Reciprocally, we may wonder if, given any triangle ABC, there are three circles centered in A, B, C and mutually tangent. The answer is "yes".

The 4th circle is defined as the

There are in general two solutions, named the "inner Soddy circle (and point)" for the one which is inside the curvilinear triangle defined by the 3 circles (A) (B) (C), hence also inside the triangle ABC.

And "outer Soddy circle (and point)" for the other one. Notice that it is the

The line through the two Soddy points is the

The common tangents in D, E, F are the radical axes of these circles pairwise,

They then concur in a single point I, radical center of these circles.

ID = IE = IF and these lines being perpendicular to the sides of ABC, I is the incenter and
DEF the contact points with incircle

Hence the existence and uniqueness of the three circles, mutually **externaly** tangent

(See appendix for internally tangent circles)

The image of circle (P) is then a circle (P') tangent to these lines in V' and W', and to (A') = (A) in U'.

Hence a circle congruent to (A) whose center P' is on the perpendicular to BC from A, with U'P' = AU'.

There are two such circles. (P') corresponds to the searched outer Soddy circle (P).
The other one (P") is the image of the inner Soddy circle.

The contact point U of (P) and (A) is the image of U' and is therefore on EU'.

And the same for the contact points V and W with (B) and (C).

Finally P, E and P' are in line, as the circles (P) and (P') are inverse of each other.

Hence a construction of the outer Soddy point and circle :

Construct the contact points DEF of the incircle, and circle (A) with center A going through D (and F)

The perpendicular from A to BC intersects this circle in U', opposite to BC, and in U" near BC.

Let P' the symmetric of A with respect to U'

Let U the other intersection of EU' with circle (A)

Lines AU and EP' intersect in P, center of the outer Soddy circle.

U is the contact point of (A) with this circle, hence the outer Soddy circle is the circle centered in P, going through U.

We do the same for the inner Soddy circle, starting from U", P".

As this construction becomes here very inaccurate, we may use the following variant :

Consider also the point V", image of the contact point with (B) in an inversion through pole F :

The perpendicular from B to AC intersects circle (B) in V", opposite to AC.

Let Q" the symmetric of B with respect to V"

Let P" the symmetric of A with respect to U" as above,

The center S of the inner Soddy circle is the intersection of EP" and FQ".

The contact point U" is sent back to U, other intersection of EU" with (A)
(not drawn on the figure)

And finally the inner Soddy circle is the circle with center S, through U

Let the circle (P), centered in P, tangent to the three circles (A) (B) (C), outer Soddy circle, which, if the circles
(A) (B) (C) are not too much unequal, surounds these circles.
The circles (A) (B) (C) being internally tangent in U, V and W.

We get : PU = PV = PW = R_{P} radius of the Soddy circle

CW = CE and BV = BE

The perimeter of PCB is then PV + PW = 2R_{P}

And the same for the two other triangles

Theorem : The outer Soddy point is the isoperimetric point

This only if the outer Soddy circle (P) surrounds the circles (A) (B) (C).

If the smaller of circles (A) (B) (C) is so small that (A) (B) (C) touch (P) externally,
there is no isoperimetric point.

The critical value is when the outer Soddy circle is a straight line :

UV² = AB² - (BU-AV)² = (r_{A} + r_{B})² - (r_{A} - r_{B})², hence UV = 2√(r_{A}r_{B})

and the same for VW, then UW = UV + VW gives a condition for existence of the isoperimetric point :

1/√r_{A} < 1/√r_{B} + 1/√r_{C}
or also :

a + b + c > 4R + r
with r the inradius and R the circumradius. |

The inner Soddy point is similarily a point of "equal detour".

Define the detour of A to B through S as the value AS + SB - AB.

If S is the inner Soddy point, the detour is 2ρ
with ρ being the radius of the inner Soddy circle, and this is the same with AB, BC and AC.

Finally, if the outer Soddy point is not an isoperimetric point, it is another point of equal détour
(when a + b + c < 4R + r).

The radii of the Soddy circles are given by the
Descartes formula, from the radii of circles (A), (B), (C) that is
(well known formulas about the contact points of incircle)

circle (A) : r_{A} = AD = AE = (AB + AC - BC)/2 = s - a with s being the semi-perimeter of ABC

and the same for the two others.

Consider the inversion through the incircle of ABC, that is of pole the incenter I, and keeping the contact points DEF unchanged

The circles centered in A, B, C, orthogonal to the incircle in D,E,F, are therefore globally unchanged

This inversion exchanges then the two Soddy circles.

Therefore the centers of these circles (the two Soddy points) are on a same line with the inversion center (the incenter I)

The Soddy line goes through the incenter |

I won't give a proof here, as it is beyond this "recreationnal" website !

This gives a variant with the same properties for nameless points "similar to Soddy points"

In this variant, we get circles centered in A,B,C and mutually tangents, but not externally !

The following applet shows this variant from the excircle in angle A, centered in Ja and its contact points, that we'll continue to name D, E, F

The point similar to the Gergonne point is Ga, the line S1S2 goes through Ja (same proof as above, by the inversion of pole Ja, through the excircle) and through the "ex-Gergonne" point Ga, and even through the de Longchamps point L (on the Euler line OGH) of ABC !

The 4 generalized Soddy lines (the normal Soddy line and the 3 lines obtained from the excircles) then go all 4 through this de Longchamps point, which then exhibits a noticeable property.