Let $AH$ be the altitude of a given triangle $ABC.$ The points $I_b$ and $I_c$ are the incenters of the triangles $ABH$ and $ACH$ respectively. $BC$ touches the incircle of the triangle $ABC$ at a point $L.$ Find $\angle LI_bI_c.$
Problem
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Tags: geometry, incenter, geometry proposed
castigioni
29.10.2010 14:08
The angle is 45 degrees
Martin N.
29.10.2010 14:42
castigioni wrote: The angle is 45 degrees What? The angle is $\frac{\pi}{2}=90^{\circ}$! I remember having seen this problem on Mathlinks, but I can't find it right now...
Amir Hossein
29.10.2010 14:52
Martin N. wrote: castigioni wrote: The angle is 45 degrees What? The angle is $\frac{\pi}{2}=90^{\circ}$! I remember having seen this problem on Mathlinks, but I can't find it right now... Really? I'm posting Russia Sharygin Geometry 2010 Problems. It will be finished soon...
Martin N.
29.10.2010 15:54
Oh uups I'm sorry, I misread the problem statement... The problem I know is to prove that $\sphericalangle{I_bLI_c}=\frac{\pi}{2}$.
castigioni
29.10.2010 16:30
The question is about angle LIbIc not angle IbLIc.The triangle LIbIc is right in L and isosceles .So the others angles are 45 degrees
oneplusone
07.11.2010 07:56
Let the feet of perpendiculars from $I_b,I_c$ onto $BC$ be $D,E$. Then note that $DI_b=DH$ and $EI_c=EH$. Also $2DH=AH+BH-AB$. But since $2LE=AC+CB-AB-AC-CH+AH=AH+BH-AB=2DH$, similarly we have $DL=HE=EI_c$. Thus $\triangle I_bDL\equiv \triangle LEI_c$, so $\angle I_bLI_c=90$.