triangle with size 6 :
This triangle have area 6² or 36 ; we can cover it with polyspidrons in several different ways :
The solution above uses only polyspidrons without head ; there are solutions with polyspidrons, which have one or several heads : one solution with 1 node of spidrons in center, another with 2 nodes, one with 3 nodes and one with 4 nodes. The above triangle, with side 6, is noted T(6). You can try T(1), T(2), and so on ... until T(12), which was Challenge 001.
T(1) of course is very easy and the difficulty grows ...
I have solutions of T(1) to T(12) ; I have put the solution of T(12) on 2007/11/22.
aire = area
and Challenge 001
If now we make, in the tray, a grid from the big triangle, we obtain :
With the previous notations, we can try cover T(r), T(2r), T(3r), T(4r), T(5r) and T(6r) ; here is for example a solution of T(2r) :
T(7r) has area 147 ; we have to take off area 3, as the area of T(r) : we obtain the triangle with hole of the tray, noted TT(7r/r), TT for "triangle troué" in french, that is "triangle with hole" :
On the tray, this hole is filled in by 3 little black triangles similar to one of the monospidrons (they are grey below) :
The position of these little triangles gives several challenges, which will appear time after time on the pages of the challenges ; here is, for example the challenge 028* :