Re: Re: in center of a triangle

*To*: mathgroup at smc.vnet.net*Subject*: [mg52652] Re: [mg52643] Re: [mg19589] in center of a triangle*From*: DrBob <drbob at bigfoot.com>*Date*: Tue, 7 Dec 2004 04:09:38 -0500 (EST)*References*: <7qvtfb$4lg@smc.vnet.net> <200412050708.CAA26793@smc.vnet.net>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

>> I'll be appreciated much if you just reply with ... or how can I calculate it. That's exactly what David did. Bobby On Sun, 5 Dec 2004 02:08:42 -0500 (EST), Hesham AM <habdelmoez at yahoo.com> wrote: > Hi there: > > I'll be appreciated much if you just reply with the rectangular > coordinates of the incenter or how can I calculate it. I will use Vb. > > Many thanks ! > > Hesham AM > > > > On 6 Sep 1999 04:19:55 -0400, David Park wrote: >>> 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. >>> >>> Thanks for any help you could offer on this!! >>> >>> Tom De Vries >>> Edmonton, Alberta, Canada >>> >> >> Hi Tom, >> >> Here is a routine which will calculate the incenter and inradius for > an inscribed >> circle in a general triangle. >> >> InscribedCircle[ptA:{_, _}, ptB:{_, _}, ptC:{_, _}] := >> Module[{AB, BC, AC, a, b, c, s, ptP, ptQ, AP, BQ, p, q, psol, qsol, > pqsol, >> center, radius}, AB = ptB - ptA; BC = ptC - ptB; AC = ptC - ptA; >> a = Sqrt[BC . BC]; b = Sqrt[AC . AC]; c = Sqrt[AB . AB]; >> AP = ptB + p*BC - ptA; BQ = ptA + q*AC - ptB; >> psol = Solve[AP . AB/c == AP . AC/b, p][[1,1]]; >> qsol = Solve[BQ . BC/a == BQ . (-AB)/c, q][[1,1]]; >> ptP = ptB + p*BC /. psol; ptQ = ptA + q*AC /. qsol; >> pqsol = Solve[ptA + p*(ptP - ptA) == ptB + q*(ptQ - ptB), {p, > q}][[1]]; >> center = ptA + p*(ptP - ptA) /. pqsol; s = (a + b + c)/2; >> radius = Sqrt[((s - a)*(s - b)*(s - c))/s]; {center, radius}] >> >> This tests it on randomly generated triangles. >> << Graphics`Colors` >> ran := Random[Real, {0, 5}]; >> >> ptA = {ran, ran}; >> ptB = {ran, ran}; >> ptC = {ran, ran}; >> {incenter, inradius} = InscribedCircle[ptA, ptB, ptC]; >> Show[Graphics[ >> {AbsolutePointSize[4], Point /@ {ptA, ptB, ptC}, >> Blue, Line[{ptA, ptB, ptC, ptA}], >> OrangeRed, Circle[incenter, inradius], >> Black, Point[incenter]}], >> >> AspectRatio -> Automatic, PlotRange -> All, Background -> Linen, >> {Frame -> True, FrameLabel -> {x, y}}]; >> >> David Park >> djmp at earthlink.net >> <a > href="http://home.earthlink.net/~djmp/";>http://home.earthlink.net/~djmp/</a> > > > > -- DrBob at bigfoot.com www.eclecticdreams.net

**References**:**Re: in center of a triangle***From:*habdelmoez@yahoo.com (Hesham AM)