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$ \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}(x_{A}-x_{B})b + (x_{C}-x_{A})c & (y_{A}-y_{B})b + (y_{C} - y_{A})c\\(x_{B}-x_{A})a+(x_{C}-x_{B})c & (y_{C}-y_{B})a+(y_{C}-y_{B})c\end{bmatrix}^{-1} \cdot \begin{bmatrix}x_{A}((x_{A}-x_{B})b + (x_{C}-x_{A})c)+y_{A}((y_{A}-y_{B})b + (y_{C} - y_{A})c)\\x_{B}((x_{B}-x_{A})a+(x_{C}-x_{B})c) + y_{B}((y_{C}-y_{B})a+(y_{C}-y_{B})c)\end{bmatrix} $

What we know about $x_{A}, y_{A}, x_{B}, y_{B},x_{C},y_{C}$ ?
They are coordinates of distinct and non colinear points
What we know about $a,b,c$

$ a^2 = (x_{C}-x_{B})^2+(y_{C}-y_{B})^2\\ b^2 = (x_{C}-x_{A})^2+(y_{C}-y_{A})^2\\ c^2 = (x_{B}-x_{A})^2+(y_{B}-y_{A})^2 $

I tried to use Python with sympy package to simplify this expression but it doesnt work I tried to use programs like Maple to simplify it but Maple only factored the denominator and failed to simplify it
I know that this expression can be simpified but I dont know how

J Doe
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  • Hint: the inverse of $2 \times 2$ matrix $\pmatrix{a&b\c&d}$ is $\frac{1}{ad-bc}\pmatrix{d&-b\-c&a}$ – Jean Marie Nov 20 '22 at 08:56
  • Wouldn't it be simpler that you explain what is the initial problem, a triangle for which we want what exactly ? – Jean Marie Nov 20 '22 at 08:58
  • Yes I know but after calculation inverse and then x and y i will get fractions which can be simplified but i dont know how Programs which i mentioned dont help – J Doe Nov 20 '22 at 09:00
  • Pleas explain the objective. What do you want in this triangle ? The coordinates of a special point like the orthocenter ? – Jean Marie Nov 20 '22 at 09:03
  • It will take me a while to write what i have got so far – J Doe Nov 20 '22 at 09:18

2 Answers2

1

There is a simpler approach.

With (your) notations $a=BC, b=CA, c=AB$ for the sidelengths of triangle $ABC$, its incenter $I$ can be expressed in the following way (known as "barycentric coordinates" representation ; you will find explanations for example in the second answer to this MathSE question):

$$I=k(aA+bB+cC) \ \text{with} \ k:=\frac{1}{a+b+c}$$

which can be explicited with cartesian coordinates in this way:

$$\pmatrix{x_I\\y_I}=ka\pmatrix{x_A\\y_A}+kb\pmatrix{x_B\\y_B}+kc\pmatrix{x_C\\y_C}$$

Jean Marie
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  • Maybe but suppose we dont have introduced barycentric coordinates Angle bisector approach should also simplfy to this result but i dont see how – J Doe Nov 20 '22 at 13:14
  • I like your "Maybe"... – Jean Marie Nov 20 '22 at 13:48
  • It is because I know nothing about barycentric coordinates When I derived equations for angle bisectors I remembered only construction and I observed that after drawing first arc later construction looks like constructon of perpendicular line – J Doe Nov 20 '22 at 14:48
  • Being a little experienced in different forms of geometry, I think that in particular for triangle geometry problems, you have to use the right tool(s) for each problem. Otherwise you loose a lot of time without benefit. Here, taking, right from the beginning, cartesian coordinates isn't the good approach. Barycentric coordinates is often a good tool, but you have also inversion, geometry with complex numbers, more or less mixed with trigonometry, etc. – Jean Marie Nov 20 '22 at 15:34
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    I found correct equations for angle bisectors because if I substitute point you gave it satisfies this system of equations – J Doe Nov 20 '22 at 18:18
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Initial problem was
Calculate coordinates of center of inscribed circle in triangle with given coordinates of vertices

I found equations of angle bisectors (two of this equations I wrote here)

With help of programs like Maple i have got following results

$ x=\frac{((ay_{B}-ay_{A}+cy_{C}-cy_{B})((x_{A}^2-x_{A}x_{B}+y_{A}^2-y_{A}y_{B})b+(x_{A}x_{C}-x_{A}^2+y_{A}y_{C}-y_{A}^2)c)+(-by_{A}+by_{B}-cy_{C}+cy_{A})((x_{B}^2-x_{A}x_{B}+y_{B}^2-y_{A}y_{B})a+(x_{B}x_{C}-x_{B}^2+y_{B}y_{C}-y_{B}^2)c))}{(c(x_{C}y_{B}-x_{C}y_{A}-x_{A}y_{B}-y_{C}x_{B}+y_{C}x_{A}+y_{A}x_{B})(b-c+a))}\\ y = \frac{((-ax_{B}+ax_{A}-cx_{C}+cx_{B})((x_{A}^2-x_{A}x_{B}+y_{A}^2-y_{A}y_{B})b+(x_{A}x_{C}-x_{A}^2+y_{A}y_{C}-y_{A}^2)c)+(bx_{A}-bx_{B}+cx_{C}-cx_{A})((x_{B}^2-x_{A}x_{B}+y_{B}^2-y_{A}y_{B})a+(x_{B}x_{C}-x_{B}^2+y_{B}y_{C}-y_{B}^2)c))}{(c(x_{C}y_{B}-x_{C}y_{A}-x_{A}y_{B}-y_{C}x_{B}+y_{C}x_{A}+y_{A}x_{B})(b-c+a))} $
But this can be simplified even more

When I draw equations of angle bisectors in Geogebra it looks lie they are written correctly but simplification of incenter coordinates is so complicated

Here is Python code for angle blsector approach

from sympy import *
xA,yA,xB,yB,xC,yC = symbols('xA yA xB yB xC yC')
a,b,c = symbols('a b c')
x,y = symbols('x y')
print('Lets find equation of bisector of angle A')
print('Equation of lines in which rays of angle are included')
eq1 = (yB-yA)*x-(xB-xA)*y-xB*(yB-yA)+yB*(yB-yA)
eq2 = (yC-yA)*x - (xC-xA)*y+yC*(xC-xA)-xC*(yC-yA)
print(eq1,'=0')
print(eq2,'=0')
print('Lets choose point B as point D')
print('Equation of circle with center at point A and radius AD')
eq1 = (x-xA)**2+(y-yA)**2-c**2
print(eq1,'=0')
eq2 = (yC-yA)*x - (xC-xA)*y+yC*(xC-xA)-xC*(yC-yA)
print('Solution of system of equations with circle and line')
print(solve((eq1,eq2),x,y))
pE = []
for p in solve((eq1,eq2),x,y):
    for q in p:
       q = q.subs(sqrt(xA**2 - 2*xA*xC + xC**2 + yA**2 - 2*yA*yC + yC**2),b)
       pE.append(simplify(q))
print('Possible points E')
print(pE)

(y-yA)=-(xE-xD)/(yE-yD)(x-xA)


(yE - yD)(y-yA)=-(xE-xD)(x-xA)


(xE-xD)(x-xA)+(yE - yD)(y-yA)


print('Lets write equation of line perpendicular to DE and passing through point A')
print('and point E is the first point found by solving system of equations above')


bA1 = (pE[0]-xB)x+(pE[1] - yB)y - xA(pE[0]-xB) - yA(pE[1] - yB)


print('Equation of perpendicular line is :',bA1,'=0')


(y-yA) = (yE-yD)/(xE-xD)(x-xA)


(yE-yD)(x-xA) = (xE-xD)(y-yA)


(yE-yD)(x-xA)-(xE-xD)(y-yA)


print('Lets write equation of line parallel to DE and passing through point A')
print('and point E is the second point found by solving system of equations above')


bA2 = (pE[3]-yB)x-(pE[2]-xB)y -xA(pE[3]-yB)+yA(pE[2]-xB)
print('Equation of parallel line is :',bA2,'=0')


print('bisector of A :',simplify(bA1),'=0')
print('bisector of A :',simplify(bA2),'=0')


print('Lets find equation of bisector of angle B')
print('Equation of lines in which rays of angle are included')
eq1 = (yB-yA)x-(xB-xA)y-xB(yB-yA)+yB(xB-xA)
eq2 = (yC-yB)x-(xC-xB)y+yC(xC-xB)-xC(yC-yB)
print(eq1,'=0')
print(eq2,'=0')
print('Lets choose point B as point D')
print('Equation of circle with center at point B and radius BD')
eq1 = (x - xB)2+(y-yB)2-c*2
print(eq1,'=0')
eq2 = (yC-yB)x-(xC-xB)y+yC(xC-xB)-xC*(yC-yB)


print('Solution of system of equations with circle and line')
print(solve((eq1,eq2),x,y))


pE = []
for p in solve((eq1,eq2),x,y):
    for q in p:
        q = q.subs(sqrt(xB2 - 2xBxC + xC2 + yB2 - 2yByC + yC2),a)
        pE.append(simplify(q))
print('Possible points E')
print(pE)


print('Lets write equation of line perpendicular to DE and passing through point B')
print('and point E is the first point found by solving system of equations above')


bB1 = (pE[0] - xA)x+(pE[1] - yA)y-xB(pE[0] - xA)-yB(pE[1] - yA)
print('Equation of perpendicular line is :',bB1,'=0')
print('Lets write equation of line parallel to DE and passing through point B')
print('and point E is the second point found by solving system of equations above')
bB2 = (pE[3] - yA)x-(pE[2] - xA)y-xB(pE[3] - yA)+yB(pE[2] - xA)
print('Equation of parallel line is :',bB2,'=0')


print('bisector of B :',simplify(bB1),'=0')
print('bisector of B :',simplify(bB2),'=0')


print('Coordinates of center of inscribed circle version 1')
print(solve((bA1,bB1),x,y))
print('Coordinates of center of inscribed circle version 2')
print(solve((bA2,bB2),x,y))
J Doe
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