212 Mr. J. J. Sylvester on Poncelet's approximate 



polygon upon the surface of the sphere. 

 Thus take X, Y, Z, each a quadrant apart 

 from the other, the points where the sur- 

 face of the sphere # 2 -f y 2 + # 2 =l is 

 pierced by the axis. If no limitation is 

 placed upon the values of x, y, z further 

 than the one throughout supposed of their 

 remaining always positive, the limiting 

 area will be XYZ. If we suppose 



we may take tan XK = k> and drawing the small c ircle KK', ZLM 

 will be the limiting area; if, again, Z <fc\/,r 2 -fy 2 , KK'YX will be 

 the limiting area ; if, again, Z < ks/x^-^y 1 , Z> lx, Z>my be the 

 limiting conditions, taking tan LX = /, tan MY = m, and drawing 

 LY, XM to intersect in 0, KK' MOL will be the corresponding 

 area, and so in general. Even so simple a set of conditions as 

 Z > Xy Z >■ y it is seen will give rise to a quadrilateral area, 

 limited in the figure by Z L M, when ZL = ZM = 45°. Thus, 

 then, we approach the preliminary question to which allusion 

 has been already made, which is to determine the least circle 

 that will cut off from a given sphere a segment containing all a 

 given system of points lying upon it. The solution is precisely 

 the same, substituting arcs of great circles for right lines, as the 

 problem of drawing upon a plane the least circle containing a 

 set of points given in the plane. 



We may, in the first place, obviously reject all those points 

 that are contained within the contour formed by arcs joining 

 the remaining points, so that the case of points lying at the 

 angles of a convex polygon alone remains to be studied. 

 Now if we confine our attention even to the simplest case of a 

 system of three points, we shall see at once that two cases arise. 

 If a circle be drawn through them, and these three points do not 

 lie in the same semicircle, no smaller circle than this can be 

 drawn to contain the three ; but if they do lie in the same semi- 

 circle, it is obvious that a circle described upon the line joining 

 the outer two as a diameter will be smaller than the circle pass- 

 ing through all three, and will contain them all. It was this 

 simple but striking fact in the geometry of situation which led 

 me to propose the question for any number of points in the 

 Quarterly Mathematical Journal ; and as Prof. Peirce's exhaust- 

 ive method of solution has not appeared in print, I may take 

 this occasion of presenting it. 



Let A, Z, B, C, D, E be the given points. Let A Z B be a 

 circle whose centre is drawn through A, Z, B, chosen so as to 

 include all the others ; then if A Z 13 are not contained in the 



