1824.] connected with the Trisection of an Arc. 359 



Prop. II. Problem. (Fig. 3.) 



In a given circle to inscribe a triangle, whose sides shall be in 

 arithmetical progression, and their common difference equal to 

 a given line. 



The solution of this problem also is obvious, from what we 

 have established in solving the former. For there we have seen 

 that A * is the common difference of the sides, and (Cor. 2) that 

 E N is equal to radius. Therefore, given the radius and common 

 difference of the sides, we can construct the right-angled triangle 

 AEN, having one of its sides inscribed in the circle, its hy- 

 potenuse E N equal to radius, and its altitude A z equal to the 

 given common difference. This line A z produced gives the 

 greatest side A a ; whence we have the other two. 



Scholium. — Whenever the given common difference is less than 

 half the radius, we may find two different triangles to solve the 

 problem ; for either of the unequal sides, A E or A N, may be 

 inscribed in the circle. It is plain that A z is a maximum, when 

 equal to half E N ; that is, that the semiradius is the greatest 

 possible common difference of the sides. 



Putting unity for radius, the triangle of whose sides the com- 

 mon difference is a maximum, is 



V i + i y i yi-\ 



2 ' 2 > 2 



the cosine of whose middle angle is -J. And it may here be 

 briefly remarked, that in the two different triangles, whose sides 

 have any smaller common difference, half the sum of the cosines 

 of the middle angles is £. Also, the sum of the two least angles 

 is the supplement of the sum of the two greatest ; and is the 

 third part of the difference between the sum of the greatest and 

 the sum of the middle angles : just as in the triangle 



V 7 + 1 -y/7 VI - 1 

 2 ' 2 * 2 ' 



the least angle is the complement of the greatest, and is third 

 part of the difference between the other two. 



Many other and curious properties of these triangles might be 

 6tated ; but we must hasten at present to other matter. 



Prop. III. (Fig. 5.) 



If on the same base A a there be two isosceles triangles, 

 whose vertical angles ABa, A E a are as three to two ; and 

 from the vertex of the smaller angle lines be drawn (E T, E t) 

 trisecting the common base; these lines will trisect (Aba) the 

 are of the circumscribing circle on which the greater angle 

 stands. 



Dem. — Produce the altitude E o, and let o e = o E. Draw 

 e A, and produce it to f. E T trisecting the altitude of the 

 isosceles triangle E A e must bisect the leg A e in 1). Draw D d 

 parallel to e L; also the radius A c, cutting E T in m, and D d 



