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

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Fig.1 Take the triangle ABC.Consider the problem solved. Call the circles S a , S b, and S c . Let their mutual contact points be a , ß , and ?. Let the points where they touch the sides AB and CA be C 1 , C 2 and ß 1 , ß 2 Fig.2 Draw the circle which is tangent to the circles S a , and S b at their contact point ß and goes through the point C 2 . Call this circle S M, and its center M. -2-
Call the intersection of S M with the side CA D. Draw two tangent lines to the circle SM at points C 2 and D. Call the intersection point of these tangent lines F. We claim that the circle S M intersects the circle S b and the side CA under equal angles. That is, the angle ∠ AC 2 F is complementary to the angle ∠ FDA. Fig.3 Proof: Theorem: Given two circles S 1, and S 2 which intersect each other at points A, and B. Any circle C which goes through A and B intersects all circles like C a , and C ß which are tangent to S 1 , and S 2 under equal angles. (Complementary angles if the circles are on opposite sides.) -3-
Page 1: Malfatti-Steiner Problem I.A.Sakmar
Page 5 and 6: In our case the circles S 1 , and S
Page 7 and 8: Fig.6 Theorem: Given two mutually t
Page 9 and 10: This shows that E is on the circle
Page 11 and 12: The circle S C which was tangent to
Page 13 and 14: This means that E is on the bisecto
Page 15 and 16: Fig.14 Since Steiner’s line is so
Page 17 and 18: A C K N O W L E D G M E N T S I ded

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

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