Given three non-overlapping circles, can we construct (via straightedge and compass) the triangle of minimum perimeter with one vertex on each circle?

Question

G. Polya "Mathematics and plausible reasoning" Chapter 9, problem 2:

Three circles in a plane, exterior to each other, are given in position. Find the triangle with minimum perimeter that has one vertex on each circle.

From the contents of the chapter it is obvious (using light reflections on three circular mirrors and rubber band methods) that the two sides of the required triangle that meet in a vertex on a given circle include equal angles with the radius.

But how can we construct (with the compass and straightedge) these vertices (A,B,C)?

UPD
Let one of the circles be an infinite radius (a straight line):

Looks like the same solution... And no idea about construction.
So let all of the circles be an infinite radius:

And we get Fagnano's problem with clear construction.
Hope this will be useful (?)

Answer 1

One more attempt:

If a triangle vertices are given, then the Fermat-Toricelli point $F$ that sees its sides isogonally $@\measuredangle\pi/3$ minimizes sum of lengths from $F$ to vertices $(A,B,C)$.

Conversely ,$(P,Q,R)$ can be slid anywhere along the radial lines $ (OA,OB,OC)$ with minimum peripheral length $2s$ with given total length of spokes.

i.e.,any triangle enclosed around the hexagonal Fermat node $F$ should minimize peripheral length sum $PQR$ as shown with required $ (AP,BQ,CR)$ radial offsets.

This solution is built on Fig2 Fermat Point Wiki reference.

Answer 2

Solution minimum triangle is determined by the given circles and angle bisectors.The following states that triangle $ABC$ has minimum perimeter for all triangles such as $EDF.$ Co-tangential circles are also included to suggest a set of problems with same inner minimum perimeter triangle $ABC$.

EDIT1/2:

Intersections of angle bisectors from vertices $(A,B,C) $ with given circles form triangle $PQR$ vertices of minimum perimeter length. This results from the ellipse property... constancy of major axis length between shown ellipse foci with their mirror reflective property considered pairwise among $(P,Q,R)$.

Answer 3

-- As addendum to Narasimham's answer --

Consider $A$ and $C$ fixed.
Then for the perimeter to be minimum, $B$ shall lie on the ellipse with foci in $A$ and $C$, minimal sum of the distances from them, so on the one tangent to the circle through $B$.

By the property of elliptical mirror, the normal to the tangent in $B$, shall halve the angle $ABC$.

Then the proof of Narasimham's answer follows.

Moreover, with $A$ and $C$ fixed, we are minimizing $p$, and $p-b$ as well.
By symmetry, the triangle has also minimal area.

Answer 4

This is not an answer but rather an extended comment.

As easy to understand the construction boils down to finding the in-center $I$ of the triangle $ABC$.

An algebraic way for this could be the following. Let the coordinates and the radius of $i$-th circle ($i=1,2,3$) be $(x_i,y_i)$ and $r_i$, respectively, and the coordinates of the point $I$ be ($x,y$). Then the coordinates of the vertex $A$ read: $$ (x_A,y_A)=(x_1,y_1)+r_1\frac{(x-x_1,y-y_1)}{\sqrt{(x-x_1)^2+(y-y_1)^2}},\tag1 $$ and similarly for the vertices $B$ and $C$.

From the condition that $I$ is the in-center of $\triangle ABC$ we have the following equations to determine $(x,y)$: $$ \frac{\frac{x-x_A}{x_B-x_A}-\frac{y-y_A}{y_B-y_A}} {\sqrt{\frac1{(x_B-x_A)^2}+\frac1{(y_B-y_A)^2}}}= \frac{\frac{x-x_B}{x_C-x_B}-\frac{y-y_B}{y_C-y_B}} {\sqrt{\frac1{(x_C-x_B)^2}+\frac1{(y_C-y_B)^2}}}= \frac{\frac{x-x_C}{x_A-x_C}-\frac{y-y_C}{y_A-y_C}} {\sqrt{\frac1{(x_A-x_C)^2}+\frac1{(y_A-y_C)^2}}},\tag2 $$ where $x_A,y_A,x_B,y_B,x_C,y_C$ are to be substituted using (1).

After expanding the equations one should end up with a system of two polynomial equation for $(x,y)$ of a rather high order. It is probable that the solution to this equation is not constructible.