Calculate the coordinates of missing point in triangle given two points and two sides and one angle

Question

I have this issue and I've been trying to solve this for hours but wasn't able to find a solution which works in all circumstances. I feel sad because this seems like a simple trigonometry problem. I hope you can help me find a solution to this.

What I try to calculate are essentially the red points (I'll call them P later on) at the end of the red lines in this picture:

The red lines are just for reference and everything is drawn by hand and might therefor not be to scale.

I have a list of coordinates (black points) building a route (black line), for example:
[ (12, 9), (13,10), (13,11) ]

In order to find the red points I iterate through the route and look at 3 consecutive points at a time. Where the two edges connect (i.e. the middle point) is a place where the red line starts.

The rules are as follows:

• given are 3 points: A for after, B for before, C for current (shorthand for each triangle and their meaning in the route)
• given is j, the length of the red line
• the angle for BCP and ACP must be the same
• if multiple answers are possible, choose the one which 'goes with the flow' and is further away from the corner points for it to form a "Y" shape as indicated by the red line
• calculate the coordinates for each P as (y,x)

To illustrate the issue I made following diagram of just one such iteration:

(The y-axis on the grid is reversed in this one, but never mind)

What I already managed to do:

1. Given the points A,B and C, I can calculate the length of the edges as well as the angle ACB=135°. For this I used this function:

def angle_between(previous, current, next):
    a = math.sqrt((next[0]-previous[0])**2 + (next[1]-previous[1])**2)
    b = math.sqrt((current[0]-previous[0])**2 + (current[1]-previous[1])**2)
    c = math.sqrt((current[0]-next[0])**2 + (current[1]-next[1])**2)
    angle_alpha = math.acos((b**2 + c**2 - a**2)/(2*b*c))
    return round(math.degrees(angle_alpha), 2)

2. I then can calculate PCB=180-(ACB/2)=112.5°.

So I know everything I should need to know except the resulting coordinates of P.
I tried to calculate them with the following code but I didn't get the results I expected:
P = (C[0] + j*math.cos(PCB_rad) , C[1] + j*math.sin(PCB_rad))

Can someone please help me? Maths is not my strength it seems.. Thx!

Answer 1

Edit

• Here we post the explicit formula which can be used to another lanugage.

• According to the Angle bisector theorem， https://en.wikipedia.org/wiki/Angle_bisector_theorem

• we can calculate the point d on the ab edge and set dir=c-d as the direction vector from c to p

Clear[a,b,c,d,ca,cb,dir];
a = {xa, ya};
b = {xb, yb};
c = {xc, yc};
ca = Sqrt[(xc - xa)^2 + (yc - ya)^2];
cb = Sqrt[(xc - xb)^2 + (yc - yb)^2];
d =  cb/(ca + cb) *a  + ca/(ca + cb) *b;
dir = c - d;

{xc - (xb *Sqrt[(-xa + xc)^2 + (-ya + yc)^2])/( Sqrt[(-xa + xc)^2 + (-ya + yc)^2] + Sqrt[(-xb + xc)^2 + (-yb + yc)^2]) - ( xa* Sqrt[(-xb + xc)^2 + (-yb + yc)^2])/( Sqrt[(-xa + xc)^2 + (-ya + yc)^2] + Sqrt[(-xb + xc)^2 + (-yb + yc)^2]), yc - (yb *Sqrt[(-xa + xc)^2 + (-ya + yc)^2])/( Sqrt[(-xa + xc)^2 + (-ya + yc)^2] + Sqrt[(-xb + xc)^2 + (-yb + yc)^2]) - ( ya* Sqrt[(-xb + xc)^2 + (-yb + yc)^2])/( Sqrt[(-xa + xc)^2 + (-ya + yc)^2] + Sqrt[(-xb + xc)^2 + (-yb + yc)^2])}

• Now,p=c+length*Normalize[dir], it can be illustrated as below.

rules = Flatten[
   Thread /@ {{xa, ya} -> {1.2, 2}, {xb, yb} -> {3, 1}, {xc, 
       yc} -> {2.1, 1}}];
rules = Flatten[
   Thread /@ {{xa, ya} -> RandomReal[{-5, 5}, 2], {xb, yb} -> 
      RandomReal[{-5, 5}, 2], {xc, yc} -> RandomReal[{-5, 5}, 2]}];
Graphics[{EdgeForm[Blue], FaceForm[], 
   Triangle[{a, c, b}], {Line[{c, d}]}, 
   Line[{c, c + .5*Normalize[dir]}], Text["a", a, {-1, -1}], 
   Text["b", b, {-1, -1}], Text["c", c, {-.5, 1}], 
   Text["d", d, {-2, -1}], 
   Text["p", c + .5*Normalize[dir], {1, 1}]}] /. rules

Original

Clear[a,b,c,line,dir];
c = {2.1, 1};
a = {1.2, 2};
b = {3, 1};
line = AngleBisector[c -> {a, b}, "Interior"];
dir = line[[2]];
λ = 0.5;
Graphics[{Triangle[{a, b, c}], Red, Line[{c, c - λ*dir}]}]

Clear[λ, bisector, pts];
λ = .5;
bisector[a_, c_, b_] := Module[{line, dir},
   line = AngleBisector[c -> {a, b}, "Interior"];
   dir = line[[2]];
   Line[{c, c - λ*Normalize@dir}]];
pts = {{0, 0}, {3, 2}, {5, 1}, {6, 4}, {10, 3}, {11, 2}, {8, -2}, {7, 
    0}, {4, -2}};
Graphics[{Line[Partition[pts, 2, 1]], Red, 
  bisector @@@ Partition[pts, 3, 1]}]