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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
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