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

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 $(P,Q,R)$.

-- 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.