Korea Third Round (FKMO) 2012 #2

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Korea Third Round (FKMO) 2012 #2

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Source: FKMO 2012

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202 posts

#1 h Mar 25, 2012, 5:34 AM • 4 Y

Y by Ikeronalio, Adventure10, Mango247, and 1 other user

For a triangle $ABC$ which $\angle B \ne 90^{\circ}$ and $AB \ne AC$ , define $P_{ABC}$ as follows ;

Let $I$ be the incenter of triangle $ABC$ , and let $D, E, F$ be the intersection points with the incircle and segments $BC, CA, AB$ . Two lines $AB$ and $DI$ meet at $S$ and let $T$ be the intersection point of line $DE$ and the line which is perpendicular with $DF$ at $F$ . The line $ST$ intersects line $EF$ at $R$ . Now define $P_{ABC}$ be one of the intersection points of the incircle and the circle with diameter $IR$ , which is located in other side with $A$ about $IR$ .

Now think of an isosceles triangle $XYZ$ such that $XZ = YZ > XY$ . Let $W$ be the point on the side $YZ$ such that $WY < XY$ and Let $K = P_{YXW}$ and $L = P_{ZXW}$ . Prove that $2 KL \le XY$ .

This post has been edited 12 times. Last edited by syk0526, Jun 27, 2012, 1:16 PM

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904 posts

#2 h Mar 26, 2012, 1:52 PM • 1 Y

Y by Adventure10

syk0526 wrote:

$T \in DE, \ TD \bot DF$

I think there's some typo error here.

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200 posts

#3 h Mar 26, 2012, 3:59 PM • 2 Y

Y by Adventure10, Mango247

Quote:

$T\in DE, TD\perp DF$

This should be $T\in DE, TF\perp DF$ .

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34 posts

#4 h Mar 29, 2012, 5:38 PM • 2 Y

Y by Adventure10, Mango247

We can see that P is the symmetry point of D through I, but apply this in XYZ, I don't have more

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202 posts

#5 h Mar 30, 2012, 2:06 PM • 2 Y

Y by Adventure10, Mango247

Nevergiveupbtw wrote:

We can see that P is the symmetry point of D through I, but apply this in XYZ, I don't have more

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349 posts

#6 h Apr 4, 2012, 9:44 AM • 2 Y

Y by Adventure10, Mango247

syk0526 wrote:

Let $T$ be the midpoint of $XW$ [/hide]

What should we do next?

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202 posts

#7 h Apr 6, 2012, 2:42 PM • 2 Y

Y by Adventure10, Mango247

KittyOK wrote:

syk0526 wrote:

Let $T$ be the midpoint of $XW$ [/hide]

What should we do next?

We get
$TK = \frac{1}{2} ( XY - YW ) , TL = \frac{1}{2} (XZ - WZ)$
$\implies 2KL \le 2 (TK + TL ) = XY$

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374 posts

#8 h Apr 22, 2012, 8:34 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

I have proved that $P$ is the symmetry point of $D$ wrt $I$ with ugly two-page calculation with coordinate plane, but does anybody solve this beautifully??

Thanks in advance. I am not so good at geometry...

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202 posts

#9 h Apr 25, 2012, 7:34 AM • 4 Y

Y by Bigwood, Adventure10, Mango247, and 1 other user

Bigwood wrote:

I have proved that $P$ is the symmetry point of $D$ wrt $I$ with ugly two-page calculation with coordinate plane, but does anybody solve this beautifully??

Thanks in advance. I am not so good at geometry...

Use Pascal's Theorem at $FFUUDE$ where $U = DI \cap I$ (incircle) and use La Hire's Theorem

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324 posts

#10 h Jun 6, 2012, 1:03 PM • 1 Y

Y by Adventure10

I think the problem is written wrong. P and Q should be switched in their definitions.

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hyperspace.rulz

287 posts

hyperspace.rulz

#11 h Apr 12, 2013, 3:43 AM • 2 Y

Y by Adventure10, Mango247

syk0526 wrote:

We get
$TK = \frac{1}{2} ( XY - YW ) , TL = \frac{1}{2} (XZ - WZ)$

Could someone please explain how we may derive these two equalities? Thanks

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1721 posts

#12 h Jul 15, 2015, 2:27 PM • 1 Y

Y by Adventure10

In $\triangle ABC$ , WLOG $AB < AC$ , and let $\omega$ be the incircle of $\triangle ABC$ .
Let $DS \cap \omega = U$ . Note that $F, U, T$ are collinear.
Let the tangent to $\omega$ at $U$ intersect $EF$ at $R'$ .
From Pascal's Theorem on $FFUUDE$ , we have $S, T, R'$ collinear, which implies $R' = R$ .
Let $P=P_{ABC}$ . $RU, RP$ are tangents to $\omega$ , so $UP$ is the polar of $R$ .
Since $R$ is on $EF$ , $R$ is on the polar of $A$ .
From La Hire's Theorem, $A$ is on the polar of $R$ . Therefore, $A, U, P$ are collinear.
Let the tangent to $\omega$ from $P$ intersect $BC$ at $M$ , and $AP \cap BC = N$ .
$MD = MP$ , and $\angle DPN = 90$ . Therefore $MD = MP = MN$ .
Let $UR$ meet $AB, AC$ at $B_1, C_1$ .
The homothety that transforms $\triangle AB_1C_1$ to $\triangle ABC$ transforms $\omega$ to the $A$ -excircle.
Therefore $N$ is the intersection of $BC$ and the $A$ -excircle, so $BD=NC$
Now $2MD=2MP=2MN=BC-BD-CN=BC-2CD=BC-(AB+BC-AC)=AC-AB$ .
Therefore, $2MP=AC-AB \implies M$ is the midpoint of $BC$ .
Let $G$ be the midpoint of $XW$ .
$2GL=XZ-ZW, 2GK=XY-YW$ , so $2KL \le 2GL + 2GK = XZ-ZW+XY-YW=XY$ $\blacksquare$

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308 posts

#13 h Oct 14, 2020, 7:06 AM • 2 Y

Y by Mango247, Mango247

We will first identify $P_{\triangle ABC}$ .
CLAIM. Let $D'$ be the antipode of $D$ with respect to the incircle. Then $P_{\triangle ABC}$ is the second intersection of $AD'$ and the incircle.
Proof.
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Applying Pascal's theorem to $FFEDD'D'$ we have $RD'$ is tangent to the incircle. Since $R$ lies on the polar $FE$ of $A$ , $A$ lies on the polar of $R$ by La Hire's theorem. Hence $A,D',P$ are collinear.
$\blacksquare$
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Let the incircle of $\triangle ZXW$ , $\triangle YXW$ touch $XW$ at $A,B$ . Let the $Z$ and $Y$ excircle touch $XW$ at $C,D$ respectively.
Now, from CLAIM 1 we see that $K$ , $L$ lies on the circle centered at $M$ with radius $MB,MA$ respectively. Hence
$2KL\leq 2(KM+ML)=BC+AD=ZX-ZW-WY+XY=XY$ as desired.

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