106. OLM - Bucuresti (geometrie).

by Virgil Nicula, Sep 9, 2010, 3:04 PM

Quote:
Let $ABCD$ be a convex quadrilateral for which $O\in AC\cap BD$ . The bisector of $\angle AOB$ meet

$AB$ , $CD$ in $E$ , $F$ respectively. Prove that $\frac {OE}{OF}\ =\ \frac {OA\cdot OB}{OA+OB}\ \cdot\ \frac {OC+OD}{OC\cdot OD}$ .
Method 1. Denote $\phi =m(\angle AOB)$ . I"ll use the length $l_a$ of the $A$ bisector - $l_a=\frac {2bc\cdot\cos\frac A2}{b+c}$ .

Thus, $\left|\ \begin{array}{c} \triangle AOB\ : \ OE=\frac {2\cdot OA\cdot OB\cdot\cos \phi}{OA+OB}\\\\ \triangle COD\ : \ OF=\frac {2\cdot OC\cdot OD\cdot\cos\phi}{OC+OD}\end{array}\ \right|\Longrightarrow\ \frac {OE}{OF}$ $=\frac {OA\cdot OB}{OA+OB}\cdot\frac {OC+OD}{OC\cdot OD}$ .

Method 2. Denote $\left\|\begin{array}{ccc} M\in BD\ ,\ AM\parallel EF\\\ N\in AC\ ,\ DN\parallel EF\end{array}\right\|$ . Observe that $OM=OA$ and $ON=OD$ . Therefore,

$\odot\ AM\parallel DN\ \Longrightarrow\ \frac {AM}{DN}=$ $\frac {OA}{ON}=\frac {OA}{OD}$ , adica $\boxed{\ \frac {AM}{DN}=\frac {OA}{OD}\ }\ (1)$ .

$\odot\ OE\parallel AM\Longrightarrow\ \frac {OE}{AM}=\frac {OB}{BM}=\frac {OB}{OB+OM}=$ $\frac {OB}{OB+OA}\Longrightarrow\ \boxed{\ OE=\frac {OB}{OA+OB}\cdot AM\ }\ (2)$ .

$\odot\ OF\parallel DN\Longrightarrow\ \frac {OF}{DN}=$ $\frac {OC}{CN}=\frac {OC}{OC+ON}=\frac {OC}{OC+OD}$ $\Longrightarrow\ \boxed{\ OF=\frac {OC}{OC+OD}\cdot DN\ }\ (3)$ .

Dividing the relation $(2)$ to the relation $(3)$ obtain $\frac {OE}{OF}=\frac {OB}{OA+OB}\cdot \frac {OC+OD}{OC}\cdot\frac {AM}{DN}$ .

Using the relation $(1)$ obtain $\frac {OE}{OF}=\frac {OB}{OA+OB}\cdot \frac {OC+OD}{OC}\cdot\frac {OA}{OD}$ , i.e. the relation from conclusion.

Metoda 3. Denote $x=m(\angle AOE)=m(\angle BOE)=m(\angle COF)=m(\angle DOF)$ . Thus,

$\odot\ [AOB]=[AOE]+[BOE]\ \Longrightarrow\ OA\cdot OB\cdot\sin 2x=OE\cdot (OA+OB)\cdot\sin x$ .

$\odot\ [COD]=[COF]+[DOF]\ \Longrightarrow\ OC\cdot OD\cdot\sin 2x=OF\cdot (OC+OD)\cdot\sin x$ a.s.o.

Generalization. Let $ABCD$ be a convex quadrilateral for which $O\in AC\cap BD$ . Consider $E\in (AB)$ and $F\in (CD)$ so that

$O\in (EF)$ . Denote $x=m(\angle AOE)$ and $y=m(\angle BOE)$ . Prove that $\frac {OE}{OF}=\frac {OA\cdot OB}{OC\cdot OD}\cdot\frac {OD\cdot\sin x+OC\cdot \sin y}{OA\cdot\sin x+OB\cdot\sin y}$ .
This post has been edited 11 times. Last edited by Virgil Nicula, Nov 23, 2015, 8:17 AM

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