Unit 'steps' along both the X- and F-axes are to be of length V2, whereas unit 'steps' along the Z-axis are to be of length y/E/2.
With these conventions, taking X = a and Y = b (with a, b being integers) will serve to define the integer point P(a, b, c). Note that actually the Z-axis is redundant, since in TTO we have c = a + b;
it is often convenient to use it, however. Finally, on occasion we shall refer to the line through Q which is perpendicular to the Z-axis as the Z'—axis.
If we take the hexagon drawn on the nearest neighbours of Q, it is evident that we can draw six hexagons around it, on its sides; then continue drawing hexagons on the outer sides, and so on, until the whole plane is tessellated with hexagons.
It will follow that by this tessellation we partition the points of 7To into two sets, namely:
B = the set of points at the hexagon centres;
H = the set of points on all the hexagons.
We may call the points in B the B-points (Bees!) of ir0; and the points in H the .ff-points (Honey points).
Definition 3.2: The .ff-points and the lines parallel to the three axes which join them constitute the Fibonacci Honey- comb, with the B-point (2(0,0,0) being the Queen Bee. (see Fig. 2)
Before embarking on a study of Fibonacci vector sequences and other geometric figures on the Honeycomb, we shall state one or two useful the- orems about B- and .ff-points. A great deal more can be said about these sets of points, and lines joining them, than we have space for here.
Algorithm: Compute jaJ3 and I&I3. Then P is a B-point if and only if both these remainders modulo 3 are equal.
Proof: We can arrive at P in two moves, travelling from Q to P as follows:
Move (i): Move a hexagon-side lengths (i.e. a\/2) in the direction QQ2 (along the X-axis); move up if a is positive, and down if a is negative.
M o v e (ii): Move b hexagon-side lengths (i.e. by/2) in the direction QQ\ (parallel to the F-axis); move up if b is positive, and down if b is negative.
There are three possibilities to be considered, for move (i) followed by move (ii). They are:
First: If |a|3 = 0, then move (i) arrives at a B-point, after which move (ii) will lead to a B-point if and only if jfo|3 — 0, because of the honeycomb structure.
Second: If |a|3 = 1, then move (i) arrives at an B-point, after which move (ii) will lead to a B-point if and only if |?>|3 = 1, because of the honeycomb structure.
Third: If |o|3 = 2, then move (i) arrives at an B-point, after which move (ii) will lead to a B-point if and only if |6|3 = 2, because of the honeycomb structure.
Since these are the only possible ways in which a B-point can be arrived at, the Theorem is proved.
Evidently, all other points in 7To are B-points. • Corollary 1: It easily follows from the theorem that P is a B-point iff \a + c\3 = 0.
Corollary 2: The above algorithm shows that we can clas- sify the B-points into three types, as follows:
(Bo, iff \a\3 = 0 ; A B-point P(a, b,c) is of type < B\, iff \a\3 = 1 ; { B2 , iff \a\3 = 2 .
Corollary 3: The set of B-points in ir0 which are of type B0 form an Abelian group under vector addition; the identity element is the Queen Bee (i.e. <3(0,0,0)!).
Proof: Q is a point of type -Bo- Take any two points of type Bo and add them. In the resulting point, the value of
|a + c\s is zero; so by Corollary 1, the new point is also of type -Bo- Hence the set is closed under addition; and vector addition is commutative; so the set forms an Abelian group
under addition. • Corollary 4: Addition of B-point types can be defined in
the obvious way, since the addition of elements of two given types always leads to an element of the same type.
Below we give the addition table for S-point types. It shows that the three types form an Abelian group under addition, with the identity element being the type B0. This group is the cyclic group of order 3.
+
Bo Bx
B2
Bo Bo Bx
B2
Bi Bi
B2
Bo B2
B2
B0
B1
Much more can be said about points and lines in TTO, and their geometry, but space does not allow it. Triangles are studied in Chapter 5.
We shall close the chapter by stating and proving a theorem about line segments in the Fibonacci plane. The objective is to show that no integer- segment (that is, no segment whose endpoints are integer-vectors) in this plane has integer length.
Theorem 3.7: Let A = (01,02,03) and B = {bi,b2,bz), with all six coordinates being integers, be any two integer- points in the plane no- Denote by AB the line-segment join- ing them; and by \AB\ the length of this segment.
Then \AB\ = v R o i integral.
A )2 + ( a2- &2)2 + (a3-&3)2] is not Proof: Let A'B' and A"B" be the projections of AB onto QZ and QZ' respectively (see Fig. 3). They have lengths
ma, and nb respectively, where a = y/6/2, b = \ / 2 / 2 , and m,n are integers (the projections are from integer-points).
B" A"
Figure 3. Projections of AB onto the z-axes.
Hence: \AB\2 = m V + n2b2 = m2- + n2-.
4 2
Suppose that \AB\ is an integer d. Then we have to show that this is impossible; namely, that the equation 3m2 +n2 = 2d2 has no integer triple (m, n, d) solution.
We begin by assuming that there is an integer solution, with gcd(m,n,d) = 1, and then derive a contradiction.
First observe that the equation implies that either (i) both m and n are even, or (ii) both m and n are odd.
In case (i), set m = 2m' and n = 2n', and note that d must be odd, since gcd(m,n,d) = 1. The equation is then:
2d2 = 12m'2 + An'2 -> d is even . This is a contradiction, therefore (i) is impossible.
In case (ii), we shall prove that 3m2 = 2d2 — n2, with m, n both odd, has no integer solution.
The right-hand side must be divisible by 3.
Assume that 3 divides both d and n ; then 3 doesn't divide m, since gcd(m, n, d) = 1. Put d = 3r and n = 3s and the equation becomes: m2 = 6r2 — 3s2 which —> 3|m . This is a contradiction, hence the assumption is wrong.
Then, for 3 to divide evenly the right-hand side of the equation, the remainders modulo 3 of 2d2 and —n2 must add to 3. But it is easy to show that both 2d2 and —n2 are either 0 or 2 modulo 3; so it is impossible for their remainders to sum to three, when at least one of d or n is not a multiple of
3. •
Fibonacci and Lucas Vector Polygons
Recall that the Fibonacci vector sequence { Fn} is:
{ . . . , ( 2 , - 1 , 1 ) , ( - 1 , 1 , 0 ) , (1,0,1), (0,1,1), (1,1,2), (1,2,3), ( 2 , 3 , 5 ) , . . . } , with the general element being F„ = (Fn-i,Fn,Fn+i).
In similar fashion, we can write down the Lucas vector sequence { Ln} . Definition 4.1: If we join the 'points' of the Fibonacci vector sequence, plotted in the Honeycomb plane, by straight lines, we obtain a geometric figure which we shall call the Fibonacci Polygon. Similarly, we can join the points of the Lucas vector sequence, to obtain the Lucas Polygon. Fig. 2 (in Chap. 3) shows these two polygons drawn in the Hon- eycomb plane.