572 THE FORMS OF TISSUES [ch. 



the manner in which they subdivide the base of the cube; in short 

 the problem of the sohd, up to a certain point, is contained in our 

 plane diagram of Fig. 221. And when our particular soUd is a 

 solid of revolution^ then it is equally obvious that a study of its 

 plane of symmetry (that is to say any plane passing through its 

 axis of rotation) gives us the solution of the whole problem. The 

 right cone is a case in point, for here the investigation of its modes 

 of symmetrical subdivision is completely met by an examination 

 of the isosceles triangle which constitutes its plane of symmetry. 



The bisection of an isosceles triangle by a Hne which shall be the 

 shortest possible is an easy problem; for it is obvious that, if the 

 triangle be low, a vertical partition will be shortest; if it be high, 

 a horizontal one ; if it be equilateral, the partition may run parallel 

 to any side; and if it be right-angled, the partition may bisect the 

 right angle or run parallel to either side equally well. 



Let ABC be an isosceles triangle of which A is the apex ; it may 

 be shewn that, for its shortest Hne of bisection, we are limited to 

 three cases: viz. to a vertical line AD, bisecting the angle at ^ and 

 the side BC\ to a transverse line parallel to the base BC\ or to an 

 obUque Hne paraUel to AB or to AC. The lengths of these partition 

 Hues follow at once from the magnitudes of the angles of our triangle. 

 We know, to begin with, since the areas of similar figures vary as 

 the squares of their linear dimensions, that, in order to bisect the 

 area, a line parallel to one side of our triangle must always have 

 a length equal to 1/a/2 of that side. If then, we take our base, 

 BC, in all cases of a length = 2, the transverse partition, EF, drawn 

 parallel to it wiU always have a length equal to 2/V2, or = a/2. 

 The vertical partition, AD, since BD = 1, will always equal tan^; 

 and the oblique partition, GH, being equal to ABjV^, = l/'v/2 cos j8. 

 If then we call our vertical, transverse and obUque partitions F, 

 T, and 0, we have F = tan j8; T == a/2; and - 1/V2 cos j8, or 



V:T:0 = tan^/V2 : 1 : 1/2 cos^S. 



And, working out these equations for various values of j8, we soon 

 see that the vertical partition (F) is the least of the three until 

 P = 45°, at which limit F and are each equal to 1/ V2 = 0-707 ; 

 that then becomes the least of the three, and remains so until 



