To construct three circles {k_{1}, k_{2}, k_{3}}
each tangent externally to the two others and also to two sides of
the triangle A_{1}A_{2}A_{3}.

Steiner gave unproven the following procedure to solve the
problem.

[1] Construct the common intersection point M of the
three bisectors of the triangle.

[2] Draw the three incircles {c_{1},
c_{2}, c_{3}} of the corresponding triangles {MA_{2}A_{3},
MA_{3}A_{1}, MA_{1}A_{2}}.

[3]
Consider the other common tangents {D_{1}E_{1},D_{2}E_{2},D_{3}E_{3}}
of the three circles (symmetric of MA_{1}, MA_{2},MA_{3}
with respect to the corresponding lines of centers of these circles)
intersecting at a point K.

[4] The requested three circles are the
incircles of the resulting quadrangles {A_{1}D_{2}KD_{3},
A_{2}D_{3}KD_{1}, A_{3}D_{1}KD_{2}}.

The following solution taken from Coolidge originates from Hart.
The proof is simple, but has some relatively difficult to distinguish
points cluttering around M the incenter of the given triangle. The
indexed labels {B, C, D, F, G, P} indicate contact points. One starts
with the analysis. Thus, assume that circles {k_{1}, k_{2},
k_{3}} have been constructed as required.

[1] Consider
then their inner common tangents {D_{1}E_{1}, D_{2}E_{2},
D_{3}E_{3}} having contact points at {P_{1},
P_{2}, P_{3}} and intersecting at K, which is their
radical center, so that KP_{1}=KP_{2}=KP_{3}.

[2]
Construct then circles {c_{1}, c_{2}, c_{3}}
inscribed in the triangles {KE_{2}E_{3}, KE_{3}E_{1},
KE_{1}E_{2}}. Next step is to show that these circles
contact the sides of the triangle precisely at the points {D_{1},
D_{2}, D_{3}}, where the inner tangents of the k_{i}'s
intersect the sides of the triangle.

[3] In fact:

E_{1}D_{2}
- E_{3}D_{2} = E_{1}B_{2} - E_{3}C_{2}
= E_{1}P_{1} - E_{3}P_{3} = E_{1}K
- E_{3}K.

This implies that D_{2} is the contact
point of c_{2} wiht line A_{1}A_{3}.
Analogous is the proof for D_{1}, D_{3}.

[4] Next
consider the other than {D_{1}E_{1}, D_{2}E_{2},
D_{3}E_{3}} inner tangents of the three circles {c_{1},
c_{2}, c_{3}}. They are three lines {L_{1},
L_{2}, L_{3}} passing also through a point M (see
Isogonal_3TangentCircles.html
). Next we identify them with the bisectors, consequently M will be
proven to be the incenter of the triangle.

[5] Denote by {F_{j},
G_{j}} the contact points of the circles {c_{i}} with
lines {D_{j}E_{j}}. Because of the symmetry relation
of lines {L_{i}} to {D_{i}E_{i}} to show that
L_{1} passes through A_{1} it suffices to show that
length F_{i}G_{i} = A_{1}D_{2} - A_{1}D_{3}
(since, by symmetry, F_{1}G_{1} is equal to the
distance of contact points of c_{2}, c_{3} and L_{1}).
But:

A_{1}D_{2} - A_{1}D_{3} =
C_{2}D_{2} - B_{3}D_{3} = P_{3}G_{3}
- P_{2}F_{2} = F_{1}G_{1}. Thus, L_{1}
passes through A_{1}. Analogous is the proof for lines L_{2}
and L_{3}.

[6] To identify L_{1} with the bisector
we show that A_{1} is on the circle of similitude of {c_{2},
c_{3}} (see CirclesSimilar.html
) hence it is viewing the two circles under equal angles. For this it
suffices to show that the chords intersected by {c_{2}, c_{3}}
on line D_{2}D_{3} are equal. This in turn is
equivalent with the equality of the powers p(D_{2}, c_{3})
= p(D_{3}, c_{2}), which amounts to the equality of
the tangents to these circles D_{2}F_{2} = D_{3}G_{3}.

[7]
To show D_{2}F_{2}
= D_{3}G_{3},
notice that D_{3}G_{3}
= D_{3}P_{3}
+ P_{3}G_{3}
= D_{3}B_{3}
+ D_{2}C_{2}
= P_{2}F_{2}
+D_{2}P_{2}
= D_{2}F_{2
}.(*) Thus line L_{1}
is identified with the bisector and analogous arguments show that L_{2},
L_{3} are also bisectors of the angles of the triangle.

These
arguments show that if there is a solution, then it is obtainable
through the procedure proposed by Steiner. But there remains the
question of the existence of a solution. How can one prove that there
is one?

(*) Thanks to Gergely Harcos for correcting the argument here (11-01-2013).

The existence proof of Hart involves a continuity argument.
Consider a small circle k_{1} tangent to sides {AB, AC} and
two other circles {k_{2}, k_{3}}, each tangent to k_{1}
and two sides of the triangle.

The three circles depend continuously on the radius r of
circle k_{1} (and the triangle data of course). For small
values of r the two circles {k_{2}, k_{3}} intersect.
For larger values of r the two circles are disjoint. Hence there is
some intermediate value r_{0} for which the two circles {k_{2},
k_{3}} are tangent. Since these circles remain all the time
tangent to k_{1} we obtain in this way a solution to
Malfatti's problem.

Isogonal_3TangentCircles.html

CirclesSimilar.html

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