To inscribe a circle in a given triangle

Question

I am trying to to inscribe a circle in a given triangle but it isn't working. I've used GeoGebra with this construction and worked but as I'm new to Mathematica, I am missing something. It can't be so complex to do. Here is my code:

pA = {0, 0}; pB = {1.1, 1}; pC = {1.5, 0};

midAB = (pA + pB)/2; midBC = (pB + pC)/2; midAC = (pA + pC)/2;

myT = Triangle[{pA, pB, pC}];

centerC = (midAB + midBC + midAC)/3

raioC = Last[centerC]

Graphics[{EdgeForm[Directive[Thick, Blue]], White, 
  myT, {PointSize[Large], Red, Point[midBC]}, {PointSize[Large], Red, 
   Point[pA]},
  {Orange, Line[{pA, midBC}]},
  {Orange, Line[{pB, midAC}]},
  {Orange, Line[{pC, midAB}]},
  {Red, Circle[centerC, Last[centerC]]},
  }, AspectRatio -> Automatic, PlotRange -> All]

I understand that I'm not doing right the radius of the circle but I've tried more than one way to do that.

Here is the original (book) construction:

To construct: The circle inscribed in $\triangle ABC$

Construction: Construct the bisectors of two of the angles of $\triangle ABC$. Their intersection is the center of the required circle, and the distance (perpendicular) to any side is the radius. (Any point on the bisector of an angle is equidistant from the sides of the angle.)

As I said, it worked in GeoGebra and it is almost working with my code. Thank you for any help.

Answer 1

As KennyColnago has observed your equations seem to be off but there is another way of determining the incircle. I'd already typed it up with sources that may be helpful so let's post it, even if it's the same thing.

{a, b, c} = {Norm[pC - pB], Norm[pA - pC], Norm[pA - pB]};
area = With[{s = (a + b + c)/2}, Sqrt[s (s - a) (s - b) (s - c)]];
r = 2 area/(a + b + c);
incenter = (a pA + b pB + c pC)/(a + b + c);

Graphics[{
  EdgeForm[Black], White, Triangle[{pA, pB, pC}],
  Black, Point[incenter],
  Circle[incenter, r]
  }]

The formula for the coordinates of the incenter can be found here, the relationship between the radius and the area is discussed here and the area is calculated with Heron's formula found here.

It's also worth noting that we can "cheat" with Mathematica 10:

incenter = {a, b, c}.{pA, pB, pC}/ArcLength[Line[{pA, pB, pC, pA}]];
r = 2 Area[Triangle[{pA, pB, pC}]]/ArcLength[Line[{pA, pB, pC, pA}]];

Although this sort of cheating is not as outrageous as that by Junho Lee of course ;)

Answer 2

version 10

pA = {0, 0}; pB = {1.1, 1}; pC = {1.5, 0};
lr = MeshRegion[{pA, pB, pC}, Triangle[Range@3]];
r1 = RegionDistance[InfiniteLine[{pA, pB}], {x, y}];
r2 = RegionDistance[InfiniteLine[{pB, pC}], {x, y}];
r3 = RegionDistance[InfiniteLine[{pC, pA}], {x, y}];
centerC = {x, y} /. NSolve[{r1 == r2, r2 == r3}, {x, y}, Reals];
centerC = Select[centerC, RegionMember[lr, #] &][[1]]
raioC = RegionDistance[Line[{pA, pB}], centerC]
Graphics[{
  EdgeForm[{Thick, Blue}], White, Triangle[{pA, pB, pC}],
  PointSize[Large], Red, Point@centerC,
  Red, Circle[centerC, raioC]}
 ]

{0.954787, 0.369127}

0.369127

Answer 3

The incircle of a triangle may be calculated as follows.

InCircle[{x1_, y1_}, {x2_, y2_}, {x3_, y3_}] := 
   With[{
      a = Norm[{x2,y2} - {x3,y3}], 
      b = Norm[{x3,y3} - {x1,y1}], 
      c = Norm[{x1,y1} - {x2,y2}]}, 
   Circle[(a {x1, y1} + b {x2, y2} + c {x3, y3})/(a + b + c), 
          1/2 Sqrt[-(((a - b - c) (a + b - c) (a - b + c))/(a + b + c))]]]

For your example points, InCircle[{0,0},{1.1,1},{1.5,0}] produces Circle[{0.954787, 0.369127}, 0.369127], which may be plotted with

Graphics[{
   Line[{{0, 0}, {1.1, 1}, {1.5, 0}, {0, 0}}],
   Red, Thick, InCircle[{0, 0}, {1.1, 1}, {1.5, 0}]}]

Answer 4

inCircle[p_] := 
  Circle[{x, y}, Abs@r] /. 
   Solve[((# - #2).{y, -x} - Det@{##})/Norm[# - #2] & @@@ 
      Subsets[p, {2}] == r {1, -1, 1}];

Manipulate[
 Graphics[{Line@p[[{1, 2, 3, 1}]], Red, inCircle@p}, PlotRange -> 6, 
  Frame -> 1], {{p, {{-2, -1}, {3, -2}, {0, 3}}}, Locator}]

Another inCircle:

inCircle[pt : {v1_, v2_, v3_}] :=
 Block[{a, b, c, p},
  {a, b, c} = Norm /@ {v2 - v3, v3 - v1, v1 - v2};
  p = (a + b + c)/2;
  Circle[{a, b, c}.pt/(2 p),  Sqrt[p (p - a) (p - b) (p - c)]/p]]

Answer 5

Although Insphere[] is now built-in, I would still like to present an alternative technique, used in this previous answer. The idea, due to Miroslav Fiedler, is to use the inverse of the Cayley-Menger matrix to determine the inradius and incenter. (Some more algebraic effort will yield the circumradius and circumcenter as well.)

The routine CayleyMengerMatrix[] in the previous answer can now be slightly improved with the new function DistanceMatrix[]:

CayleyMengerMatrix[pts_?MatrixQ] /; Subtract @@ Dimensions[pts] == 1 :=
     ArrayFlatten[{{0, 1},
                   {1, DistanceMatrix[pts, DistanceFunction -> SquaredEuclideanDistance]}}]

CircumInSphere[pts_?MatrixQ] /; Subtract @@ Dimensions[pts] == 1 := 
       Module[{cv, icm, iv, rc},
              icm = -2 Inverse[CayleyMengerMatrix[pts]]; cv = icm[[1, 2 ;;]];
              {rc, iv} = Through[{First, Rest}[Sqrt[Tr[icm, List]]]];
              {{cv.pts/Total[cv], rc/2}, {iv.pts, 1}/Total[iv]}]

Now, generate the incircle:

tri = {{0, 0}, {11/10, 1}, {3/2, 0}};
is = Circle @@ Last[CircumInSphere[tri]] // FullSimplify
   Circle[{1/20 (15 - 2 Sqrt[29] + Sqrt[221]), 15/(15 + 2 Sqrt[29] + Sqrt[221])},
          15/(15 + 2 Sqrt[29] + Sqrt[221])]

Graphics[{Directive[{EdgeForm[Black], FaceForm[]}], Polygon[tri], is}]

---

Displaying the angle bisectors is not too hard:

bisectors = With[{ab = 1.5 (* arbitrary scaling *)
                  Normalize[Total[Map[Composition[Normalize,
                                                  TranslationTransform[-#[[2]]]], 
                                      Delete[#, 2]]]]},
                 Arrow[{#[[2]], #[[2]] + ab}]] & /@ 
            Partition[tri, 3, 1, 1];

Graphics[{Directive[{EdgeForm[Black], FaceForm[]}], Polygon[tri], is, bisectors}]

Answer 6

(For completeness, an answer using Insphere)

As mentioned by @JM, in M10.2+ you can use Insphere:

pA = {0, 0}; pB = {1.1, 1}; pC = {1.5, 0};

Insphere[{pA, pB, pC}]

Sphere[{0.954787, 0.369127}, 0.369127]

in agreement with the other answers.