## 1. Incircle in trilinears

The prerequisites for this file are in Inconics_In_Trilinears.html .  There it is proved that the equation of an inconic with
respect to the triangle of reference ABC has the form:
(1)                                                m2x2 + n2y2 + p2z2 - 2mnxy - 2npyz - 2pmzx = 0.
The equation of the incircle in trilinears results by finding the constants {m,n,p}  of the equation in (4). This results from the
fact that the contact point A' with BC has known trilinear coordinates which satisfy
(3)                                                                                          ny - pz = 0,
and analogously for the other contact points: pz - mx = mx - ny = 0.

z = A'G = A'E + EG = r + rcos(IA'E) = r + rcos(B) = r(1+cos(B)) = 2rcos2(B/2). Analogous identities for y = A'H = 2rcos2(C/2)
and the equation  n/p = z/y = cos2(B/2)/cos2(C/2). Thus the equation of the incircle results from (4) by substituting
{m,n,p} correspondingly with {cos2(A/2), cos2(B/2), cos2(C/2)}, giving corresponding norm equation:

Another form of the equation results using the lengths instead of the angles of the triangle of reference. In fact, using the equation
z = A'G = BA'sin(B) and taking into account that BA' = s-b, where s = (a+b+c)/2 is half the perimeter of the triangle and {a,b,c}
denote the lengths of its sides (see Bisector1.html ). Since by the sinus theorem sin(B) = b/(2R), where R is the circumcircle radius,
we obtain finally that  n/p = z/y = ((s-b)b)/((s-c)/c)  and consequently the coefficients {m,n,p} in the form {(s-a)a, (s-b)b, (s-c)c}
and the norm equation in trilinears (equivalent to the previous one):

In barycentric coordinates (x',y',z') (see BarycentricCoordinates.html ), which are related to trilinears (x,y,z) through
x' = ax,        y' = by,         z' = cz
the previous equation obtains the form [Bottema, p. 81]:
(s-a)2x'2 + (s-b)2y'2 + (s-c)2z'2 - 2(s-a)(s-b)x'y' - 2(s-b)(s-c)y'z' - 2(s-c)(s-a)z'x' = 0,
or the norm equation in barycentrics:

## 2. Excircle in trilinears

The method is the same with the previous one, in determining the coefficients {m, n, p} from the equations
ny - pz = pz - mx = mx - ny = 0, defining the contact points with the sides of the triangle. In this case the contact points
are at known places. For example the excircle opposite to A.

1) Point A' on BC(x=0) with coordinates z = A'F, y = A'H.
z = A'G = IC'-IE = r- rcos(B) = r(1-cos(B)) = 2rsin2(B/2), y=A'H = 2rsin2(C/2) (analogously).
2) Point B' on AC(y=0) with coordinates x=B'O, z=B'K.
x = B'O = r(1-cos(C)) = 2rsin2(C/2), z = B'K = r+rsin(LIB') = r(1+cos(A))=2rcos2(A/2).
3) Point C' on AB(z=0) with coordinates x=C'N, y=C'M.
x=C'N=...= r(1-cos(B)) = 2rsin2(B/2), y=C'M=r(1+cos(A))= 2r(cos2(A/2)). (B'K=C'M).

Taking into account the location of points we have  m/p=z/x=B'K/(-B'O) = -cos2(A/2)/sin2(C/2).
n/m=x/y=(-C'N)/C'M=-sin2(B/2)/cos2(A/2). p/n = y/z = A'H/A'G = sin2(C/2)/sin2(B/2).
It follows (m: n: p) = (-cos2(A/2) : sin2(B/2) : sin2(C/2)) and the normed equation

An equivalent equation using the side-lengths of the triangle is deduced in a way analogous to the one for the incircle: