424 Dr. Eankine on Rational Approximations to the Circle. 



9. Proposition III. Problem. — Upon a given circle to construct 

 an inscribed polygon and the corresponding circumscribed polygon, so 

 that the perimeters of both shall be commensurable with the diameter 

 of the circle, and shall not differ from each other by more than a given 



fraction (— ) of the perimeter of the circumscribed polygon. 



First solution. — By Lemma I., find the sine, cosine, and 

 tangent of an angle (0) which shall be rational fractions, and 

 such that the difference between the tangent and sine shall not 



exceed — of the tangent. 



m ° 



Then by the ordinary rules (referred to in Lemma II.) com- 

 pute successively the sines of the multiples of that angle, sin 20, 

 sin 30, &c, until a sine is arrived at, say sin n0, less than sin 0*. 



Then the inscribed and circumscribed polygons will each have 

 n + 1 sides ; n of the sides of the inscribed polygon will be equal 

 to each other, and to the diameter of the circle x sin 0, and the 

 remaining side to the diameter x sin n0 ; and 2n of the half- 

 sides of the circumscribed polygon will be equal to each other, 



and to the diameter of the circle x t> , and the remaining 



two half-sides will be equal to each other, and to the diameter 

 tanf— n0) 



Thus the ratios of the perimeters of the inscribed and circum- 

 scribed polygons to the diameter of the circle will be respectively 



inscribed polygon • />•/» ,~\ 



T . V—Lb — =wsm0 + smw0, .'.--.' <5) 



diameter * ' 



circumscribed polygon Q ; , n . , a . 

 diamet J »»tong + t m (— tf).; ; . (6) 



both of which ratios are rational fractions ; and the difference be- 

 tween them is less than — of the greater of them. Q. E. P. 



m ° 



Second solution.— Proceed as before until sin n0 is found 

 less than sin#; then find sin (n-\-l) 0, which will also be less 

 than sin 0. The inscribed and circumscribed polygons will each 

 now have n + 2 sides ; n -f 1 of the sides of the inscribed polygon 



* The value of n might be found with the aid of trigonometrical tables, 



TV 



by taking the greatest whole number in the quotient -q\ but this would be 



a departure from the priociple of using none but the elementary processes 

 of arithmetic. 



