Malfatti-Steiner Problem I. A. Sakmar, University of ... - MAA Sections

Thus we have the following segments of S t which are all equal.
C 1 C 2 = EL = C 2 K = XY = ßZ
Consequently we can draw a circle which is tangent to all those segments. Let us
call this circle S v . Obviously S v has the same center as the circle S t . this circle
is also seen to be the in-circle of the triangle ABO where O is the intersection
point of the bisectors of the triangle ABC.
Moreover the tangent to the circles S a , and S c at ß is also tangent to this circle
S v .
Finally this very tangent is also tangent to the in-circle of the triangle OBC,
which we will call S w . The reason for this is that the roles of S v and S w with
respect to the circles S a , and S c are similar. Had we focused on S w instead of
on S v we would find that the tangent line to S a and S c at ß is also tangent to
the circle S w . This finally proves Steiner’s conjecture.
Fig.13 Steiner’s Conjecture:
Suppose the problem of finding three Malfatti circles is solved. Then the tangent
line to two of these three circles , say S a and S c at their contact point ß is
tangent to both of the circles S v , and S w inscribed into the triangles ABO and
OBC where O is the intersection point of the bisectors of the angles of the
given triangle ABC.
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