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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 to construct (with the compass and straightedge) these vertices (A,B,C)?
enter image description here

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

enter image description here
Looks like the same solution... And no idea about construction.
So let all of the circles be an infinite radius:
enter image description here
And we get Fagnano's problem with clear construction.
Hope this will be useful (?)

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    $\begingroup$ Setting it differently, is the common point of $AD, BE, CF$ constructible by straightedge and compass ? You must know that there are many cases of remarkable points that are not contructible by straightedge and compass... $\endgroup$ – Jean Marie Mar 4 at 21:21
  • $\begingroup$ @Jean Marie, such a case is possible! But I would also like to prove this, at least heuristically, in the spirit of Polya's book. $\endgroup$ – lesobrod Mar 5 at 5:01
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    $\begingroup$ @lesobrod: I have already drawn the triangle $PQR$ in my comment and it applies to all circles with arbitrary radii. So I am certain that this generalization with Fagnano's triangle (when the the straight lines are replaced by an arena of 3 circles) would hold good in the sought future solution here. $\endgroup$ – Narasimham Mar 7 at 17:48
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A comment only (in the answer area, to include image) fwiw Comment

The same solution triangle $ABC$ holds for all co-tangential brown circles. in-center of this solution triangle $I$ ( red in-circle) coincides with ortho-center of outer triangle $PQR$.

I was wondering if Area $\Delta= s.r$ of $ABC$ would be also an extremum.. as a product of (semi) perimeter and in-radius.

EDIT1:

From theory of mechanisms the 3 links $(AB,BC,CA)$ are rotating about their common instantaneous center of rotation $O$ in a way that extremizes some (to me now unknown) geometric property.

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