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Given a triangle I want to construct three tangent circles inscribed in the triangle (every two sides of the triangle are tangent lines of one of the circles). For better understanding of the problem I tried to draw desired result (picture is only approximative, I do not know exact construction). Could you give me some hint, how exact construction can be done?

enter image description here

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  • $\begingroup$ The circle tangent to $A B$ and $A C$ lies on the line bisecting angle $B A C$. From there I'd write equations for the circles, and look for the conditions that make the two intersection points between the circles centered at $S_1$ and $S_2$ coincide. $\endgroup$ – vonbrand Feb 5 '13 at 22:20
  • $\begingroup$ Are you looking for a construction using straightedge and compass? $\endgroup$ – Sgernesto Feb 5 '13 at 23:16
  • $\begingroup$ This is a special case of a circle packing, so I know how a solution can be computed using a variational principle. But that won't be of much use for a construction with straightedge and compass. $\endgroup$ – MvG Feb 5 '13 at 23:50
  • $\begingroup$ Yes, I am looking for a construction using straightedge and compass. However if direct construction is not possible I am also interested in some computational solution. $\endgroup$ – mcihak Feb 6 '13 at 5:41
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It seemed to me that since there are many different centers in a triangle, one of them might be useful for your construction. Going through the links of various centers, I stumbled upon a name for the problem you describe: these three circles are known as Malfatti circles.

The Wikipedia article has some details on a construction by Steiner. ℝⁿ was kind enough to add a comment below, providing links to details about some of the steps in that construction, namely the construction of bitangents and incircles to quadrilaterals. The latter seems obvious in hindsight…

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I put a detail steps to show how to construct circles by straight and compass. the principle is provided by MvG.

enter image description here

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  • $\begingroup$ Have you noticed that $A_2A_3 = B_2B_3 = C_2C_3 = s+r-IA-IB-IC$, where $s$ is the half-perimeter and $r$ the inradius of the triangle? This shortens the construction. $\endgroup$ – chizhek Sep 25 '17 at 13:39

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