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Difference between revisions of "2017 USAMO Problems/Problem 3"
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==Problem==
 
Let <math>ABC</math> be a scalene triangle with circumcircle <math>\Omega</math> and incenter <math>I.</math> Ray <math>AI</math> meets <math>BC</math> at <math>D</math> and <math>\Omega</math> again at <math>M;</math> the circle with diameter <math>DM</math> cuts <math>\Omega</math> again at <math>K.</math> Lines <math>MK</math> and <math>BC</math> meet at <math>S,</math> and <math>N</math> is the midpoint of <math>IS.</math> The circumcircles of <math>\triangle KID</math> and <math>\triangle MAN</math> intersect at points <math>L_1</math> and <math>L.</math> Prove that <math>\Omega</math> passes through the midpoint of either <math>IL_1</math> or <math>IL.</math>
 
Let <math>ABC</math> be a scalene triangle with circumcircle <math>\Omega</math> and incenter <math>I.</math> Ray <math>AI</math> meets <math>BC</math> at <math>D</math> and <math>\Omega</math> again at <math>M;</math> the circle with diameter <math>DM</math> cuts <math>\Omega</math> again at <math>K.</math> Lines <math>MK</math> and <math>BC</math> meet at <math>S,</math> and <math>N</math> is the midpoint of <math>IS.</math> The circumcircles of <math>\triangle KID</math> and <math>\triangle MAN</math> intersect at points <math>L_1</math> and <math>L.</math> Prove that <math>\Omega</math> passes through the midpoint of either <math>IL_1</math> or <math>IL.</math>
  
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==Solution==
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Let <math>X</math> be the point on circle <math>\Omega</math> opposite <math>M \implies  \angle MAX = 90^\circ, BC \perp XM.</math>
  
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<math>\angle XKM = \angle DKM = 90^\circ \implies</math> the points <math>X, D,</math> and <math>K</math> are collinear.
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Let <math>D' = BC \cap XM \implies DD' \perp XM \implies S</math> is the ortocenter of <math>\triangle DMX \implies</math> the points <math>X, A,</math> and <math>S</math> are collinear.
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Let <math>\omega</math> be the circle centered at <math>S</math> with radius <math>R = \sqrt {SK \cdot SM}.</math> We denote <math>I_\omega</math> inversion with respect to <math>\omega.</math>
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<math>I_\omega (K) = M \implies</math> circle <math>\Omega = KMCXAB \perp \omega \implies C = I_\omega (B), X = I_\omega (A).</math>
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<math>I_\omega (K) = M \implies</math> circle <math>KMD \perp \omega \implies D' = I_\omega (D) \in KMD \implies \angle DD'M = 90^\circ \implies AXD'D</math> is cyclic <math>\implies</math>  the points <math>X, D',</math> and <math>M</math> are collinear.
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Let <math>F \in AM, MF = MI.</math> It is well known that <math>MB = MI = MC \implies \Theta = BICF</math> is circle  centered at <math>M. C = I_\omega (B) \implies \Theta \perp \omega.</math>
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Let <math>I' =  I_\omega (I ) \implies I' \in \Theta \implies \angle II'M =  90^\circ.</math>
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<math>I' =  I_\omega (I ), X =  I_\omega (A ) \implies AII'X</math> is cyclic.
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<math>\angle XI'I = \angle XAI =  90^\circ \implies</math>  the points <math>X, I' ,</math> and <math>F</math> are collinear.
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<math>I'IDD'</math> is cyclic <math>\implies \angle I'D'M = \angle I'D'C + 90^\circ =  \angle I'ID + 90^\circ, \angle XFM = \angle I'FI = 90^\circ – \angle I'IF = 90^\circ – \angle I'ID  \implies</math>
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<math>\angle XFM +  \angle I'D'M = 180^\circ \implies I'D'MF</math> is cyclic.
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Therefore point <math>F</math> lies on  <math>I_\omega (IDK).</math>
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In <math>\triangle FSX</math> <math>FA \perp SX, SI' \perp FX \implies I</math> is orthocenter of <math>\triangle FSX.</math>
  
 
Contact v_Enhance at https://www.facebook.com/v.Enhance.
 
Contact v_Enhance at https://www.facebook.com/v.Enhance.

Revision as of 17:11, 20 September 2022

Problem

Let $ABC$ be a scalene triangle with circumcircle $\Omega$ and incenter $I.$ Ray $AI$ meets $BC$ at $D$ and $\Omega$ again at $M;$ the circle with diameter $DM$ cuts $\Omega$ again at $K.$ Lines $MK$ and $BC$ meet at $S,$ and $N$ is the midpoint of $IS.$ The circumcircles of $\triangle KID$ and $\triangle MAN$ intersect at points $L_1$ and $L.$ Prove that $\Omega$ passes through the midpoint of either $IL_1$ or $IL.$

Solution

Let $X$ be the point on circle $\Omega$ opposite $M \implies \angle MAX = 90^\circ, BC \perp XM.$

$\angle XKM = \angle DKM = 90^\circ \implies$ the points $X, D,$ and $K$ are collinear.

Let $D' = BC \cap XM \implies DD' \perp XM \implies S$ is the ortocenter of $\triangle DMX \implies$ the points $X, A,$ and $S$ are collinear.

Let $\omega$ be the circle centered at $S$ with radius $R = \sqrt {SK \cdot SM}.$ We denote $I_\omega$ inversion with respect to $\omega.$

$I_\omega (K) = M \implies$ circle $\Omega = KMCXAB \perp \omega \implies C = I_\omega (B), X = I_\omega (A).$

$I_\omega (K) = M \implies$ circle $KMD \perp \omega \implies D' = I_\omega (D) \in KMD \implies \angle DD'M = 90^\circ \implies AXD'D$ is cyclic $\implies$ the points $X, D',$ and $M$ are collinear.

Let $F \in AM, MF = MI.$ It is well known that $MB = MI = MC \implies \Theta = BICF$ is circle centered at $M. C = I_\omega (B) \implies \Theta \perp \omega.$

Let $I' = I_\omega (I ) \implies I' \in \Theta \implies \angle II'M = 90^\circ.$

$I' = I_\omega (I ), X = I_\omega (A ) \implies AII'X$ is cyclic.

$\angle XI'I = \angle XAI = 90^\circ \implies$ the points $X, I' ,$ and $F$ are collinear.

$I'IDD'$ is cyclic $\implies \angle I'D'M = \angle I'D'C + 90^\circ = \angle I'ID + 90^\circ, \angle XFM = \angle I'FI = 90^\circ – \angle I'IF = 90^\circ – \angle I'ID \implies$

$\angle XFM + \angle I'D'M = 180^\circ \implies I'D'MF$ is cyclic. Therefore point $F$ lies on $I_\omega (IDK).$

In $\triangle FSX$ $FA \perp SX, SI' \perp FX \implies I$ is orthocenter of $\triangle FSX.$

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