I asked this question on MSE [here][1].


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Given a non-regular pentagon $A_1B_1C_1D_1E_1$ with no two adjacent angle having a sum of 360 degrees, from the pentagon  $A_nB_nC_nD_nE_n$ construct the pentagon  $A_{n+1}B_{n+1}C_{n+1}D_{n+1}E_{n+1}$ as follows:

 - $A_{n+1}$ is the intersection between the angle bisectors of $\angle
   C_n $ and $\angle D_n$. 
 - $B_{n+1}$ is the intersection between the angle bisectors of $\angle D_n$ and $\angle E_n$.
 - $C_{n+1}$ is the intersection between the angle bisectors of $\angle E_n$ and $\angle A_n$.
 -  $D_{n+1}$ is the intersection between the angle bisectors of $\angle A_n$ and $\angle B_n$.
 - $E_{n+1}$ is the intersection between the angle bisectors of $\angle B_n$ and $\angle C_n$.

 **(the two opposite angles)** I am allowing self-intersecting polygons in this constructions.  

 >The reason why I chose this construction is that the point $A_{n+1}$ is the only point that doesn't depend on $A_n$. 

My question is if this process is repeated indefinitely would the sequences ${A_n}$
, ${B_n}$
, ${C_n}$ , $D_n $ and $E_n$
 converge ? and if they converge what is the limit? if they din't always converge what are the necessary conditions that the pentagon  $A_1B_1C_1D_1E_1$ should satisfy so the five sequences converge ?


There are only four possible scenarios:

 

 1. The points will converge.
 2. The points will eventually trapped  on a loop.
 3. The points will diverge completely.
 4. Two or more points coincide, or two adjacent angle have a sum of 360 degrees ending the sequence.






 

 
I tried to draw the first few pentagons to  see if the point will converge or not.

[![enter image description here][2]][2]
[![enter image description here][3]][3]

Here is the first 50 pentagons in a zoom-in animation: 
[![enter image description here][4]][4]

I conjecture that the sequences converge to a single point for all non-regular pentagons. If convergence occurs, how can we determine the limit point based on the initial pentagon? If the limit point exist what interesting properties does it have for the pentagon?   

---

[Here][5] is a Geogebra file that have the first 50 pentagons. 





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This is a python code than can calculate $A_n, B_n , C_n ,D_n ,E_n$
```python
import math 
# enter your coordinates here
A=[0,0]
B=[1,1]
C=[3,5]
D=[2,7]
E=[2,6]

X=[0,0]
Y=[0,0]
Z=[0,0]
W=[0,0]
T=[0,0]

def v(a,b):
    result= [b[0]-a[0], b[1]-a[1]]

    return result 


def crs(a, b):
    result = [-0.5*((a[0]*b[1])-(a[1]*b[0]))]

    return result


def dis(a,b):
    result =[math.sqrt((a[0]-b[0])**2+(a[1]-b[1])**2)]
    return result 


for i in range (2,100+1):
    AB= dis(A,B)[0]
    BC= dis(B,C)[0]
    CD= dis(C,D)[0]
    DE= dis(D,E)[0]
    EA= dis(E,A)[0]



    #Here to calculate A_n
    x=   CD*crs(v(D,C),v(D,E))[0]
    y=  -CD*crs(v(E,D),v(E,B))[0]+  2*DE*crs(v(C,B),v(C,D))[0]
    z=  -CD*crs(v(C,B),v(C,E))[0]+  2*BC*crs(v(D,C),v(D,E))[0]
    t=   CD*crs(v(C,B),v(C,D))[0]
    X[0]=(x *B[0]+y*C[0]+z*D[0]+t*E[0])/(x+y+z+t)
    X[1]=(x *B[1]+y*C[1]+z*D[1]+t*E[1])/(x+y+z+t)



    #Here to calculate B_n
    x=   DE*crs(v(E,D),v(E,A))[0]
    y=  -DE*crs(v(A,E),v(A,C))[0]+  2*EA*crs(v(D,C),v(D,E))[0]
    z=  -DE*crs(v(D,C),v(D,A))[0]+  2*CD*crs(v(E,D),v(E,A))[0]
    t=   DE*crs(v(D,C),v(D,E))[0]
    Y[0]=(x *C[0]+y*D[0]+z*E[0]+t*A[0])/(x+y+z+t)
    Y[1]=(x *C[1]+y*D[1]+z*E[1]+t*A[1])/(x+y+z+t)



    #Here to calculate C_n
    x=   EA*crs(v(A,E),v(A,B))[0]
    y=  -EA*crs(v(B,A),v(B,D))[0]+  2*AB*crs(v(E,D),v(E,A))[0]
    z=  -EA*crs(v(E,D),v(E,B))[0]+  2*DE*crs(v(A,E),v(A,B))[0]
    t=   EA*crs(v(E,D),v(E,A))[0]
    Z[0]=(x *D[0]+y*E[0]+z*A[0]+t*B[0])/(x+y+z+t)
    Z[1]=(x *D[1]+y*E[1]+z*A[1]+t*B[1])/(x+y+z+t)



    #Here to calculate D_n
    x=   AB*crs(v(B,A),v(B,C))[0]
    y=  -AB*crs(v(C,B),v(C,E))[0]+  2*BC*crs(v(A,E),v(A,B))[0]
    z=  -AB*crs(v(A,E),v(A,C))[0]+  2*EA*crs(v(B,A),v(B,C))[0]
    t=   AB*crs(v(A,E),v(A,B))[0]
    W[0]=(x *E[0]+y*A[0]+z*B[0]+t*C[0])/(x+y+z+t)
    W[1]=(x *E[1]+y*A[1]+z*B[1]+t*C[1])/(x+y+z+t)



    #Here to calculate E_n
    x=   BC*crs(v(C,B),v(C,D))[0]
    y=  -BC*crs(v(D,C),v(D,A))[0]+  2*CD*crs(v(B,A),v(B,C))[0]
    z=  -BC*crs(v(B,A),v(B,D))[0]+  2*AB*crs(v(C,B),v(C,D))[0]
    t=   BC*crs(v(B,A),v(B,C))[0]
    T[0]=(x *A[0]+y*B[0]+z*C[0]+t*D[0])/(x+y+z+t)
    T[1]=(x *A[1]+y*B[1]+z*C[1]+t*D[1])/(x+y+z+t)
    A=X
    B=Y
    C=Z
    D=W
    E=T

    
    print(f"A_{i}= {A}")
    print(f"B_{i}= {B}")
    print(f"C_{i}= {C}")
    print(f"D_{i}= {D}")
    print(f"E_{i}= {E}")
```

















  [1]: https://math.stackexchange.com/questions/4937560/does-the-sequence-formed-by-intersecting-angle-bisector-in-a-pentagon-converges
  [2]: https://i.sstatic.net/8M3RUVFT.png
  [3]: https://i.sstatic.net/2fbkfWiM.png
  [4]: https://i.sstatic.net/Ddnk9Px4.gif
  [5]: https://www.geogebra.org/classic/tywbvnyg