Re: in center of a triangle

• To: mathgroup at smc.vnet.net
• Subject: [mg19628] Re: in center of a triangle
• From: phbrf at t-online.de (Peter Breitfeld)
• Date: Mon, 6 Sep 1999 04:20:43 -0400
• Organization: das ist ein breites Feld ...
• References: <7qqbur\$4gv@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Tom De Vries <tdevries at mail2.westworld.ca> schrieb:
> Hello!
>
> I was wondering if anyone has seen or has written a Mathematica  procedure to
> generate the incenter of a triangle given the vertices of the triangle.  The
> incenter of a triangle is the point at which the angle bisectors meet.  From
> this point you can draw an inscribed circle in the triangle.
>
>
> Tom De Vries
>
Maybe you look for something like this:

Winkelhalb[A_List,B_List,C_List]:=
Module[{Wa,Wb,Wc,W,u,v,s,t},
u=(A-C)/Sqrt[(A-C).(A-C)];
v=(B-C)/Sqrt[(B-C).(B-C)];
Wc=Flatten[A+(B-A)s/.Solve[C+(u+v)/2 t==A+(B-A)s,{s,t}]];
u=(A-B)/Sqrt[(A-B).(A-B)];
v=(C-B)/Sqrt[(C-B).(C-B)];
Wb=Flatten[A+(C-A)s/.Solve[B+(u+v)/2 t==A+(C-A)s,{s,t}]];
u=(C-A)/Sqrt[(C-A).(C-A)];
v=(B-A)/Sqrt[(B-A).(B-A)];
Wa=Flatten[B+(B-C)s/.Solve[A+(u+v)/2 t==B+(B-C)s,{s,t}]];
W=Flatten[A+t(Wa-A)/.Solve[A+t(Wa-A)==B+s(Wb-B),{s,t}]];
{W,Wa,Wb,Wc}
]

This function returns the incenter W of the triangel with vertices A, B
and C and the points Wa, Wb, Wc where the where the bisectors meet the
opposite sides of the triangle, so Wa ist the section point of the
bisector through A with BC.

Peter
--
=--=--=--=--=--=--=--=--=--=--=--=--=  http://home.t-online.de/home/phbrf  =--=
=--= Peter Breitfeld, Saulgau, Germany        PGP public key: 08548045  =--=--=

```

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