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2017 USAMO Problems/Problem 3: Difference between revisions

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==Solution==
==Solution==
[[File:2017 USAMO 3.png|400px|right]]
Let <math>X</math> be the point on circle <math>\Omega</math> opposite <math>M \implies  \angle MAX = 90^\circ, BC \perp XM.</math>
Let <math>X</math> be the point on circle <math>\Omega</math> opposite <math>M \implies  \angle MAX = 90^\circ, BC \perp XM.</math>


Line 11: Line 12:
<math>S</math> is the orthocenter of <math>\triangle DMX \implies</math> the points <math>X, A,</math> and <math>S</math> are collinear.
<math>S</math> is the orthocenter of <math>\triangle DMX \implies</math> the points <math>X, A,</math> and <math>S</math> are collinear.


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>
Let <math>\omega</math> be the circle centered at <math>S</math> with radius <math>R = \sqrt {SK \cdot SM}.</math>


Note that the circle <math>\Omega</math> has diameter <math>AX</math> and contain points <math>K, M, C,</math> and <math>B.</math>
We denote <math>I_\omega</math> inversion with respect to <math>\omega.</math>
 
Note that the circle <math>\Omega</math> has diameter <math>MX</math> and contain points <math>A, B, C,</math> and <math>K.</math>


<math>I_\omega (K) = M \implies</math> circle <math>\Omega \perp \omega \implies C = I_\omega (B), X = I_\omega (A).</math>  
<math>I_\omega (K) = M \implies</math> circle <math>\Omega \perp \omega \implies C = I_\omega (B), X = I_\omega (A).</math>  

Revision as of 03:24, 21 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

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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 orthocenter 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 $\omega.$

Note that the circle $\Omega$ has diameter $MX$ and contain points $A, B, C,$ and $K.$

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

$I_\omega (K) = M \implies$ 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 $\triangle FSX.$

$N$ is midpoint $SI, M$ is midpoint $SI, I$ is orthocenter of $\triangle FSX, A$ is root of height $FA \implies$ circle $AMN$ is the nine-point circle of $\triangle FSX \implies I'$ lies on circle $AMN.$

Let $N' = I_\omega (N) \implies R^2 = SN \cdot SN' = SI \cdot SI' \implies \frac {SN'}{SI'} = \frac {SI}{SN} =2 \implies$ $\angle XN'I' = \angle XSI' = 90^\circ – \angle AXI' = \angle IFX \implies N'XIF$ is cyclic.

$I_\omega (NAI'M) = N'XIKF \implies I_\omega(F) = L \implies$ the points $F, L,$ and $S$ are collinear.

Point $I$ is orthocenter $\triangle FSX \implies XI \perp SF, \angle ILS = \angle SI'F = 90^\circ \implies$ the points $X, I, E,$ and $L$ are collinear.

$SAI'F = I_\omega (XIF) \implies AI \cdot IF = SI \cdot II' = AI \cdot \frac {IF}{2} = \frac {SI}{2} \cdot II' = AI \cdot IM = EI \cdot IX \implies AEMX$ is cyclic.

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