436 
ON A PARTICULAR CASE OF CASTILLON’s PROBLEM. 
[282 
Hence observing that the distance of one of the angles of the circumscribed triangle 
is sec ^ (/3 — 7), and that the projection of this line perpendicular to the corresponding 
side of the inscribed triangle is equal to sec £ (/3 — 7) cos (a' - a'), which is equal to the 
projection of the radius perpendicular to the same side, or to cos ^(6— c), we have 
cos (a — a') = cos £ (/3 — 7) cos ^ (b — c), 
cos (¡3' — b') = cos ^ (7 — a.) cos £ (c — a), 
cos (7' — c') = cos ^ (a — /3) cos ^ (a — b). 
Write 
b — c = 4f, b + c — 2a = 4iX, 
c — a = 4g, c + a —W=4>y, 
a — b = 4h, a + b — 2y = 4z ; 
therefore 
cos(y + 0 + g — h) = cos 2/cos(f+y— z), 
cos (z + x + h —f) = cos 2 g cos (g + z — x), 
cos (x + y + f— g) = cos 2h cos (h + x — y), 
or, since f+g + h=Q, 
cos {(y - h) + (z+ g)} = cos 2/cos {(y - h) - (z + g)), 
cos {{z —f) + (x + h)} = cos2g cos {(z—f) — (x+ /1)}, 
cos {(a-g) + (y +/)] = cos 2h cos {(x -g)-(y +/)}, 
or 
tan 2 / = tan (y — h) tan (.z + g), 
tan 2 g = tan (z —f) tan (x 4- h), 
tan 2 h = tan (x — g) tan (y+f ). 
Put 
x + h = g, x-g = £ li 
y +/= ?» y-h = Zi, 
z+g = £, z -/=%, 
tan f =1 , tan £ = x, tan = x 1 , 
tan g = m, tan g — y, tan g x = y lf 
tan h = n , tan £ = z, tan = z 2 . 
We have then 
tan 2 / = tan | tan £, 
tan 2 g = tan r; tan g x , 
tan 2 A = tan £ tan £i, 
or, what is the same thing, 
l~ = xXj, mr — yy l , n- = zzj.