162. Generalization of proposed problem from Mongolia, 2000.

by Virgil Nicula, Oct 20, 2010, 6:33 PM

Let $ABC$ be a triangle. Consider a point $P$ with the normalized barycentrical coordinates $(x,y,z)$ w.r.t. $\triangle ABC$ . Denote $D\in AP\cap BC$ ,

$E\in BP\cap CA$ , $F\in CP\cap AB$ . Prove that $AEDF$ is a cyclical quadrilateral $\iff$ $\frac {a^2}{x(1-x)}=\frac {b^2}{y(1-y)}+\frac {c^2}{z(1-z)}$ .

Particular cases.

$\blacktriangleright$ If $P:=I\left(\frac {a}{2s},\frac {b}{2s},\frac {c}{2s}\right)$ (incenter), then the quadrilateral $AEDF$ is cyclically $\iff$ $\frac {a}{b+c}=\frac {b}{a+c}+\frac {c}{a+b}$ (Mongolia, 2000).

$\blacktriangleright$ If $P:=N\left(\frac {s-a}{2s},\frac {s-b}{2s},\frac {s-c}{2s}\right)$ (Nagel's point), then the quadrilateral $AEDF$ is cyclically $\iff$ $\frac {a}{b+c-a}=\frac {b}{a+c-b}+\frac {c}{a+b-c}$ .

$\blacktriangleright$ If $P:=S\left(\frac {a^2}{\sum a^2},\frac {b^2}{\sum a^2},\frac {c^2}{\sum a^2}\right)$ (Lemoine's point) , then the quadrilateral $AEDF$ is cyclically $\iff$ $\frac {a^2}{b^2+c^2}=\frac {b^2}{a^2+c^2}+\frac {c^2}{a^2+b^2}$

Remark. The quadrilateral $BFEC$ is cyclically $\iff$ $\frac {b^2}{y(1-y)}=\frac {c^2}{z(1-z)}$ .


Proposed problem. Let $ABC$ be a triangle. Denote $\left\|\begin{array}{c}
\{M,D,L\}\subset BC\\\\
MB=MC\ ,\ AD\perp BC\\\\
\widehat{LAB}\equiv\widehat{LAC}\end{array}\right\|$ . Prove that $a^2\cdot \overrightarrow{AD}+4s(s-a)\cdot\overrightarrow{AM}=(b+c)^2\cdot\overrightarrow{AL}$ .
This post has been edited 10 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:44 PM

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