THE FIGURES OF EQUILIBRIUM OF A LIQUID MASS 265 



Let. us first demonstrate that the three centres f, d, and g, are in a right line. 

 For this purpose, repeating what has been done in Fig. 11, let us take on *c a 

 length 5w = * 6-' = />', and join c' w; we know that this last line will be paral- 



lei to s d, and consequently we may assume -z~- =^ ■ = — ,-. 1 or the same 



■^ ac siv [j' 



reason, in considering the two portions of circumferences which have for centres 

 c and c\ and which intersect each other at m, we shall have, by simply reversing 



fc" p" 

 the two ratios, - — == • ; and the two portions of circumferences having for 



y^' P 



centres c' and c", and which intersect at v, will also give -^-— - =—7—. Multi- 



• gc" i> 



dcfc"gc' 

 plying these three equalities, member by member, there results—— — ■* 



dc'J'c gc" 



Let it be remarked now, 1st, that the three centres, d, f, and g, are on the 

 prolongations of the sides of the triangle cc' c"; 2d, that the six quantities, dc, 

 fc" ,g c',dc',J'c,gc" diVG the distances, reckoned on these same prolongations, from 

 the three points d, f, and g, to the three summits c, c , and c ; od, that, in the last 

 of the above formulas, the three factors of the numerator represent right lines, 

 no two of .which have a common extremity, and that the same holds with regard 

 to the three factors of the denominator. Now we know, by a theorem of the 

 theory of transversals, that when the condition expressed by this formula is 

 fulfilled in regard to any triangle, the three points in question, taken on the pro- 

 longations of the three sides, are necessarily in q^ right line. Our three centres 

 d,J\ and g partake, theiefore, of this property. 



This first point established, let us demonstrate the others. Let us consider 

 the point o as bcang simply the intersection of the two arcs u o and v o, having 

 for centres f and g ; let us join o d, of, and o g, and without inquiring at first 

 whether o d i& really the radius of the arc having d for its centre, and proceed- 

 ing from the point s, let us show that the angles^'o d and g o d are each of 60'^, 

 or what amounts to the same, that the angle^ g is of 120°, and that the right 

 line o ^ is its bisector. 



Let us seek first to determine the lengths yd' and g d, and, to this end, let us 

 consider them as pertaining, respectively, to the triangles ^c d and g c' d, in 

 which we can calculate the sides f c, d c, g c', and d c', as well as the angles 

 which they comprise. To arrive at these last values, let us calculate the sides 

 of the triangle c c' c". By means of the triangle esc, in which the sides c s 

 andc' s are, respectively,/) and /s', and comprise between them an angle of 60°, 

 we find without diffic»lty c dz:z.y/ p^-{-p''^ — pp' ; the triangles c u c" and c' v c" 

 likewise give c c"^=^^ i?-\-p"'^ — pp", and c' c^rz:; V/Z^-f//'^ — p' p" . "Whence, by 

 the known formula, we deduce 



/ „ . . P'^+{P—P'){p—P") 

 cos c' cc"— cos dcf— ^ ^' '^ ')' ' ' ; 



2 ^/ p-i+pi^'—pp' ^f p^Jrp"'—pp" 

 in the same way we find 



p'^Hp-p'^'-p' 



cos cc c znzcos gc d 



'2y/ p^-\-p'^—pp' y/ pi-^-YpX^—p'pii ' 

 On another side we have 



fd— '^fc^+dP—2dcfc cos dcfimdi gd= ^1 dc" '^+^—2dc'.gc'. cos dc' g 

 formulas in which the lines dc,fc, do', gc' still remain to be determined ; but in 

 the triangle c- s d, in which we know that wc' is parallel to s d, we have 



dc dc' cc': 



