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Thébault's Problem III, Proof

 

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.


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The applet illustrates a synthetic solution for Thébault's Problem III. See if you can surmise what is it about before proceeding to the explanation.

Copyright © 1996-2009 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

 

 

The applet illustrates a solution to Thébault's Problem III by Jean-Louis Ayme (2003).

 

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.


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What if applet does not run?

By the construction, PE and QG are both perpendicular to BC; so PE||QG. By Sawayama's Lemma, both EF and GH pass through the incenter I of ΔABC. Triangles DHG and QGH being isosceles in D and Q, DQ is

  1. the perpendicular bisector of GH,
  2. the D-internal bisector of ΔDHG.

Mutatis mutandi, DP is

  1. the perpendicular bisector of EF,
  2. the D-internal bisector of ΔDEF.

As the bisectors of two adjacent and supplementary angles are perpendicular, DQ ⊥ DP. Therefore, GH||DP and DQ||EF. In the hexagon PEIGQD the sides are pairwise parallel, i.e. concur Point at infinityat infinity and therefore the points of incidence all lie on a line at infinity. By the converse of Pappus' Theorem, the points P, I, Q are collinear.

Michel Cabart found an alternative ending:

D is barycenter of E and G, with coefficients DG and DE, thus

(1) ID = (DG/GE)·IE + (DE/GE)·IG

Let α be the angle of DP with respect to DE

  IE = GE·sinα  thus IE = (GE sinα)·(QD/QD)
IG = GE·cosα  thus IG = (GE cosα)·(PD/PD)

Replacing in (1) yields

 ID= (DG sinα/QD)·QD + (DE cosα/PD)·PD
  = (sin²α)·QD + (cos²α)·PD.

thus I is barycenter of Q and P with coefficients sin²α and cos²α and, as such, is collinear with both.

References

  1. J.-L. Ayme, Sawayama and Thébault's Theorem, Forum Geometricorum, v 3 (2003), 225-229
  1. Thébault's Problem I
  2. Thébault's Problem II
  3. Thébault's Problem III
  4. Circles Tangent to Circumcircle

Copyright © 1996-2009 Alexander Bogomolny

34222694Page copy protected against web site content infringement by Copyscape


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