[alogo] 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.

[0_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:

[0_0]

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):

[0_0]

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:

[0_0]

[alogo] 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.

[0_0] [0_1] [0_2]

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

[0_0]

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

[0_0] [0_1] [0_2] [0_3] [0_4]

See Also

Inconics_In_Trilinears.html
Bisector1.html
BarycentricCoordinates.html

Bibliography

[Bottema] O. Bottema Topics in Elementary Geometry Springer Verlag Heidelberg 2007

Return to Gallery


Produced with EucliDraw©