@Kitonum I tried

L:=[[0,2],[2,0],[4,4]]:

C:=LargestIncircle(L);

A:=plottools[disk](op(C), color=green):

B:=plottools[polygon](L, style=line, thickness=2):

plots[display](A,B, scaling=constrained);

And got a circle with center and radius not integer numbers. A problem is, if I want a circle with center I(1,1) and radius r = 3. How to choose a convex polygon, triangle example with the coordiantes of vertices are integral numbers?

@Carl Love I got a RootOf. I used your line and got the correct ansewr. Thank you very much.

@Kitonum I can not get your result.

@acer Thank you very much.

@pagan How about expr:=sin(Pi*(3*x+1)/(4*x^2+4*x+2))=cos(Pi*(5*x-2)/(4*x^2+4*x+2))?

@acer Yes, I used Mathematica. I don't know how to insert a picture. 

@brian bovril The roots are approximate, not exact.

@Kitonum I also use Mathematica. 

@Carl Love 

My problem is, with two given points O(0,0) and A, find the numbers a, b, c, d of the function y = (a*x + b)/(c*x + d) so that there is four points B1, B2, B3, B4 and the mesure of the angle BOA of the triangle OB_iA = 45 degrees (i = 1..4).

@Carl Love 

I added some conditions for x1, y1, x2, y2. This is my code

> restart:

 ListTools[Categorize]:

N:=10:

L:=[]:

for x1 from -5 to N do

for y1 from -5 to N do

for x2 from -5 to N do

for y2 from -5 to N do

a:=x1^2+y1^2:

b:=x2^2+y2^2:

c:=a+b-2*sqrt(a*b)*cos(Pi/4):

 if type(a, positive) and type(b, positive) and   c=(x2-x1)^2 + (y2-y1)^2 and x1*(x2-x1)+y1*(y2-y1)<>0 and x1<> 0 and x2<>0 and y1<>0 and y2<>0 and x1+y1<>0 and x1-y1<>0 and  x2+y2<>0 and x2-y2<>0 and x2*(x2-x1)+y2*(y2-y1)<>0 then L:=[op(L), [[0, 0], [x1, y1], [x2, y2]]]: fi:

od: od: od: od:

nops(L);

L;

With Maple out put, I notice the triangles has two vertices (0, 0) and (10, 5), and select three points (1,3), (2,6), (3, -1), next,  I write the code

restart:
f:=x->(a*x + b)/(c*x + d):
solve([f(2)=6,f(1)=3,f(3)=-1],[a,b,c,d]);

And then, I got

[[a = 9/5*c, b = -6*c, c = c, d = -12/5*c]]

I choose c = 5, then a = 9, b = -30, d = -12. The function I got is y = (9*x -30)/(5*x - 12). I tested

restart:
with(LinearAlgebra):
a:=<10,5>:
b:=<x,(9*x -30)/(5*x - 12)>:
sol:=solve(VectorAngle(a,b)=Pi/4,x);

and got 

sol := 1, 2, 3. Then, I get three points (1,3), (2,6), (3, -1). 

Clearly, I did by hand.

Four points which I want to get does not contain the point (0, 0) and the numbers a, b, c, d in the functions are integer numbers. I only got three points, not four.

@mehdi jafari I don't know solvee it by hand.

@Carl Love My problem is "I want to find constants a, b, c, d such that there exists a triangle ABC with centroid O such that all four points have integer coordinates and all are on the curve y = (a*x+b)/(c*x+d)".

N:=Student:-Calculus1:-Roots(sin(2*x)=1/2,x=-2*Pi..2*Pi);
nops(N);

N:=Student:-Calculus1:-Roots(sin(2*x)=1/2,x=-2*Pi..2*Pi);
nops(N);

@ Kitonum, your command is not correct with the equation x^4 -(3*m+2)*x^2 + 3*m+1 = 0.