91. JBMO 2010, Problem 3.

by Virgil Nicula, Aug 27, 2010, 5:29 PM

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1915250&sid=5954534452a76146ace61a65ec21fa85#p1915250
Quote:
Let $AL$ and $BK$ be angle bisectors in the non-isosceles triangle $ABC$ , where $L\in (BC)$ and $K\in (AC)$ . The perpendicular bisector of $BK$ intersects the line $AL$ at point $M$. Point $N$ lies on the line $BK$ such that $LN$ is parallel to $MK$. Prove that $LN = NA$.
Remark. An easy and nice proposed problem ! Observe that always $AB>AK$ . Indeed, $m(\widehat{AKB})=\frac B2+C>\frac B2=$ $m(\widehat{ABK})$

$\implies$ $m(\widehat{AKB})>m(\widehat{ABK})\implies AB>AK$ . Otherwise, $AB>AK\Longleftrightarrow c>\frac {bc}{a+c}\Longleftrightarrow a+c>b$ what is truly always !
Quote:
An easy extension. Let $AL$ be the angle bisector in the non-isosceles triangle $ABC$ , where $L\in (BC)$ and let $X\in (AL)$ be a point for which denote $K\in BX\cap AC$ . The perpendicular bisector of $BK$ intersects the line $AL$ at the point $M$. The point $N$ lies on the line $BK$ such that $LN\parallel MK$. Prove that $\frac {NA}{NL}=\frac {XA}{XL}\cdot\frac {IL}{IA}$ .
Proof Is well-known that "in any nonisosceles triangle the intersection point between the bisector of an angle of triangle and the perpendicular bisector of the opposite side belongs to the circumcircle of the given triangle". Apply this property to the triangle $ABK$ and the vertex $A$ . Thus obtain that the quadrilateral $ABMK$ is cyclically $\implies$ $\widehat{BAM}\equiv\widehat {BKM}\equiv\widehat{BNL}\implies$ $\widehat{BAL}\equiv\widehat{BNL}\implies$ the quadrilateral $BANL$ is cyclically $\implies$ $\frac {NA}{NL}=\frac {\sin\widehat {NBA}}{\sin\widehat{NBL}}=\frac {\sin\widehat {XBA}}{\sin\widehat{XBL}}$ . Since $\frac {XA}{XL}=$ $\frac {BA}{BL}\cdot\frac {\sin \widehat{XBA}}{\sin\widehat{XBL}}=$ $\frac {IA}{IL}\cdot\frac {\sin \widehat{XBA}}{\sin\widehat{XBL}}$ obtain $\frac {\sin \widehat{XBA}}{\sin\widehat{XBL}}=\frac {XA}{XL}\cdot\frac {IL}{IA}$ . In conclusion $\frac {NA}{NL}=\frac {XA}{XL}\cdot\frac {IL}{IA}$ .

Remark. In the particular case when $X:=I$ obtain the proposed problem. Another interesting particular case obtain such : denote $P\in AB\cap CI$ , $R\in AC\cap BI$ , $L\in BC\cap AI$ and $S\in PR\cap AI$ . Then for $X:=S$ obtain $NA=2\cdot NL$ . This results from the well-known relation $\frac {IA}{IL}=2\cdot\frac {SA}{SL}$ .
This post has been edited 3 times. Last edited by Virgil Nicula, Nov 23, 2015, 2:02 PM

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