Admin

Dec 23 2006, 04:13 AM

Note: A1 means A sub 1.

Draw three angle bisectors IA1, IA2, IA3.

Find the incircle ck of triangle IAiAj. Note that the angle bisector IAi serves as a common internal tangents for circles cj and ck.

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

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

Assuming Malfatti's problem solved, two of the circles touch AiAj. There point of contact is denoted Pk. Their common tangent through Pk meets AiAj in Dk. The points of contact with AiAj are Bk, Ck, as in the applet below. The line DkPk 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 DkPk crosses the perimeter of triangle A1A2A3 in a point Ek different from Dk. Which side Ek lies on -- either Aki or Ajk -- depends on a specific configuration. To be specific, we shall assume that E1 and E3 lie on A1A3. Then

E1D2 - E3D2 = E1B2 - E3C2
= E1P1 - E3P3
= E1L - E3L.


It therefore appears that D2 is the point of contact of the incircle of A1LA3 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 LEi, LEk, and EkEi. The Malfatti circle inscribed into this triangle touches LEi in Fi, LEk in Gk, and, as we agreed before, EkEi in Dj. Note that

A1D2 - A1D3 = C2D2 - B3D3
= P3G3 - P2F2
= P1F1 - P1G1
= F1G1.


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

D2F2 = D2P2 + P2F2
= D2C2 + D3B3
= P3G3 + D3P3
= D3G3.


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 DjFiGk is inscribed into triangle AiLAk. Thus, if Malfatti's problem has a solution, the solution can be obtained by Steiner's construction.

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.