60 Solution of a Problem of Col. Silas Til us. [Jan. 



equations (1), (2), (3), Wallls has substituted them in the place of 

 a, h, c, by which lie obtained the following result : — 



16 — a" + b c = 16,000,000,000,000,000 (52) 



IJ = h^ + ac= 17,000,000,000,000,000 (53) 



18 = c- + ab = 17,999,999,999,959,997 (54) 



However near Wallis's approximation may be, the equations 

 which he obtained are not for that reason the less absurd, nor are 

 they the less 



Equations, whose two meinhers are not equal and never can be- 

 come so. 



In order that the two members should be equal, it is necessary 

 that the inequality of the units compensate the inequality of the 

 numbers, as in the equation 2/. = 40.?. ; this compensation takes 

 place if each member and its unit are in an inverse ratio to each 

 other. This is the case in my solution, where the height of each 

 ring, mtiUipUed by the mean circumference between the two extreme 

 circumferences, givefor the product the constant area of the central 

 circle, as I thus demonstrate, for we have for each ring the follow- 

 ing proportion ; 



■K- The sum of the radii of the extreme circumferences of the rings 

 : To the tangent drawn to the smallest of these circuvferences 



:: This tangent, (which is = the radius of the central circle,) 

 : The difference of the radii of the extreme circumferences of the 

 rings. (This difference being the height of the ring, 185.) 



From wTience we have the following theorem. 



In a series of concentric rin^s, each of ivhose areas are equal to 

 the central circle. The rectangle formed by the sum of the radii of 

 the extreme circumferences of any one of the rings and its height, is 

 equal to the square of the central circles. (56.) Now the area of 

 each ring is equal the area of a trapezium, which has for its base 

 the height of the ring, and for the mean height half the sum of 

 the extreme circumferences. But this half is equal to a mean cir- 

 cumference between the two extreme circumferences. We can then 

 transform theorem (56) into the tollovving. 



In a series of concentric ri?igs whose areas are ecjiial to the area 

 of the central circle, the rectangle formed of the height of each 

 ring and the circu>nfcrcnce, luhich is a mean lelncen the tivo ex- 

 treme circumferences, is equal the area of the central circle, which 

 is the proposition I had to demonstrate. 



General Corollary. 



The preceding resolution of equations (I), (2), (3), gives a com- 

 plete solution of Gauss's problem, viz. " To divide a circumference 

 into 17 equal parts." The dotted isosceles triangle O I 1 J (fig- 1) 



17 16 17 



whose summit is tiie centre Q, has for its base the continued line 

 1 I 1 i it is this base whose extremities are 1 1 , which divides the 



17 iC 1: w 17 



