Sections on this page: — Knightly Triangles — Knightly Quadrangles — Intersections

Some other material relevant to the geometry of the knight's move will be found on the Theory of Moves and Theory of Journeys pages, which deal with properties of leapers in general that are not specific to the knight alone. This includes such properties as classification of angles between two moves. Ways in which the knight differs from other leapers are also considered there.

This article is based on Puzzle Question 32 in *Games and Puzzles Journal* (1999) issue 16 with Answer in
issue 17. We will prove or solve the following results and problems:

(a) Despite what may appear on some hand-drawn knight-tour diagrams it is impossible to have three knight-lines
concurrent,

except at the centre of a cell.

(b) If three knight-lines form a triangle, it is __always__ a 3:4:5 right-angled triangle, i.e. with sides
in the proportions 3 to 4 to 5 units.

(c) Taking a square of the board to be unit area, and numbering the triangles from the smallest area upwards,
the area of a size *k* triangle is *k*²/120.

(d) To find the area of the triangle formed by three successive knight moves.

(e) To draw a triangle of knight-lines whose area is as near as possible to a unit area.

The fact that knight moves make the same angles with each other as the angles in 3:4:5 Pythagorean triangles was noted by H. J. R. Murray (1942 ms p.3), in fact he evaluated the acute angles as 36° 52' 11.4" and 53° 7' 48.6".

He has a diagram of such a triangle (p.18) formed of 3, 4 and 5 knight moves, as shown here, which is sufficient to prove the point. Since a knight move has length Ö5 (taking the sides of the squares of the board as unit), this triangle has area ½ (base×height) = ½(4Ö5)(3Ö5) = 30 square units. To make this complete triangle the knight requires a 9×11 board, somewhat larger than its usual domain.

Any other knight lines drawn on this board will be parallel or perpendicular to one of these three lines.
In particular a line in the fourth direction, perpendicular to the hypotenuse, and passing through the right-angled
vertex splits the right angle into two parts equal to the above two acute angles, so any such lines will not create
any new angles. We can therefore say that __all__ triangles formed of lines of knight moves are 3:4:5 triangles.

[This is a type of **fractal** property of knight's tours: all triangles, on whatever scale of magnification,
are geometrically similar.]

Theorem: Proof: A set of close-spaced parallel knight-lines cuts a knight move crossing it into either 5, 4 or 3 equal parts, depending on the angle (see the inset diagram). Thus the distance between the two points where knight-lines at different angles cross another knight-line will be a multiple of sixtieths of a knight move, since the least common multiple of 3, 4 and 5 is 60. Denoting a 60th of a knight move by u = Ö5/60, the sides of the size k triangle are 3ku, 4ku, 5ku and the area ½ (base×height) is thus 6k²u². Inserting the value for u and a little arithmetic gives the required k²/120. [I communicated this result to Prof. D. E. Knuth in a letter of 4 December 1992, in response to his idea of a 'celtic tour', which he defined as a tour having 'no three lines nearly concurrent'; in other words no size-one triangles.] |

The diagrams below show all the possible sizes of knight triangles from k = 1 to 24. The long heavy lines are those of the most acute angle.

The answers to parts (d) and (e) of the question asked are:

(d) A triangle of three successive knight moves, being size 12, has area 6/5 = 1 + 1/5.

(See the bold cross-line in the third diagram.)

(e) The size 11 triangle is the closest to unit area, since 11²/120 = 121/120 = 1 + 1/120,

whereas the size 10 triangle has area 10²/120 = 100/120 = 1 − 1/6.

The size number k for the big 3:4:5 triangle shown above, calculated from k²/120 = 30, is k = 60, as might be expected. The number 60 has other resonances with knight moves. T. R. Dawson's manuscripts in the British Chess Problem Society's Library include a chart of 'The Unit of the Nightrider's Two-Move Domain 60×60' which shows that any square can be reached in one, two or three knight-rider moves.

This article is based on Puzzle Question 41 in *Games and Puzzles Journal* (1999) issue 17 with Answer
in issue 18. The puzzle was: Draw a quadrilateral of knight's moves which encloses a unit area (i.e. equal to
the area of a square of the board). A more general recreation is to construct knight's tours showing triangles
and squares of all possible sizes, alone or in combination. This question provides an excuse to present some
more results on geometry of knight moves. First note that the areas of the three types of rhomb formed by four
knight moves, known as diamond, lozenge and square, are 3, 4 and 5 units respectively, as is readily seen from
the diagrammed dissections.

Thus the lozenge provides a way of making knight quadrangles with area 1 unit, solving the stated problem. [The right-hand figure (not in the original article) shows the various different shapes and sizes of rhombs that can be formed by knight lines. The areas of lozenges possible are 1/4, 4/4, 9/4, 16/4, and so on. The areas of squares are 1/5, 4/5, 9/5, 16/5, 25/5 and so on. The areas of diamonds are 1/3, 4/3, 9/3 and so on. It is possible to form parallelograms of unit area from 3 smallest lozenges and 5 smallest squares and 4 smallest lozenges, but only the lozenges will form a unit rhomb (4 being a square number).]

The complete network of knight moves divides up a board (except near an edge) into five basic **irreducible**
shapes (irreducible in the sense that no knight move can cut across them).

These are: the smallest triangle (A), three sizes of kite (B, C, D) and an octagon (E). The areas of these are 1/120, 1/40, 1/15, 1/10, 1/6 respectively, all 'aliquot parts' of the unit square.

In terms of triangles of size k whose area was proved in the previous section to be k^2/120, each kite can be
seen to be the difference of two triangles:

B = T2 – T1 area (4 – 1)/120 = 1/40.

C = T3 – T1 area (9 – 1)/120 = 1/15.

D = T4 – T2 area (16 – 4)/120 = 1/10.

That the area of the octagon is 1/6 can be proved by noting that the square containing it, as proved earlier, is of area 1/5 = 24/120, from which four corners of size 1/120 are removed. In terms of 120ths the five areas are therefore 1, 3, 8, 12, 20.

The first part of this article is based on my piece in *Chessics*, issue 19, 1984.

A single knight's move can be intersected by other knight moves in nine different places, as shown here. However, the maximum number of intersections that can occur on one move of a knight's path is seven (since a move must be removed at each of the cells where three moves meet in this diagram).

If we consider the types of crossovers that are possible between a pair of knight moves, there are four
geometrically distinct cases. Working from the centre of the reference move outwards:

(a) central cross, both moves divided 1:1,

(b) right cross one move 2:3 and the other 1:5,

(c) diagonal cross, both moves divided 1:3,

(d) lateral cross, both moves divided 1:4.

The fifth intersection repeats the right cross with the roles of the two moves transposed.

T. R. Dawson constructed a number of short paths with intersection properties, as shown here.

(A1) is a shortest closed path, of 12 moves, with a maximum intersected move (*Sphinx at Play* December 1934).
I found another, symmetric, solution (A2). Dawson also constructed four examples of knight-move chains with single
intersection of every link, the 'Lover's Knot' problem: B1-B3 open paths (*British Chess Magazine* December
1930) and B4 closed path (*Evening Standard* 14 August 1933). He also found a circuit with double intersection
of every link C (1937, cited in Murray's 1942 ms).

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