Greece Junior National Olympiad 1997-98 Pronlem 4

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#1 h Aug 10, 2015, 8:04 PM • 1 Y

Y by Adventure10

Let $K(O,R)$ be a circle with center $O$ and radious $R$ and $(e)$ to be a line thst tangent to $K$ at $A$ . A line parallel to $OA$ cuts $K$ at $B, C$ , and $(e)$ at $D$ , ( $C$ is between $B$ and $D$ ). Let $E$ to be the antidiameric of $C$ with respect to $K$ . $EA$ cuts $BD$ at $F$ .

i)Examine if $CEF$ is isosceles.
ii)Prove that $2AD=EB$ .
iii)If $K$ si the midlpoint of $CF$ , prove that $AB=KO$ .
iv)If $R=\frac{5}{2}, AD=\frac{3}{2}$ , calculate the area of $EBF$

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#2 h Aug 11, 2015, 1:02 PM • 2 Y

Y by Adventure10, Mango247

$(i)$
Bacause $OA$ is parallel with $BCDK$
We get $\triangle{EOA} \cong \triangle{ECF}$
And because $OE=OA$ we get $CE=CF$
So $\triangle{CEF}$ is isoscele triangle
Moreover we get $EA=AF$
$(ii)$
Because $AD$ is parallel with $EB$ ( $\angle{EBF}=90=\angle{ADF}$ )
So $\triangle{FAD} \cong \triangle{FEB}$
And because $EA=AF$ we get $\frac{AF}{EF}=\frac{1}{2}=\frac{AD}{EB}$
So $2AD=EB$

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#3 h Aug 11, 2015, 1:07 PM • 2 Y

Y by Adventure10, Mango247

$(iii)$
Let $T$ is the midpoint of $BC$
We get $AB=\sqrt{AD^2+BD^2}$ and $OK=\sqrt{OT^2+TK^2}$
And we have $AD=OK$
So $AB=OK \Leftrightarrow BD=TK$
But $BD=r+BT$
And from result in $(i)$ we get $CF=2OA$ give us $CK=\frac{OA}{2}=r$
So $TK=TC+CK=TC+r$ but $BT=TC$
So $BD=OK$ give us $AB=OK$

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#4 h Aug 11, 2015, 1:11 PM • 2 Y

Y by Adventure10, Mango247

$(iv)$
From $r=\frac{5}{2}$ we get $CE=2r=5$
And because $AD=\frac{3}{2}$ we get $BE=3$
So $BC=4$
$\text{Area of} \triangle{BEF} =\frac{1}{2} (BE)(BF)=\frac{3*(4+5)}{2}=13.5$

This post has been edited 1 time. Last edited by ThE-dArK-lOrD, Aug 11, 2015, 1:11 PM

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