Let $ABC$ be a triangle with $AB \neq BC$ and let $BD$ the interior bisectrix of $ \angle ABC$ with $D \in AC$ . Let $M$ be the midpoint of the arc $AC$ that contains the point $B$ in the circumcircle of the triangle $ABC$ .The circumcircle of the triangle $BDM$ intersects the segment $AB$ in $K \neq B$ . Denote by $J$ the symmetric of $A$ with respect to $K$ .If $DJ$ intersects $AM$ in $O$ then prove that $J,B,M,O$ are concyclic.
Problem
Source: Romania JBMO TST 2015 Day 2 Problem 4
Tags: geometry, Concyclic, circumcircle, angle bisector
AdithyaBhaskar
14.05.2015 11:09
I think we can apply sine rule in triangles $ ABD$ and $BDJ$ after finding $\angle AKD$ (which I have not yet found), to get $ \angle KJD = C$.....
THVSH
14.05.2015 11:18
My solution: Let $N$ be the midpoint of $AC$. Then $B, M, N, D, K$ lie on $\odot (MD)$. $\Longrightarrow AK.AB=AD.AN \Longrightarrow AJ.AB=2AK.AB=2AD.AN=AD.AC \Longrightarrow B,J,D,C$ are concyclic. $\Longrightarrow \angle{ADO}=\angle{JBC}\equiv \angle{ABC}=\angle{AMC} \Longrightarrow D,O,M,C$ are concyclic. $\Longrightarrow \angle{JOM}=\angle{AOD}=\angle{ACM}=180^{\circ}-\angle{JBM}$ $\Longrightarrow J, B, M, O$ are concyclic. Q.E.D
TelvCohl
14.05.2015 11:24
See AZE JBMO TST
tranquanghuy7198
15.05.2015 20:27
Let $\odot{(MD)}\cap{BC} = L$ We have $\triangle{MAK} = \triangle{MCL}$ $\Rightarrow$ $AK = CL, MK = ML$ $\Rightarrow$ $JK = CL, DK = DL$ $\Rightarrow$ $\triangle{DKJ} = \triangle{DLC}$ $\Rightarrow$ $\angle{DJK} = \angle{DCL}$ $\Rightarrow$ $\angle{OJB}+\angle{OMB} = 180$ Q.E.D
aditya21
15.05.2015 20:54
what do you mean by this Denote by $J$ the symmetric of $A$ with respect to $K$?? is $J$ reflection of $A$ wrt $K$ that is $K,A,J$ are collinear with $AK=KJ$?
samithayohan
02.06.2015 06:17
First part of my solution is similar to THVSH. Quote: [/Let $N$ be the midpoint of $AC$. Then $B, M, N, D, K$ lie on $\odot (MD)$. $\Longrightarrow AK.AB=AD.AN \Longrightarrow AJ.AB=2AK.AB=2AD.AN=AD.AC \Longrightarrow B,J,D,C$ are concyclic.] But the rest is very easy than that.Since JDCB are cyclic,$/angleAJD=/angleDCB=/angleAMB$ Hence we are done.