a Truncated Triangular Pyramid. 169 



p, q, and by the collocation of the symbols for three points 

 understanding the plane passing through them, it is clear 



1. That the six planes, 



/S,7; 7, a; a,b; y,a; u, b ; b, c y a, /3 ; ft c; c, a; 



y,/3; fta; a,c, u,y; y,b; b,a; fta; u,c; c, ft 



will meet in a single point which may be called the cross-centre, 

 being the true analogue of the intersection of the two diagonals 

 of a quadrilateral figure in the plane. 



2. That if we join this cross-centre (say Q) with the mid- 

 centre, and produce QO to G making OG = JQO, G will be the 

 centre of the frustum abcafiy. 



It may be satisfactory to some of my readears to have a direct 

 verification of the above. 

 Let, then, 



a*bc — a*Pv 



abc—a/3y ' 



B. 



ab^c—a^y 



c= 



abc^—afty 2 



abc—a/3y' abc—a/3y ' 



A moment's reflection will serve to show that A, B, C are the 

 coordinates of the centre of the frustum. 



Again, the first three of the six planes last referred to will be 

 found to have for their equations respectively, 



fiyx + yay + abz = 2ay{b -f ft , 

 bcx + yay + abz = 2ba(c +7), 

 (3 ex + cay + af3z = 2cfta + a) . 



The determinant 











The determinant 



fty 



be 

 /3c 



ya 



ya 

 ca 



ab 1 



ab =(abc—a/3y)' 2 



«13 \ 





ya 

 ya 

 ca 



ab 

 ab 

 a/3 



2ay{b + /3) 

 2ba(c + y) 

 2c/3(a + a) 





= 2a,a(bc— /3y)(abc—a/3y), 



= 2\{a 2 /3y-a%c){abc-a/3y) + (a + a){abc-a/3y) 2 }. 



Hence if x } y, z be the coordinates of the intersection of the 

 above-mentioned three planes, 



x=— 2A + 2(« + a), 



2/=-2B + 2(Z> + ft, 



*=-2C + 2(c + 7); 

 and the same will evidently be true of the other ternary system 

 of planes ; so that all six planes intersect in a single point Q, of 



