respect to the triangle of reference ABC has the form:

(1) m

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)) = 2rcos

and the equation n/p = z/y = cos

{m,n,p} correspondingly with {cos

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

x' = ax, y' = by, z' = cz

the previous equation obtains the form [Bottema, p. 81]:

(s-a)

or the norm equation in barycentrics:

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)) = 2rsin

2) Point B' on AC(y=0) with coordinates x=B'O, z=B'K.

x = B'O = r(1-cos(C)) = 2rsin

3) Point C' on AB(z=0) with coordinates x=C'N, y=C'M.

x=C'N=...= r(1-cos(B)) = 2rsin

Taking into account the location of points we have m/p=z/x=B'K/(-B'O) = -cos

n/m=x/y=(-C'N)/C'M=-sin

It follows (m: n: p) = (-cos

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

Bisector1.html

BarycentricCoordinates.html

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