Three tangent Circles

by Klaus-Anton, May 27, 2017, 7:09 PM

Perhaps you know that here is a problem seen.

And indeed. There is more than only one problem.

Three circles may be tangent. And in the midpoint there is is the origin with the coordinates (x,y)=(0,0).

$A_0=(\sqrt3/1.5,0)$ .

$A_0$ is first rotated by 90 + 120 degree around the origin $O=(0,0)$ to give $A_1$ . $A_2$ now is given by the rotation of $A_0$ by 90 + 240 degree. And finally $A_3$ is given by the rotation of $A_0$ by 90 + 360 degree.

You see. Things could have been explained some easier, but so it is as it was. And so i have overtaken it.
$[asy] pair O; pair[] A; O=(0,0); dot("(0,0)",O); real my_x=sqrt(3)/1.5; pair my_dot=(my_x,0); dot(my_dot); pair A0=my_dot; //label("$A_0$",A0, dir(O--A2)); draw(circle(O, my_x), dotted); pair A1; A1= rotate(90+120*1)*(my_x,0); A[1]=A1; dot("$A_1$",A1, dir (O--A1)); draw(Circle(A1,1)); draw("$c_1$",Circle(A[1],1),red); pair A2=rotate(90+120*2)*(my_x,0); dot("$A_2$",A2, dir(O--A2)); draw(Circle(A2,1)); A[2]=A2; draw("$c_2$",Circle(A[2],1),dir(-40),red); label("$A_0$",A0, dir(O--A2)); pair A3=rotate(90+120*3)*(my_x,0); dot("$A_3$",A3,N); draw(Circle(A3,1)); A[3]=A3; draw("$c_3$",Circle(A[3],1),red); for(int i=1; i<=3; ++i) { A[i]=rotate(90+120*i)*(my_x,0); dot(A[i],blue); draw(Circle(A[i],1),blue); } draw(A1--A2--A3--cycle, blue+dotted); [/asy]$
You can see that $A_0$ is positioned on the east-point of the circumcircle from the triangle $(A_1,A_2,A_3)$ . This triangle is a regular triangle. Alle the sides have length 2. From $A_1$ to $A_2$ the length is 2. From $A_2$ to $A_3$ the length is 2. And also from $A_3$ to $A_1$ the side-lenth is 2.

This is therefore so, because the radii of the blue circles around $A_1,\ A_2, \text{and}\ A_3$ always is 1.

Now you may ask how long is the distance from the point $A_3$ to the baseline of the triangle $(A_1,A_2,A_3)$ .

The distance from $A_3$ to (0,0) already is given. It is the same as from (0,0) to $A_0$ . This is $\sqrt3/1.5$ . $A_3.y=\sqrt3/1.5$ .

So still is open the distance from the origin $O=(0,0)$ to the point where i positioned the labell of the circle around $A_2$ with $\color{red}c_2$ . This is the midpoint of $A_1--A_2$ , in Asymptote-code defined by the statement "midpoint(A1--A2);".

As the regular, equilateral triangle $A_1,A_2,A_3$ has side4lenght $a=2$ its hight $h$ is given by
$h=\frac{\sqrt{3}}{2}a =\frac{\sqrt{3}}{2}\ast 2 =\sqrt{3}$ .

There from we can follow that the distance from the origin $O=(0,0)$ to the midpoint( $A_1--A_2$ ) is:
$\sqrt{3} - \frac{\sqrt{3}}{1.5} =\frac{\sqrt{3}}{3}= 0.577350\dots$ .

$[asy] size(6.3cm); pair O; pair[] A; O=(0,0); dot("(0,0)",O); real my_x=sqrt(3)/1.5; pair my_dot=(my_x,0); dot(my_dot); pair A0=my_dot; //label("$A_0$",A0, dir(O--A2)); //draw(unitcircle); draw(circle(O, my_x), dotted); pair A1; A1= rotate(90+120*1)*(my_x,0); A[1]=A1; dot("$A_1$",A1, dir (O--A1),blue); draw(Circle(A1,1)); draw("$c_1$",Circle(A[1],1),red); pair A2=rotate(90+120*2)*(my_x,0); dot("$A_2$",A2, dir(O--A2),blue); draw(Circle(A2,1)); A[2]=A2; draw("$c_2$",Circle(A[2],1),dir(-40),red); label("$A_0$",A0, dir(O--A2)); pair A3=rotate(90+120*3)*(my_x,0); dot("$A_3$",A3,N,blue); draw(Circle(A3,1)); A[3]=A3; draw("$c_3$",Circle(A[3],1),red); for(int i=1; i<=3; ++i) { A[i]=rotate(90+120*i)*(my_x,0); dot(A[i],blue); draw(Circle(A[i],1),blue); } draw(A1--A2--A3--cycle, blue+dotted); pair c4_label=rotate(150)*(my_x,0); label("$c_4$", c4_label, dir(O--c4_label)); //draw(shift(A3)*unitcircle); path Unitcircle=Circle((0,0),1,n=400); //draw(Unitcircle); pair A5=(A1.x-sqrt(3),A1.y); pair A6=(A5.x,A3.y); //dot(A5^^A6); draw(A5--A6, blue, Arrows,Bars); label("$\sqrt{3}$", midpoint(A5--A6),1.5*W,blue); pair A7=(0,-1.15*sqrt(3)); pair A8=(A2.x,A7.y); //dot(A7^^A8); draw(A7--A8, blue, Arrows,Bars); label("$1$", midpoint(A7--A8),1.5*S,blue); //draw( (A1.x-1.5,A1.y)--(,A3.y) ); pair A9=(0,my_x+1+.25*1.5); pair A10=(A0.x,A9.y); //dot(A9^^A10); draw(A9--A10, Arrows,Bars); //label("$\sqrt3/1.5$", midpoint(A9--A10),2.5*N); //label("$\displaystyle{\frac{\sqrt3\ast 2}{3}}$", midpoint(A9--A10),2.0*N); label("$\displaystyle{\sqrt3\ast \frac{2}{3}}$", midpoint(A9--A10),2.0*N); //dot(relpoint(A3--A1,(2/3))); [/asy]$

If the sidelength is given you can construct geometrically the equlateral triangle as shown in Marry me!.

The the circumcenter is given you can construct geometrically the equlateral triangle like this (url):

$[asy] draw(circle((0,0),1)); draw(W--E^^S--N, dashed+gray); label("$C$",N,N); label("$D$",S,S); path southcircle=circle(S,1); draw(circle(S,1)); pair AB [] =intersectionpoints(unitcircle,southcircle); //dot(AB[0]); label("$A$", AB[0],W); label("$B$", AB[1],E); draw(AB[0]--AB[1]--N--cycle); [/asy]$