September, 1913. 



KNOWLEDGE. 



339 



Method 3. Let A B (see Figure 365) be one side 

 of the required polygon. Bisect A B in M, and at 

 M draw M P perpendicular to A B and of indefinite 

 length. With centre M and radius M A (or M B) 

 describe an arc cutting M P in C. With centre A 

 and radius AC describe an arc cutting B A produced 

 in D. With centre M and radius M D describe an 

 arc cutting M P in 8. It can be easily shown that 

 this point is the centre of the circle circumscribing 

 the regular octagon drawn on A B as base {i.e., y 

 — 2-414 when * = 8). Trisect MB (or AM) and, 

 commencing from 8, mark off divisions 9, 10, 11, 

 12, and so on, along 8 P, and 7, 6, 5 along 8 M, each 

 equal to J M B. These points give the centres of the 

 circumscribing circles. The polygons are then 

 drawn as in the former methods. In Figure 365 are 

 shown polygons containing 7, 9, 12, and 14 sides 



thus obtained. The first three are very accurate; 

 the last is affected by a slight error. The equation 

 connecting x and y for this method can be obtained 

 thus : Let y = mx +- c. Then clearly 



tn -3, and 8m + c = 2-414; 

 whence c = —-253. 



Therefore y=-3x — 253. The values of y thus 

 calculated are given in Column 8 of Table 59, and 

 the differences between them and the true values of 

 y are shown in Column 9. In every case after x=6, 

 these differences are less than with Method 1 ; but 

 after x=10 they are greater than with Method 2. 

 For polygons containing 7, 8, 9, or 10 sides this 

 method is the most accurate. The graph of the 

 equation y=-3x— -253 cannot be shown distinct 

 from that of _y=-3.r — 268 in Figure 367, since the 

 two lines practically coincide. 



Table 59. 

 6 



10 



11 



