Difference between revisions of "2017 USAMO Problems/Problem 3"

Revision as of 21:47, 18 October 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 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 $KMD \perp \omega \implies D' = I_\omega (D) \in KMD \implies$ $\angle DD'M = 90^\circ \implies$ 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 $I_\omega (IDK).$

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

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

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.

Therefore point $F$ lies on $I_\omega (AMN) \implies I_\omega(F) = L \implies$

The points $F, L,$ and $S$ are collinear, $AXFL$ is cyclic.

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.

$AXFL$ is circle $\implies AI \cdot IF = IL \cdot XI\implies$

$AI \cdot \frac {IF}{2} = \frac {IL}{2} \cdot IX \implies AI \cdot IM = EI \cdot IX \implies AEMX$ is cyclic. $E \in \Omega.$ vladimir.shelomovskii@gmail.com, vvsss

Contact

Contact v_Enhance at https://www.facebook.com/v.Enhance.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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Difference between revisions of "2017 USAMO Problems/Problem 3"

Revision as of 21:47, 18 October 2022

Problem

Solution

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