Malfatti's Problem, Hart's Proof
Malfatti's 200 years old problem
To draw within a given triangle three circles each of which is tangent to the other two and to two sides of the triangle. 
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 A_{1}A_{2}A_{3}, 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 ΔA_{1}A_{2}A_{3} plays an important role:
Draw three angle bisectors IA_{1}, IA_{2}, IA_{3}.
Find the incircle c_{k} of triangle IA_{i}A_{j}. Note that the angle bisector IA_{i} serves as a common internal tangents for circles c_{j} and c_{k}.
For each pair of the circles consider the second internal tangents. The latter concur in a point (L) and cross the sides A_{i}A_{j} in points D_{k}, as shown in the applet.
The quadrilaterals A_{i}D_{j}LD_{k} are inscriptible. Their incircles solve Malfatti's problem.
What if applet does not run? 
Proof
Assuming Malfatti's problem solved, two of the circles touch A_{i}A_{j}. There point of contact is denoted P_{k}. Their common tangent through P_{k} meets A_{i}A_{j} in D_{k}. The points of contact with A_{i}A_{j} are B_{k}, C_{k}, as in the applet below. The line D_{k}P_{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_{k}P_{k} crosses the perimeter of triangle A_{1}A_{2}A_{3} in a point E_{k} different from D_{k}. Which side E_{k} lies on  either A_{k}_{i} or A_{j}_{k}  depends on a specific configuration. To be specific, we shall assume that E_{1} and E_{3} lie on A_{1}A_{3}. Then

It therefore appears that D_{2} is the point of contact of the incircle of ΔA_{1}LA_{3} with A_{1}A_{3}. This fact actually does not depend on whether points E_{1} and E_{3} lie between A_{1} and A_{3} 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_{k}E_{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_{k}E_{i} in D_{j}. Note that

Hence, the second common tangent of the two circles goes through A_{1}, and similar considerations apply to the other two vertices. Further,

We see then that the points D_{2} and D_{3} on the incircles D_{2}F_{1}G_{3} and D_{3}F_{2}G_{1} 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 D_{2}F_{1}G_{3} and D_{3}F_{2}G_{1}. The tangents being A_{1}D_{2} and A_{1}D_{3}, we conclude that A_{1} lies on the circle of similitude of the two circles D_{2}F_{1}G_{3} and D_{3}F_{2}G_{1}. By a property of the circle of similitude, the line from A_{1} to the internal center of similarity of the two circles bisects the vertex angle A_{1}. But, as we have seen, this line is tangent to the circles D_{2}F_{1}G_{3} and D_{3}F_{2}G_{1}. We conclude that the bisector of the vertex angle at A_{1} is one of the common tangents the circles D_{2}F_{1}G_{3} and D_{3}F_{2}G_{1}. It follows that the circle D_{j}F_{i}G_{k} is inscribed into triangle A_{i}LA_{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):
Conversely, if a very small circle be drawn tangent to two sides of the triangle, the two circles each touching this little circle and two other sides will surely intersect. But if the little circle swells up, always touching the two sides until it becomes the incircle, the other two circles shrinking in the process, are eventually separated it. Hence, for some intermediate value of the little circle, the three will touch. Hart's solution is thus complete. 
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.
References
 J. L. Coolidge, A Treatise On the Circle and the Sphere, AMS  Chelsea Publishing, 1971
 F. G.M., Exercices de Géométrie, Éditions Jacques Gabay, sixiéme édition, 1991, pp. 726728
 D. O. Shklyarsky, N. N. Chentsov, I. M. Yaglom, Selected Problems and Theorems of Elementary Mathematics, v 2, Moscow, 1952 (in Russian)
Activities Contact Front page Contents Geometry
Copyright © 19962018 Alexander Bogomolny63580286 