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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 can we 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
    Commented Mar 4, 2019 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
    Commented Mar 5, 2019 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
    Commented Mar 7, 2019 at 17:48

4 Answers 4

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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 $ABC$.

Comment

EDIT1/2:

In-Center based Construction

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 $(P,Q,R)$.

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    $\begingroup$ Thank you for your diagram. I don't understand your construction, though: how are $A, B$ and $C$ determined? Or, to put it another way, how is $I$ determined? (Once $I$ is found, it's easy to find $A, B$ and $C$.) $\endgroup$
    – Rosie F
    Commented May 1, 2020 at 19:32
  • $\begingroup$ @Rosie F May: The three given non-overlapping circles are centered at $(A,B,C)$ $\endgroup$
    – Narasimham
    Commented May 5, 2020 at 20:40
  • $\begingroup$ @user: No. It may be for special cases like three given circles of same radius and when their centers are placed at vertices of given equilateral triangle. $\endgroup$
    – Narasimham
    Commented May 5, 2020 at 21:13
  • $\begingroup$ Thank you for your second diagram. That answers the question in my earlier comment. (Note that in the first diagram and my comment, the centres are $D, E, F$ and the vertices are $A, B, C$, but in the second diagram, the centres are $A, B, C$ and the vertices are $P, Q, R$.) Further question: do the angle-bisectors in the second diagram also show the vertices of that triangle of greatest perimeter with one vertex on each circle? $\endgroup$
    – Rosie F
    Commented May 6, 2020 at 5:19
  • $\begingroup$ @user: In-center $I$ of $ABC$ need have no relation to any center of inner triangle $PQR$. $\endgroup$
    – Narasimham
    Commented May 6, 2020 at 12:18
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-- 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.

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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.

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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.

Fermat Pt Logic

This solution is built on Fig2 Fermat Point Wiki reference.

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  • $\begingroup$ From the construction it is obvious that $F$ is Fermat point both for $\triangle ABC$ and $\triangle PQR$. But we are looking for the in-center of $\triangle PQR$, which generally does not coincide with the Fermat point (except for equilateral triangle). $\endgroup$
    – user
    Commented May 7, 2020 at 11:14
  • $\begingroup$ Not considering any in-center in the Fermat point option. $\endgroup$
    – Narasimham
    Commented May 7, 2020 at 11:32
  • $\begingroup$ Previously the construction was around the in-center, and now a fresh approach is adopted around the Fermat point. Upto now it appears both are incorrect ! $\endgroup$
    – Narasimham
    Commented May 7, 2020 at 11:42

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