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IN RELATION TO TWO TETRAHEDRA. 
203 
(P, G) is U + Lx O = 0, and similarly those of circles through (C, A) and through 
(A, B) are U + My 0 = 0 and U + Nz 0 = 0 respectively. But we require the interpre 
tation of the coefficients L, M, N which enter into these equations. 
3. Considering the triangle ABC, if through B, G we have a circle, this is by 
the side BG divided into two segments, and I consider that lying on the same side 
with A as the positive segment, and define the angle of the circle to be the angle 
in this positive segment. It is clear that if we have within the triangle a point P, 
and, through this point and (B, C), (G, A), (A, B) respectively, three circles, then if 
a, A, 7 be the angles of these circles, we have a + A + 7 = 27r; and conversely, if the 
circles through (B, G), (G, A), (A, B) are such that their angles a, fi, y satisfy the 
relation a + A + 7 = 27r, then the three circles meet in a point. But it is further to 
be noticed, that if, producing the sides of the triangle so as to divide the plane into 
seven spaces, the triangle, three trilaterals, and three bilaterals, we take the point P 
within one of the bilaterals, we still have a + /3 + 7 = 27r; but taking it within one 
of the trilaterals, we have a + A + 7 = 7r. And the converse theorem is, that if the 
three circles (P, G), (G, A), (A, B) are such that a + /3 + y= tt or 27t, then the circles 
meet in a point; viz. if the sum is 27t, then this point lies in the triangle or one 
of the bilaterals; but if the sum is = 7r, then this point lies in a trilateral. 
4. I seek for the equation of a circle through the points B, G, and containing 
the angle L. The equation in rectangular coordinates is easily seen to be 
(X - a 2 ) (X - a s ) + (Y—/3 2 )(Y— fi 3 ) - cot L {(& -¡3 3 )X- (a 2 - a 3 ) Y + a 2 /3 3 - a 3 /3 2 | = 0. 
In fact this is the equation of a circle through (P, C); and taking for a moment the 
origin at B, and axis of X to coincide with BG, or writing a 2 , (3 2 = 0, 0; a 3 , /3 3 = a, 0, 
the equation is 
X (X — a) + Y 2 — aY cot P = 0, 
viz. the equation of the tangent at B is — aX — aY cot L = 0, that is, F = — X tan L, 
or the angle in the positive segment is = L. 
If for a moment X, ¡a, v are the inclinations of the sides of the triangle ABG 
to the axis of X, then A, B, G being the angles, we may write 
fi — v = 7t — A, 
v — A = 7r — B, 
X — /A = — 7T — G, 
and 
X — oc 2 = {a x x + a 2 y + a 3 z) — (x + y + z) = c cos v . x — a cos X . z, 
X — a 3 = {a x x + OL 2 y + or 3 z) — a 3 (x + y + z) = — b cos /x . x + a cos X . y, 
Y - /3 2 = fta? + - /3. 2 (x + y + z) = c sin v . x - a sin X . z, 
Y — fis = fiiX + /3 2 y + ft 3 z — fi 3 (x + y + z) = — b sin v . x + a sin X. z ; 
whence 
(Z - a 2 ) (X - a 3 ) + (F-&) (F - A) 
= — a 2 yz — b 2 zx — c 2 xy + be cos A . x 2 + (b 2 — ab cos G) zx + (c 2 — ac cos B) xy ; 
26—2