Given an acute triangle $ ABC$. The incircle of triangle $ ABC$ touches $ BC,CA,AB$ respectively at $ D,E,F$. The angle bisector of $ \angle A$ cuts $ DE$ and $ DF$ respectively at $ K$ and $ L$. Suppose $ AA_1$ is one of the altitudes of triangle $ ABC$, and $ M$ be the midpoint of $ BC$. (a) Prove that $ BK$ and $ CL$ are perpendicular with the angle bisector of $ \angle BAC$. (b) Show that $ A_1KML$ is a cyclic quadrilateral.
Problem
Source: Indonesian MO (INAMO) 2009, Day 2, Problem 8
Tags: geometry, trigonometry, circumcircle, angle bisector, perpendicular bisector, geometry unsolved
plane geometry
08.08.2009 15:07
according to http://www.mathlinks.ro/viewtopic.php?p=1525553#1525553 we have $ AK\bot BK$ $ AK\bot CL$ Notice ML=MK=MD=c-b => ML//BA done
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wangsacl
08.08.2009 16:32
How do you get $ MK=ML=MD$?
babylon
08.08.2009 16:56
$ MK||AC$ $ \Rightarrow$ $ \angle CED=\angle MKD$ and $ \angle CDE=\angle MDK$ so $ MD=MK$ $ ML||AB$ $ \Rightarrow$ $ \triangle MLD\sim \triangle BDF$ $ \Leftrightarrow$ $ MD=ML$
plane geometry
08.08.2009 17:21
wangsacl wrote: How do you get $ MK = ML = MD$? AC'=AC => ML=c-b AB=AB' => MK=c-b
plane geometry
08.08.2009 17:31
babylon wrote: $ MK||AC$ $ \Rightarrow$ $ \angle CED = \angle MKD$ and $ \angle CDE = \angle MDK$ so $ MD = MK$ $ ML||AB$ $ \Rightarrow$ $ \triangle MLD\sim \triangle BDF$ $ \Leftrightarrow$ $ MD = ML$ No, in this step we have not proved AB//ML because we want use ML=MK=c-b to prove AB//ML MK//AC
wangsacl
09.08.2009 09:01
Wow, nice. I use trigonometry to prove $ MK=ML$ (which is quite long and complicated)...
jayme
09.08.2009 13:36
Dear Mathlinkers, 1. for (a) see : http://perso.orange.fr/jl.ayme/ vol. 4, An unlikely concurrence p. 5-6 2. for (b) see : http://perso.orange.fr/jl.ayme/ vol. 1, Feuerbach's theorem p. 3 this is the "Calabre's circle". Sincerely Jean-Louis
jgnr
10.08.2009 14:17
Extend $ LM$ until it intersects $ BK$ at $ X$. Triangle $ BMX$ is congruent to triangle $ CML$. Note that $ KM$ is a median of right triangle $ KLX$, thus $ KM=LM$. With some angle chasing, we get $ \angle LA_1M=\angle MA_1K$. So $ M$ is the intersection of the angle bisector of $ \angle LA_1K$ and the perpendicular bisector of $ KL$. Clearly $ M$ is on the circumcircle of triangle $ LA_1K$, as desired.
stergiu
10.08.2009 18:45
An easy hint for b): We have a) $ \angle MLK = \angle A / 2$, since $ LM // AB$ b) $ \angle MA_1K = \angle BA_1K = \angle BAK = \angle A /2$, since $ BKA_1A$ is insribed($ \angle AA_1B = 90 = \angle AKB$). Hence $ \angle MLK = \angle MA_1K$ . Babis