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What an interesting observation i did not notice before, though i like
AltOrth chess and other twists with shape. It seems to me as being able to
open more paths of making curious chess interpretations.
The first idea that came in my head is to make a variant wich is
intermediate between Glinski and AltOrth hexchess - in same way as Betza's
Rectahex chess is intermediate between FIDE chess and my "Octagonal"
chess (hex of hex!)... But then i realized that such game would be
asymmetrical and ambiguous, just after noticing that difference between
numbers of directions of rook and forerook is odd.

I don't understand the difference between your two statements.  What
exactly do you conjecture to be true?  Why does 49 not give a
counterexample?

Do you mean something like multi-path?  Then the dababbah (thought of as a
1,1 diagonal-orthogonal leaper) fails, doesn't it?

 ___     ___     ___     ___     ___
/ . \___/ . \___/ . \___/ . \___/ # \
\___/ . \___/ . \___/ # \___/ . \___/
/ . \___/ . \___/ d \___/ . \___/ b \
\___/ . \___/ . \___/ . \___/ . \___/
/ . \___/ . \___/ . \___/ c \___/ b \
\___/ . \___/ . \___/ d \___/ . \___/
/ . \___/ . \___/ . \___/ . \___/ b \
\___/ # \___/ . \___/ . \___/ c \___/
/ . \___/ b \___/ . \___/ d \___/ b \
\___/ . \___/ b \___/ . \___/ . \___/
/ . \___/ . \___/ b \___/ . \___/ a \
\___/ . \___/ . \___/ b \___/ d \___/
/ # \___/ c \___/ c \___/ a \___/ a \
\___/ . \___/ . \___/ . \___/ a \___/
/ d \___/ d \___/ d \___/ d \___/ @ \
\___/ . \___/ . \___/ . \___/ . \___/

To clarify for other than Charles and Ben, unit 1 is centre to centre. 
Each regular hexagon has six equilateral triangles, and each height is 1/2,
so each hexagon side is 1/root-3 by 30-60-90 triangle. That makes each
orthogonal step in hexagons the 1.0 and each diagonal step 3/root-3 seen by
inspection and counting.

[Examples with even SOLL irrelevantly: (0,10) and (5,5) have same LL and SOLL 10 and 100; also (0,12) and (6,6); also (0,6) and (3,3)]

What would be second case with odd SOLL? Many, (2,3) of 31 SOLL for one.

So is your conjecture that in the hex-grid (or triangular grid depending on
how you look at it), there are no two right triangles that share a
hypotenuse with odd squared length?

EDIT: I should probably add "where the legs of the triangles are along the orthogonal and diagonal directions".

More edit: and to avoid the trivial swapping of the order of diagonal/orthogonal steps, make it "nonisomorphic triangles"

I think the 4-coloring of the hex board can help to prove this. See
wikipedia's 4-coloring image

If a leap has odd SOLL, then in any right angle diagonal-orthogonal path--say m diagonal and n orthogonal as before--we must have that m and n have different parities (else 3m^2+n^2 is even). Then the starting and landing cells have different colors in the 4-coloring.

But each orthogonal-diagonal pair of directions at right angles involve exactly two colors, one of which is the starting cell's color. So traveling along distinct orthogonal-diagonal directions lands at distinct color cells.

No, as I have already cited a counterexample to that conjecture if you include triangles of side zero. The orientation of the hypotenuse on the board matters as well as its length. I will have a think to see whether I can formulate the conjecture with purely numeric variables.

"I think the 4-coloring of the hex board can help to prove this. See wikipedia's 4-coloring image. If a leap has odd SOLL, then in any right angle diagonal-orthogonal path--say m diagonal and n orthogonal as before--we must have that m and n have different parities (else 3m^2+n^2 is even). Then the starting and landing cells have different colors in the 4-coloring. But each orthogonal-diagonal pair of directions at right angles involve exactly two colors, one of which is the starting cell's color. So traveling along distinct orthogonal-diagonal directions lands at distinct color cells."

I will have to think about this offline and reply at a later date.

Charles: which example?  the 49?  Those triangles do not share a hypotenuse (aha, when I say "share a hypotenuse", I mean that geometrically, not just that they have the same length).  And, the orientation of the hypotenuse is determined by the choice of right-angled orthogonal-diagonal pair.

(I may still be misunderstanding your question, so the rephrasing and proof may yet be incorrect.)

That is certainly part of what I was hoping someone could prove or disprove. It is self-evident for two orthogonals at right angles on a square-cell board, or even three on a cubic one. It is self-evident for a diagonal and an orhogonal with a 45° turn on a square-cell board. It is even self-evident for two orhogonals with a 60° turn on a hex board. It is however not only not self-evident for a diagonal and an orhogonal at right angles on a hex board, but untrue in the case of an even SOLL. Again taking m diagonal and orthogonal steps, m=1, n=1 gives the same destinations, not just leaps of the same length as by m=0, n=2. Likewise m=1, n=5 gives the same destinations as m=3, n=1. For odd-SOLL leaps I suspect from lack of counterexamples, and for prime-SOLL leaps I strongly suspect, that such duplicate vaules of m and n for the same destination are impossible, but cannot yet prove either. I have now checked all values of m up to 20 and n up to 40.

On a hex board, a dabbabah (a 2-square orthogonal leaper) can only reach
one-fourth of the spaces on the board.

Color a hex board in four colors, such that a dabbabah can only move to
hexes of the same color.

Claim: given two hexes, if there are two different paths of the type
Charles Gilman described between those hexes, then those hexes are the same
color. And every leap between two hexes of the same color is an even-SOLL
leap. Thus, Gilman is correct.

Proof left to the reader. :-D

I am now more certain that I understand what you're saying, and that my proof from before works.  (Joseph has the same idea.)  Let me give some more details:

Given a cell, an orthogonal direction, and a diagonal direction at a right
angle to the orthogonal direction, the set of cells reachable from the
given cell in those directions has a rectangular geometry (it is a proper
subset of the entire board, consisting of all cells of two colors in the
4-coloring).

So, given starting and destination cells, there are at most three ways to
write the leap as a combination of orthogonal and diagonal (at right
angles) steps.  These three ways correspond to the three orthogonal (or
diagonal) directions.  But the 4-coloring has the property that these three
rectangular sub-geometries (not using that as a technical term) each
consist of the color of the starting cell and one other color, and these
three other colors are distinct among the three sub-geometries.

Furthermore, the 4-coloring has the following property: the destination
cell's color is the same as the starting cell's color if and only if the
number of diagonal steps and the number of orthogonal steps have the same
parity.  (This can be seen by noting that the 4-coloring induces the usual 2-coloring on any of the rectangular subgeometries.)

So, if the SOLL is odd, the target cell is of different color than the
starting cell, and so there is only one of these subgeometry paths.
If the SOLL is even, then the target cell is of the same color as the
starting cell, and there are three distinct subgeometry paths (except for
SOLL=0).  These three are distinct in that they use different
ortho-diagonal subgeometries, but they may have the same number of
orthogonal and diagonal steps.

For example, m=1,n=5; m=2,n=4; m=3,n=1 all have SOLL 28 and have a common
destination cell.  (Perhaps this is the smallest with all three distinct?)

In the first case m must be even, in the second n must be even, and in the third m+n must be even. With an odd SOLL exactly one of these is the case and so only one of the three pairs of equations holds true. With an even SOLL all three are true and so all three pairs work, and not necessarily for the same values of m and n.

In the special case of 3m=n, p and q are equal and the first two pairs of equations are identical. In the special case of m=n, all equations except the second hold and q is zero and m and n both p/2. The missing equation may hold for different values of m and n if p and q are even.

I think a got a proof for the hex geometry.

We orient the hexes such that there is a horizontal line of rook movement,
and denote that direction by 1. The other directions of rook movement are
denoted by \omega and (\omega-1) [the use of the letter \omega is
inspired by Eisenstein numbers]. The centre of a hex is given by a+b\omega
with a,b integer numbers.

First step is a drawing: When we go horizontally firs and vertically as a
hex bishop second, we can reach only one half of the hexes (a+2b\omega).
We repeat this for the other rook directions and mark the hexes
accordingly. They fall in two classes: (i) hexes which can be reached in
one way only (ii) hexes that can be reached in all three way.

The second class forms a grid described by 2a+2b\omega (both coordinates
must be even.

Finally we map these to rook and bishop moves. The path to a three-way
reachable hex (2a+2b\omega) using horizontal and vertical moves
(elementary vertical bishop step: (2\omega -1)) consists of b bishop steps
and b+2a rook steps. Therefore the number of rook and bishop steps are both
odd or both even, giving an even SOLL.

The other direction: Take r rook steps and s bishop steps and demand that
r+s is even. Then we go to r+s*(2\omega -1) = (r-s) +2s\omega. This is a
three-way reachable square again, because (r+s) even implies (r-s) even.

Would the Rectahex board help or hinder visualizing Charles Gilman's
latest theorems?  http://www.chessvariants.org/diffmove.dir/rectahex.html.

We find: 'Hexagon -> Rectahex -> Square -> Triangle (all reversible)'!


Triangles: http://www.chessvariants.org/index/displaycomment.php?commentid=26710. 

 Squares: http://www.chessvariants.org/index/displaycomment.php?commentid=26708, 
http://www.chessvariants.org/index/displaycomment.php?commentid=26706.


This present topic of Chess Geometry is meaning Hexagonal chess geometry,
but Ralph Betza asks in 2003, "Is hexagonal chess really hexagonal or
merely a rectangular dream?" All the hex-board adjacencies are intact,
looking only at Squares once the Rectahex slide is performed.  There are
differences in some corners, but not edges which have all same adjacencies
cell-to-cell Hex and Rectahex.  However, LL and SOLL may be problematic
because of inconsistent perpendiculars from the skewing.
http://www.chessvariants.org/index/displaycomment.php?commentid=26724.

A-A-A-A-A-A-A-A
-B---C---C---B-
---D-------E---
-------F-------

  ---------------
 |       k       |
  ---------------
 |   q   |   c   |
  ---------------
 | r | n | n | r |
  ---------------
 |p|p|p|p|p|p|p|p|
 -----------------
| | | | | | | | | |
 -----------------
| | | | | | | | | |
 -----------------
 |P|P|P|P|P|P|P|P|
  ---------------
 | R | N | N | R |
  ---------------
 |   Q   |   C   |
  ---------------
 |       K       |
  ---------------

Has anyone got any good ideas for a name for this third pentagonal geometry?

I was thinking of something alike, while trying to figure out (not quite
succesful), how septagonal chess would look like. Game with progressive (or
wavy) number of cells per rank. Perhaps, even with cells, being
orthogonally-adjecent to three cells on next rank. While these three cells
could be orthogonally-adjecent to three further cells each (three different
for each (9), or external ones being shared by adjecent (7)), and so on.

Name? Progressively-shrinking pentagons, perhaps?

Pritchard mentions in CECV a game with a similar topologically-pentagonal
but geometrically-"doubling" board.  It is called the Fourth Dimension,
but has no 4d characteristics.  It is played on a round board with
concentric ranks of 4, 8, 16, and 32 cells, a generic cell having one
neighbor in the next rank inward, two neighbors on its own rank, and two
neighbors in the next rank outward.  The pieces don't seem to be
particularly chesslike.

(Maybe this round version could be realized as almost a pentagonal tiling
of the hyperbolic plane?)

CECV doesn't suggest whether the word Xyrixa has some meaning.  Since we
now know that this geometry can be viewed as the rhombic dodecahedron
tiling, perhaps a name based on that is appropriate.  A clever name
lacking, I'll use RD for the rest of this message.

On RD geometry: it's been noted that it can be viewed as the standard 3D
cubic tiling, but using the cubic diagonal as the RD orthogonal (and
restricting to one color of the cubic 2-coloring, say black).  Charles's
RD bishop direction seems to correspond to the cubic dabbabah.  Perhaps
another interesting direction is the cubic triagonal, which must skip over
one white cell.  This direction corresponds to passing through a degree-3
vertex of the RD.  (The RD orthogonal passes through the faces, and
Charles's RD bishop direction passes through the degree-4 vertices.)  I'm
not quite sure, but I think passing through the edges of the RDs would
correspond to a (2,1,1)-leap in the cubic tiling.  I haven't thought yet
about: do these directions have meaning in the hex-plane parts of RD
geometry?