Find the smallest sum distance among three points on three circles There are three separate circles on a plane.  How does one find three points, $a$, $b$ and $c$, one on each circle, such that the sum distance $\bar{ab} + \bar{ac} + \bar{bc}$ is minimal?
 A: Here's how to solve this (draw a figure!):
Without loss of generality, circle $C_1$ (with radius $r_1$), can be centered at the origin; circle $C_2$ (with radius $r_2$) can be placed on the $x$-axis a distance $l_{2x}$ from the origin; circle $C_3$ (with radius $r_3$) will be centered at coordinates $(l_{3x}, l_{3y})$. 
The position of point $a$ on $C_1$ can be specified by the single angle variable $\theta_1$, and likewise for $b$ and $\theta_2$ and $c$ and $\theta_3$.  Thus your problem reduces to finding the three angles that minimize the sum of the inter-point distances.
The distance $\bar{ab}$ is
$$\sqrt{(r_1 \cos \theta_1 - (l_{2x} + r_2 \cos \theta_2))^2 + (r_1 \sin \theta_1 - r_2 \sin \theta_2)^2}$$
I'll let you fill in the two other distances.
Then express the sum $L_{total} =\bar{ab} + \bar{ac} + \bar{bc}$, which will depend upon the three angles.
For $L_{total}$ to be minimal, the three derivatives must be zero:


*

*${\partial L_{total} \over \partial \theta_1} = 0$

*${\partial L_{total} \over \partial \theta_2} = 0$

*${\partial L_{total} \over \partial \theta_3} = 0$


This will give you a set of simultaneous equations, whose solution minimizes $L_{total}$.
Actually, the above will give you the maximum and the minimum distance $L_{total}$, so you choose the minimum by simple comparison.  (You could also impose conditions on the second derivative, but that is too much work for such a simple problem.)
In pathological cases (e.g., all circle centers are co-linear), there will be multiple minimizing solutions.
A: This cannot be solved by straightedge and compass since it leads to a cubic equation, but assuming that $\Gamma_A,\Gamma_B,\Gamma_C$ are three disjoint circles through $A,B,C$, there is a simple iterative approach for finding an approximate solution:


*

*Move $A$ on $\Gamma_A$ until the ellipse through $A$ with foci at $B,C$ is tangent to $\Gamma_A$;

*Move $B$ on $\Gamma_B$ until the ellipse through $B$ with foci at $A,C$ is tangent to $\Gamma_B$;

*Move $C$ on $\Gamma_C$ until the ellipse through $C$ with foci at $A,B$ is tangent to $\Gamma_C$;

*Iterate until the required accuracy is reached.

