Concurrency in the Intouch Triangle
What is this about?
A Mathematical Droodle


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Explanation

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Copyright © 1996-2017 Alexander Bogomolny


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

The applet may suggest the following statement [Savchev, pp. 141-141]:

In ΔABC, the points A', B', C' of tangency of the incircle with the sides ΔABC form the intouch ΔA'B'C'. From the midpoints Am, Bm, Cm of the sides of the latter drop perpendiculars AmAh, BmBh, CmCh on the opposite sides of ΔABC. The three lines AmAh, BmBh, CmCh are concurrent.

The proof is remarkably short and reminds one of that employed to prove the concurrency of maltitudes in an inscribed quadrilateral.

Recollect that the medians in a triangle are divided in ratio 1:2 by the center of the triangle. We apply this fact to the intouch triangle A'B'C'. (I'll use G' to denote its center, although the point is not named in the applet.) This means that points A', B', C' can be obtained from points Am, Bm, Cm by a homothety with coefficient -2 and center G'. A homothety maps straight lines onto parallel lines. Therefore, say, A'Am is mapped onto a line through A' parallel to A'Am and thus perpendicular to BC. Since BC is tangent to the incircle of ΔABC, that line passes through its incenter I. The same holds of the images of B'Bm and C'Cm. The three images therefore concur (in the incenter I of ΔABC.) Since homothety is a reversible transformation, the lines themselves are also concurrent. Furthermore, the point of concurrency S and the incenter I are collinear with G' and 2·SG' = G'I.

References

  1. S. Savchev, T. Andreescu, Mathematical Miniatures, MAA, 2003

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Copyright © 1996-2017 Alexander Bogomolny

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