Construction of three tangent circles in a triangle

Question

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

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. — vonbrand, Feb 05 '13 at 22:20
Are you looking for a construction using straightedge and compass? — Sgernesto, Feb 05 '13 at 23:16
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. — MvG, Feb 05 '13 at 23:50
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. — mcihak, Feb 06 '13 at 05:41

score 6 · Accepted Answer · edited Apr 13 '17 at 12:21

6

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…

edited Apr 13 '17 at 12:21

Community

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answered Feb 06 '13 at 01:43

MvG

42,596

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Certainly you can construct bitangents using a straightedge and compass, and the incircle to a tangential quadrilateral is centered on the point of concurrency of its angle bisectors. – Feb 06 '13 at 04:12

score 4 · Answer 2 · answered Mar 18 '13 at 06:05

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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

answered Mar 18 '13 at 06:05

chenbai

7,581

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. – chizhek Sep 25 '17 at 13:39

Construction of three tangent circles in a triangle

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