1. Malfatti's problem

To construct three circles {k1, k2, k3} each tangent externally to the two others and also to two sides of the triangle A1A2A3.

2. Steiner's solution idea

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 {c1, c2, c3} of the corresponding triangles {MA2A3, MA3A1, MA1A2}.
[3] Consider the other common tangents {D1E1,D2E2,D3E3} of the three circles (symmetric of MA1, MA2,MA3 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 {A1D2KD3, A2D3KD1, A3D1KD2}.

3. Hart's proof

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 {k1, k2, k3} have been constructed as required.
[1] Consider then their inner common tangents {D1E1, D2E2, D3E3} having contact points at {P1, P2, P3} and intersecting at K, which is their radical center, so that KP1=KP2=KP3.
[2] Construct then circles {c1, c2, c3} inscribed in the triangles {KE2E3, KE3E1, KE1E2}. Next step is to show that these circles contact the sides of the triangle precisely at the points {D1, D2, D3}, where the inner tangents of the ki's intersect the sides of the triangle.
[3] In fact:
E1D2 - E3D2 = E1B2 - E3C2 = E1P1 - E3P3 = E1K - E3K.
This implies that D2 is the contact point of c2 wiht line A1A3. Analogous is the proof for D1, D3.
[4] Next consider the other than {D1E1, D2E2, D3E3} inner tangents of the three circles {c1, c2, c3}. They are three lines {L1, L2, L3} 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 {Fj, Gj} the contact points of the circles {ci} with lines {DjEj}. Because of the symmetry relation of lines {Li} to {DiEi} to show that L1 passes through A1 it suffices to show that length FiGi = A1D2 - A1D3 (since, by symmetry, F1G1 is equal to the distance of contact points of c2, c3 and L1). But:
A1D2 - A1D3 = C2D2 - B3D3 = P3G3 - P2F2 = F1G1. Thus, L1 passes through A1. Analogous is the proof for lines L2 and L3.
[6] To identify L1 with the bisector we show that A1 is on the circle of similitude of {c2, c3} (see CirclesSimilar.html ) hence it is viewing the two circles under equal angles. For this it suffices to show that the chords intersected by {c2, c3} on line D2D3 are equal. This in turn is equivalent with the equality of the powers p(D2, c3) = p(D3, c2), which amounts to the equality of the tangents to these circles D2F2 = D3G3.
[7] To show D
2F2 = D3G3, notice that D3G3 = D3P3 + P3G3 = D3B3 + D2C2 = P2F2 +D2P2 = D2F2 .(*) Thus line L1 is identified with the bisector and analogous arguments show that L2, L3 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).

3. Hart's proof of existence

The existence proof of Hart involves a continuity argument. Consider a small circle k1 tangent to sides {AB, AC} and two other circles {k2, k3}, each tangent to k1 and two sides of the triangle.

The three circles depend continuously on the radius r of circle k1 (and the triangle data of course). For small values of r the two circles {k2, k3} intersect. For larger values of r the two circles are disjoint. Hence there is some intermediate value r0 for which the two circles {k2, k3} are tangent. Since these circles remain all the time tangent to k1 we obtain in this way a solution to Malfatti's problem.

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