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 k_i'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 {F_j, G_j} the contact points of the circles {c_i} with lines {D_jE_j}. Because of the symmetry relation of lines {L_i} to {D_iE_i} to show that L₁ passes through A₁ it suffices to show that length F_iG_i = 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 D₂F₂ = 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.

See Also

Isogonal_3TangentCircles.html
CirclesSimilar.html

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