Malfatti's Problem, Hart's Proof

Malfatti's 200 years old problem

has a curious history, with a milestone in 1826 when J. Steiner published his solution without proof. There are many proofs of Steiner's construction, see, for example, references at the bottom of the page. Below, I shall follow Hart's solution (1856) found in [Coolidge].

The proof introduces an unusual (for discussions at this site) amount of points, lines, and circles. Which suggests that there may be a benefit in changing notations. The base triangle will be denoted A₁A₂A₃, with a heavier use of indices compared to a more standard ABC. In Steiner's construction and Hart's solution, geometric elements come in triples related to the three vertices. Three distinct indices involved will be denoted i, j, k, with an implied order. In Steiner's construction, the incenter I of ΔA₁A₂A₃ plays an important role:

1. Draw three angle bisectors IA₁, IA₂, IA₃.

2. Find the incircle c_k of triangle IA_iA_j. Note that the angle bisector IA_i serves as a common internal tangents for circles c_j and c_k.

3. For each pair of the circles consider the second internal tangents. The latter concur in a point (L) and cross the sides A_iA_j in points D_k, as shown in the applet.

4. The quadrilaterals A_iD_jLD_k are inscriptible. Their incircles solve Malfatti's problem.

Proof

Assuming Malfatti's problem solved, two of the circles touch A_iA_j. There point of contact is denoted P_k. Their common tangent through P_k meets A_iA_j in D_k. The points of contact with A_iA_j are B_k, C_k, as in the applet below. The line D_kP_k is the radical axes of the two circles; three such lines meet in the radical center of the three circles, which we denote L. Each of the lines D_kP_k crosses the perimeter of triangle A₁A₂A₃ in a point E_k different from D_k. Which side E_k lies on -- either A_ki or A_jk -- depends on a specific configuration. To be specific, we shall assume that E₁ and E₃ lie on A₁A₃. Then

It therefore appears that D₂ is the point of contact of the incircle of ΔA₁LA₃ with A₁A₃. This fact actually does not depend on whether points E₁ and E₃ lie between A₁ and A₃ and holds equally well in other configurations. Thus it holds for each of the three triangles with the side lines LE_i, LE_k, and E_kE_i. The Malfatti circle inscribed into this triangle touches LE_i in F_i, LE_k in G_k, and, as we agreed before, E_kE_i in D_j. Note that

Hence, the second common tangent of the two circles goes through A₁, and similar considerations apply to the other two vertices. Further,

We see then that the points D₂ and D₃ on the incircles D₂F₁G₃ and D₃F₂G₁ have equal tangent segments with respect to the other circle. The tangents at these points therefore meet on the circle of similitude of the two circles D₂F₁G₃ and D₃F₂G₁. The tangents being A₁D₂ and A₁D₃, we conclude that A₁ lies on the circle of similitude of the two circles D₂F₁G₃ and D₃F₂G₁. By a property of the circle of similitude, the line from A₁ to the internal center of similarity of the two circles bisects the vertex angle A₁. But, as we have seen, this line is tangent to the circles D₂F₁G₃ and D₃F₂G₁. We conclude that the bisector of the vertex angle at A₁ is one of the common tangents the circles D₂F₁G₃ and D₃F₂G₁. It follows that the circle D_jF_iG_k is inscribed into triangle A_iLA_k. Thus, if Malfatti's problem has a solution, the solution can be obtained by Steiner's construction.

Up to this point, the derivation was relatively elementary, if not simple. Now I quote verbatim from Coolidge (p. 176):

This argument, although simple, is well beyond elementary. The argument depends on the notion of continuity and the Intermediate Value Theorem to be made rigorous.

1. J. L. Coolidge, A Treatise On the Circle and the Sphere, AMS - Chelsea Publishing, 1971
2. F. G.-M., Exercices de Géométrie, Éditions Jacques Gabay, sixiéme édition, 1991, pp. 726-728
3. D. O. Shklyarsky, N. N. Chentsov, I. M. Yaglom, Selected Problems and Theorems of Elementary Mathematics, v 2, Moscow, 1952 (in Russian)

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