64 GEOMETRICAL RECREATIONS [CH. IV 



by putting them at points whose coordinates are (2, 0), (2, 2), 

 (2, 4), (4, 0), (4, 2), (4, 4), (0, 0), (3, 2), (6, 4); another way 

 is by putting them at the points (0, 0), (0, 2), (0, 4), (2, 1), 

 (2, 2), (2, 3), (4, 0), (4, 2), (4, 4) ; more generally, the angular 

 points of a regular hexagon and the three points (at infinity) of 

 intersection of opposite sides form such a group, and therefore 

 any projection of that figure will give a solution. At present it 

 is not possible to say what is the maximum number of rows of 

 three which can be formed from re counters placed on a plane. 



Extension to p-in-a-row. The problem mentioned above at 

 once suggests the extension of placing n counters so as to form 

 as many rows as possible, each of which shall contain p and only 

 p counters. Such problems can be often solved immediately by 

 placing at infinity the points of intersection of some of the lines, 

 and (if it is so desired) subsequently projecting the diagram 

 thus formed so as to bring these points to a finite distance. One 

 instance of such a solution is given above. 



As examples I may give the arrangement of 10 counters in 

 5 rows, each containing 4 counters; the arrangement of 16 

 counters in 15 rows, each containing 4 counters; the arrange- 

 ment of 18 counters in 9 rows, each containing 5 counters ; and 

 the arrangement of 19 counters in 10 rows, each containing 5 

 counters. These problems I leave to the ingenuity of my readers. 



Tesselation. Another of these statical recreations is known 

 as tesselation, and consists in the formation of geometrical 

 designs or mosaics covering a plane area by the use of tiles of 

 given geometrical forms. 



If the tiles are regular polygons, the resulting forms can be 

 found by analysis. For instance, if we confine ourselves to the 

 use of like tiles each of which is a regular polygon of n sides, we 

 are restricted to the use of equilateral triangles, squares, or hex- 

 agons. For suppose that to fill the space round a point where 

 one of the angles of the polygon is situated we require m poly- 

 gons. Each interior angle of the polygon is equal to (re— 2) irjn. 

 Hence m (re - 2) irjn - 2w. Therefore (m - 2) (re — 2) = 4. Now 

 from the nature of the problems m is greater than 2, and so 

 is re. If m = 3, re = 6. If m > 3, then re < 6, and since re > 2, 



