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Source: 2016 KJMO #6

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583 posts

khyeon

#1 h Nov 13, 2016, 4:41 AM • 1 Y

Y by Adventure10

circle $O_1$ is tangent to $AC$ , $BC$ (side of triangle $ABC$ ) at point $D, E$ .
circle $O_2$ include $O_1$ , is tangent to $BC$ , $AB$ (side of triangle $ABC$ ) at point $E, F$
The tangent of $O_2$ at $P(DE \cap O_2, P \neq E)$ meets $AB$ at $Q$ .
A line passing through $O_1$ (center of $O_1$ ) and parallel to $BO_2$ ( $O_2$ is also center of $O_2$ ) meets $BC$ at $G$ , $EQ \cap AC=K, KG \cap EF=L$ , $EO_2$ meets circle $O_2$ at $N(\neq E)$ , $LO_2 \cap FN=M$ .
IF $N$ is a middle point of $FM$ , prove that $BG=2EG$

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583 posts

khyeon

#2 h Nov 13, 2016, 5:09 AM • 2 Y

Y by Adventure10, Mango247

very hard to draw and read..
but it is easy when you get a idea.

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583 posts

khyeon

#3 h Nov 13, 2016, 12:18 PM • 2 Y

Y by Adventure10, Mango247

no ideas?

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583 posts

khyeon

#4 h Nov 13, 2016, 12:54 PM • 1 Y

Y by Adventure10

here's my drawing

Attachments:

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373 posts

#5 h Nov 13, 2016, 2:22 PM • 1 Y

Y by Adventure10

I think Similarity and Menelaus will kill this problem
$KG // QB$ since $\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ$
Then Menelaus, done.

This post has been edited 3 times. Last edited by Hypernova, Nov 13, 2016, 4:35 PM
Reason: Latex

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583 posts

khyeon

#6 h Nov 14, 2016, 1:42 PM • 2 Y

Y by Adventure10, Mango247

Hypernova wrote:

I think Similarity and Menelaus will kill this problem
$KG // QB$ since $\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ$
Then Menelaus, done.

nope

How can you prove $KG$ is parallel to $QB$ ?
One more, if you proved $KG//QB$ , just done

This post has been edited 2 times. Last edited by khyeon, Nov 14, 2016, 1:45 PM

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#7 h Nov 14, 2016, 1:43 PM • 2 Y

Y by Adventure10, Mango247

khyeon wrote:

Hypernova wrote:

I think Similarity and Menelaus will kill this problem
$KG // QB$ since $\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ$
Then Menelaus, done.

nope

How can you prove $KG$ is parallel to $QB$ ?

read again

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583 posts

khyeon

#9 h Nov 14, 2016, 1:54 PM • 2 Y

Y by Adventure10, Mango247

Skravin wrote:

khyeon wrote:

Hypernova wrote:

I think Similarity and Menelaus will kill this problem
$KG // QB$ since $\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ$
Then Menelaus, done.

nope

How can you prove $KG$ is parallel to $QB$ ?

read again

nothing changed

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763 posts

#10 h Nov 14, 2016, 2:08 PM • 1 Y

Y by Adventure10

Hypernova wrote:

I think Similarity and Menelaus will kill this problem
$KG // QB$ since $\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ$
Then Menelaus, done.

I assume that $RQ$ means $PQ$

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373 posts

#11 h Nov 14, 2016, 2:10 PM • 1 Y

Y by Adventure10

Skravin wrote:

Hypernova wrote:

I think Similarity and Menelaus will kill this problem
$KG // QB$ since $\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ$
Then Menelaus, done.

I assume that $RQ$ means $PQ$

Same. Let $QP\cap BC=R$ .

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763 posts

#12 h Nov 14, 2016, 2:10 PM • 2 Y

Y by Hypernova, Adventure10

Hypernova wrote:

Skravin wrote:

Hypernova wrote:

I think Similarity and Menelaus will kill this problem
$KG // QB$ since $\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ$
Then Menelaus, done.

I assume that $RQ$ means $PQ$

Same. Let $QP\cap BC=R$ .

Ah got it

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khyeon

#14 h Nov 14, 2016, 2:24 PM • 2 Y

Y by Adventure10, Mango247

What I mean is
$\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ \to KG // QB$
I can't understand..

This post has been edited 1 time. Last edited by khyeon, Nov 14, 2016, 2:26 PM

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#15 h Nov 14, 2016, 2:45 PM • 1 Y

Y by Adventure10

khyeon wrote:

What I mean is
$\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ \to KG // QB$
I can't understand..

Full solution
Let $QP\cap BC=R$ . Since $CE=CD, RE=RP$ (Tangent), $\angle ECD=\angle ERP$ . So $CK//RQ$ .
And $E, O_1, O_2$ collinear, so $\triangle EO_{1}D \sim EO_{2}P (AA)$ .
Hence $\frac{EK}{EQ}=\frac{ED}{EP}=\frac{O_1D}{O_2P}=\frac{EO_1}{EO_2}=\frac{EG}{EB}$ , so $KG//QB$ .
Then use Menelaus in $\triangle EFN$ , secant $LO_2M$ . It gives $\frac{LF}{EL}*\frac{MN}{FM}*\frac{O_2E}{NO_2}=\frac{r_2}{r_1}*\frac{1}{2}*\frac{r_1}{r_2}=1$ . So $r_2=2r_1$ .
Then $\frac{BG}{EG}=\frac{EO_2}{EO_1}=\frac{r_2}{r_1}=2$ , Q.E.D.

This post has been edited 1 time. Last edited by Hypernova, Nov 14, 2016, 2:52 PM
Reason: typo

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#16 h Nov 14, 2016, 2:46 PM • 1 Y

Y by Adventure10

khyeon wrote:

What I mean is
$\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ \to KG // QB$
I can't understand..

It was very simple except the drawing
(I didn't take the Junior exam)

This post has been edited 1 time. Last edited by Hypernova, Nov 14, 2016, 2:48 PM

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583 posts

khyeon

#19 h Nov 14, 2016, 2:55 PM • 2 Y

Y by Adventure10, Mango247

Hypernova wrote:

khyeon wrote:

What I mean is
$\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ \to KG // QB$
I can't understand..

It was very simple except the drawing
(I didn't take the Junior exam)

Thank you for your kindness...
At my test, I proved $L$ is on the $O_1$

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170 posts

#20 h Nov 4, 2024, 7:14 AM • 1 Y

Y by Roh_Hoi_Chan

One of the best geometry questions I have ever seen.

It is easy to assume that $\triangle KGC \sim \triangle QBR$
The proof of this is the following :
proof

So $\bar{GL} \parallel \bar{BF}.$
Because of Menelaus, $\frac{\bar{EL}}{\bar{LF}} \times \frac{\bar{FM}}{\bar{NM}} \times \frac{\bar{N{O}_{2}}}{\bar{E{O}_{2}}} =1$
So, $\frac{\bar{EL}}{\bar{FL}}=\frac{2}{1} .$
So, $\bar{BG}= 2 \times \bar{EG}$

This post has been edited 2 times. Last edited by hanulyeongsam, Nov 4, 2024, 7:18 AM

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