Vietnamese national Olympiad 2007, problem 6

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Vietnamese national Olympiad 2007, problem 6

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645 posts

hien

#1 h Feb 8, 2007, 9:22 AM • 2 Y

Y by Adventure10, Mango247

Let ABCD be trapezium that is inscribed in circle (O) with larger edge BC. P is a point lying outer segment BC. PA cut (O) at N(that means PA isn't tangent of (O)), the circle with diameter PD intersect (O) at E, DE meet BC at N. Prove that MN always pass through a fixed point.

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#2 h Feb 8, 2007, 10:10 AM • 2 Y

Y by Adventure10, Mango247

How is the trapezium, $AD\parallel BC$ or $AB\parallel CD$ ?

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3595 posts

#3 h Feb 8, 2007, 10:17 AM • 2 Y

Y by Adventure10, Mango247

young_boy wrote:

$AD\parallel BC$

That's right

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#4 h Feb 8, 2007, 1:31 PM • 2 Y

Y by Adventure10, Mango247

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gemath

#5 h Feb 8, 2007, 1:55 PM • 2 Y

Y by Adventure10, Mango247

You have true and nice solution hikaru123

here this problem is still true when $M\in$ segment $BC$ thus it is true for all $M\in BC$

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#6 h Feb 8, 2007, 2:04 PM • 2 Y

Y by Adventure10, Mango247

Thanks. I cant believe that problem is so easy.

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gemath

#7 h Feb 8, 2007, 2:08 PM • 2 Y

Y by Adventure10, Mango247

hikaru123 wrote:

I cant believe that problem is so easy.

Because it is for VMO

I think so!
Sorry for spam

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hien

#8 h Feb 8, 2007, 3:02 PM • 2 Y

Y by Adventure10, Mango247

hien wrote:

Let ABCD be trapezium that is inscribed in circle (O) with larger edge BC. P is a point lying outer segment BC. PA cut (O) at N(that means PA isn't tangent of (O)), the circle with diameter PD intersect (O) at E, DE meet BC at M. Prove that MN always pass through a fixed point.

I have a mistake typing, i have edited

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645 posts

hien

#9 h Feb 9, 2007, 9:10 AM • 2 Y

Y by Adventure10, Mango247

Yes, this is a the easiet problem in seven problem that i have to solve. But we waste at least quater to discover that solution

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gemath

#10 h Feb 9, 2007, 3:43 PM • 2 Y

Y by Adventure10, Mango247

hien wrote:

Yes, this is a the easiet problem in seven problem that i have to solve. But we waste at least quater to discover that solution

I'm not good at other object, but through geometry I think other problems are the same

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103 posts

#11 h Jan 17, 2010, 2:39 PM • 3 Y

Y by makar, Adventure10, Mango247

spam

:

I think the corret problem must be :

Let $ABCD$ be a quadrilateral $(AD \nparallel BC )$ inscribed in circle $(O)$ with larger edge $BC$ . $P$ is a point lying outer segment $BC$ . $PA$ cut $(O)$ at N(that means $PA$ isn't tangent of $(O)$ ), the circle with diameter $PD$ intersect $(O)$ at $E$ , $DE$ meet $BC$ at $M$ . Prove that $MN$ always pass through a fixed point.

Solution :

Image not found

Let $K$ , $L$ and $J$ respective is the intersections of $MN$ and $(ABCD)$ ; diameter $PD$ and $BC$ ; $KL$ and $(ABCD)$

$\bullet$ Because $DL \perp BC$ then $\boxed{L \ is \ fixed \ point}$ .

$\bullet ML.MP = ME.MD = MK.MN \Longrightarrow KLMP$ is concyclic $\Longrightarrow \widehat{JLC} = \widehat{KNA} = \widehat{KJA}$ . It's mean $AJ \parallel BC \Longrightarrow \boxed{J \ is \ fixed \ point}$ .

$\blacksquare$ $K \equiv JL \cap (ABCD)$ ... is a fixed point

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#12 h May 30, 2019, 7:18 AM • 1 Y

Y by Adventure10

Vietnam NMO 2007 P6 wrote:

Let $ABCD$ be trapezium that is inscribed in circle $(O)$ with larger edge $BC$ . $P$ is a point lying outer segment $BC$ . $PA$ cut $(O)$ at $N$ (that means $PA$ isn't tangent of $(O)$ ), the circle with diameter $PD$ intersect $(O)$ at $E$ , $DE$ meet $BC$ at $N$ . Prove that $MN$ always pass through a fixed point.

Solution: Assume $PB>PC$ and $DE \cap BC=M$ . Let $F$ be the $A-$ antipode WRT $\odot (ABC)$ . Let $DF \cap BC=G$ , $FN \cap BC=L$ and $AP$ $\cap$ $DF$ $=$ $K$
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$G \in \odot (DEP)$ , and $L$ is the orthocenter WRT $\Delta KFP$
$GL \cdot LP = LN \cdot LF \implies L \in DE \implies L \equiv M$ Hence, $MN$ always passes through fixed point $F$

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#13 h Oct 9, 2019, 1:44 AM • 1 Y

Y by Adventure10

Let $D'$ be the point diametrically opposite to $D$ on the circle. Notice then that $E = PD' \cap (O)$ . Let $D^*$ be the point on $(O)$ so that $D'D^* || BC.$ Then, we've by Pascal on $ADED'D^*N$ that $ND^* \cap DE$ is on $BC.$ This implies that $M = ND^* \cap DE \Rightarrow D^* \in MN.$ Hence, as $D^*$ is clearly fixed, we're done.

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