Problem 4 IMO 2009 (Day2)

Soit ABC un triangle isocel en A. le bissectrice interieur de l'angle CAB coupe [BC] en D. et le bissectrice interieur de l'angle ABC coupe [AC] en E.Soit K le centre du cercle inscrit au triangle ADC.

On suppose que BEK=45°, trouvez toutes les valeurs possibles de l'angle CAB.

j'ai lu le lien : il donne deux solutions : 60° ou 90°


Ou est l'ERREUR ?

je prends CAB = 40°

====> BEK = 45° , CEK = 30° , BEA = 105°

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ce que je trouve c'est une relation : CEK = 3/4.CAB

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Salut,

Soit M le point d'intersection de EK avec BC
vous avez choisit CAB=40°

et comme ABC est isocel alors ABC=70 et ACB=70

donc EBC=35 et puique BEK=45 alors

BME=100
d'ou EMC=80 et comme CEK=35

alors MCE=65° ce ki né pas correcte puisk MCE=70°

CEK = 30° et non pas 35°

je vois pas ta déduction : MCE = 65°

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je tiens à ma relation ( je peux poster la démo.)

comme 0 < CEK < 135° ( = 180°-45°)

=====> 0 < CAB < 140°

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Let K,H be the incenter of triangle ADC, ABC.AK\cap HE = O. Put AB = AC = x, BC = y
\angle AKH = \angle HEK = 45^o then HK^2 = HO.HE
But \frac {HO}{HE} = \frac {AH}{AH + AE}, \frac {HK^2}{HC^2} = \frac {HD^2}(HD + DC)^2}
We get \frac {HC^2.HD^2}{(HD + DC)^2} = \frac {HE^2.AH}{AH + AE}
\Rightarrow \frac {HD^2}{(HD + DC)^2} = \frac {HE^2}{HC^2}.\frac {AH}{AH + AE} = \frac {AE^2}{AB^2}.\frac {AH}{AH + AE}
\Rightarrow \frac {HD}{(HD + DC)^2} = \frac {AE^2}{AB^2}.\frac {AH}{HD(AH + AE)}
\Rightarrow \frac {HD}{(HD + DC)^2} = \frac {AE^2}{AB^2}.\frac {AB}{BD(AH + AE)} = \frac {AE^2}{AB.BD.(AH + AE)}
\Rightarrow AE^2.(HD + DC)^2 = HD.AB.BD.(AH + AE)
\Rightarrow \frac {x^4}{(x + y)^2}.(\frac {1/2y.AD}{1/2y + x} + 1/2y)^2 = \frac {1/2y.AD}{1/2y + x}.x.1/2y.(\frac {x.AD}{1/2y...
\Rightarrow \frac {x^2}{(x + y)^2}.(\frac {1/2AD}{1/2y + x} + 1/2)^2 = \frac {1/2AD}{1/2y + x}.1/2.(\frac {AD}{1/2y + x} + \f...
Then set x = ky and note that AD = \sqrt {k^2 -1/4}y we can easy find k.
\Rightarrow \angle A = 60^o or 90^o.

This is my solution, I don't no if it is true

First, we prove that n devide a1(ak-1)
then we prove that n don't devide ak(a1-1)

We have
n devide a1(a2-1) then there exist x1 that a1(a2-1)=x1.n ==>a1a2=x1.n+a1

multiplaying by (a3-1) then a1a2(a3-1)=x1.n.(a3-1)+a1(a3-1)

do a1(a3-1) is devided by n because a1a2(a3-1) is.

We repeat the same thing with a1(a3-1)

a1(a3-1) is devided by n then exist y such :
a1(a3-1)=y.n ==> a1a3=y.n+a1

multiplaying by (a4-1) .....(1)

a1a3(a4-1)=y.n(a4-1)+a1(a4-1)

a1a3(a4-1) is devided by n because a3(a4-1) is devided by n

then a1(a4-1) .........(2)


We continue, until we get that a1(ak-1) is devided by n

2) To prove that ak(a1-1) is not devided by n

we have ak(a1-1) = a1(ak-1) - (ak-a1)

a1(ak-1) s devided by n then :

ak(a1-1) is devided by n if (ak-a1) is devided by n

However ak-a1 < n then is not devided by n

Do ak(a1-1) is not devided by n..


Finally ,reply me if there is an error or a remark,please

Thank you