Difference between revisions of "2017 USAMO Problems/Problem 3"
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==Solution== | ==Solution== | ||
− | Let <math>X</math> be the point on circle <math>\Omega</math> opposite <math>M | + | [[File:2017 USAMO 3.png|400px|right]] |
+ | [[File:2017 USAMO 3a.png|400px|right]] | ||
+ | [[File:2017 USAMO 3b.png|400px|right]] | ||
+ | Let <math>X</math> be the point on circle <math>\Omega</math> opposite <math>M</math>. This means <math>\angle MAX = 90^\circ, BC \perp XM.</math> | ||
<math>\angle XKM = \angle DKM = 90^\circ \implies</math> the points <math>X, D,</math> and <math>K</math> are collinear. | <math>\angle XKM = \angle DKM = 90^\circ \implies</math> the points <math>X, D,</math> and <math>K</math> are collinear. | ||
− | Let <math>D' = BC \cap XM \implies DD' \perp XM \implies | + | Let <math>D' = BC \cap XM \implies DD' \perp XM \implies</math> |
− | + | <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> | + | Let <math>\omega</math> be the circle centered at <math>S</math> with radius <math>R = \sqrt {SK \cdot SM}.</math> |
− | <math>I_\omega | + | We denote <math>I_\omega</math> inversion with respect to <math>\omega.</math> |
− | 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> | + | 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>KMD \perp \omega \implies D' = I_\omega (D) \in KMD \implies</math> | ||
+ | <math>\angle DD'M = 90^\circ \implies</math> the points <math>X, D',</math> and <math>M</math> are collinear. | ||
+ | |||
+ | Let <math>F \in AM, MF = MI.</math> It is well known that <math>MB = MI = MC \implies</math> | ||
+ | |||
+ | <math>\Theta = BICF</math> is circle centered at <math>M.</math> <math>C = I_\omega (B) \implies \Theta \perp \omega.</math> | ||
Let <math>I' = I_\omega (I ) \implies I' \in \Theta \implies \angle II'M = 90^\circ.</math> | Let <math>I' = I_\omega (I ) \implies I' \in \Theta \implies \angle II'M = 90^\circ.</math> | ||
− | |||
<math>I' = I_\omega (I ), X = I_\omega (A ) \implies AII'X</math> is cyclic. | <math>I' = I_\omega (I ), X = I_\omega (A ) \implies AII'X</math> is cyclic. | ||
<math>\angle XI'I = \angle XAI = 90^\circ \implies</math> the points <math>X, I' ,</math> and <math>F</math> are collinear. | <math>\angle XI'I = \angle XAI = 90^\circ \implies</math> the points <math>X, I' ,</math> and <math>F</math> are collinear. | ||
− | <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> | + | <math>I'IDD'</math> is cyclic <math>\implies \angle I'D'M = \angle I'D'C + 90^\circ = \angle I'ID + 90^\circ,</math> |
+ | <math>\angle XFM = \angle I'FI = 90^\circ – \angle I'IF = 90^\circ – \angle I'ID \implies</math> | ||
− | <math>\angle XFM + \angle I'D'M = 180^\circ \implies I'D'MF</math> is cyclic. | + | <math>\angle XFM + \angle I'D'M = 180^\circ \implies I'D'MF</math> is cyclic. |
+ | |||
Therefore point <math>F</math> lies on <math>I_\omega (IDK).</math> | Therefore point <math>F</math> lies on <math>I_\omega (IDK).</math> | ||
− | + | <math>FA \perp SX, SI' \perp FX \implies I</math> is orthocenter of <math>\triangle FSX.</math> | |
+ | |||
+ | <math>N</math> is midpoint <math>SI, M</math> is midpoint <math>FI, I</math> is orthocenter of <math>\triangle FSX, A</math> is root of height <math>FA \implies AMN</math> is the nine-point circle of <math>\triangle FSX \implies I' \in AMN.</math> | ||
+ | |||
+ | Let <math>N' = I_\omega (N) \implies R^2 = SN \cdot SN' = SI \cdot SI' \implies</math> | ||
+ | <cmath>\frac {SN'}{SI'} = \frac {SI}{SN} =2 \implies</cmath> | ||
+ | <math>\angle XN'I' = \angle XSI' = 90^\circ – \angle AXI' = \angle IFX \implies N'XIF</math> is cyclic. | ||
+ | |||
+ | Therefore point <math>F</math> lies on <math>I_\omega (AMN) \implies I_\omega(F) = L \implies</math> | ||
+ | |||
+ | The points <math>F, L,</math> and <math>S</math> are collinear, <math>AXFL</math> is cyclic. | ||
+ | Point <math>I</math> is orthocenter <math>\triangle FSX \implies XI \perp SF, \angle ILS = \angle SI'F = 90^\circ</math> | ||
+ | <math>\implies</math> The points <math>X, I, E,</math> and <math>L</math> are collinear. | ||
+ | |||
+ | <math>AXFL</math> is circle <math>\implies AI \cdot IF = IL \cdot XI\implies</math> | ||
+ | |||
+ | <math>AI \cdot \frac {IF}{2} = \frac {IL}{2} \cdot IX \implies AI \cdot IM = EI \cdot IX \implies AEMX</math> is cyclic. | ||
+ | <cmath>E \in \Omega.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Contact== | ||
Contact v_Enhance at https://www.facebook.com/v.Enhance. | Contact v_Enhance at https://www.facebook.com/v.Enhance. | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 18:24, 8 May 2023
Problem
Let be a scalene triangle with circumcircle
and incenter
Ray
meets
at
and
again at
the circle with diameter
cuts
again at
Lines
and
meet at
and
is the midpoint of
The circumcircles of
and
intersect at points
and
Prove that
passes through the midpoint of either
or
Solution
Let be the point on circle
opposite
. This means
the points
and
are collinear.
Let
is the orthocenter of
the points
and
are collinear.
Let be the circle centered at
with radius
We denote inversion with respect to
Note that the circle has diameter
and contain points
and
circle
circle
the points
and
are collinear.
Let It is well known that
is circle centered at
Let
is cyclic.
the points
and
are collinear.
is cyclic
is cyclic.
Therefore point lies on
is orthocenter of
is midpoint
is midpoint
is orthocenter of
is root of height
is the nine-point circle of
Let
is cyclic.
Therefore point lies on
The points and
are collinear,
is cyclic.
Point is orthocenter
The points
and
are collinear.
is circle
is cyclic.
vladimir.shelomovskii@gmail.com, vvsss
Contact
Contact v_Enhance at https://www.facebook.com/v.Enhance.
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