If we divide the isoceles triangle in two right triangles we get a triangular relation based on the top angle (I called theta). Sin(theta)=1/(1+L1)=r/(r+2+2/(r-1)). This gives us L1=2/(r-1).
Then we use the right triangle from the center of the circle with radius r, to the center of one of the circles of the botom and the quadrant between circles of radius 1, this gives us the relation L2=√(r*(r+2))-r. Finally using the first triangular relation, we get the transient equation:(r-1)*(cos(theta)+1)=cos(theta)*((r-1)*(tan(theta)+√(r*(r+2))+r+2)+2) where theta = arcsin(r*(r-1)/(2+(r-1)*(r+2))). Solving the mentioned equation, the radius of the circle in the middle is r=1.51067547264cm
Posted by Antonio
on 2003-12-02 01:11:36