Difference between revisions of "2017 USAMO Problems/Problem 3"
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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. | <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 S</math> is the | + | Let <math>D' = BC \cap XM \implies DD' \perp XM \implies</math> |
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+ | <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> We denote <math>I_\omega</math> inversion with respect to <math>\omega.</math> | ||
− | <math>I_\omega (K) = M \implies</math> circle <math>\Omega | + | Note that the circle <math>\Omega</math> has diameter <math>AX</math> and contain points <math>K, M, C,</math> and <math>B.</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 \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. | <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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<math>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</math> is cyclic. | <math>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</math> is cyclic. | ||
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==Contact== | ==Contact== | ||
Contact v_Enhance at https://www.facebook.com/v.Enhance. | Contact v_Enhance at https://www.facebook.com/v.Enhance. |
Revision as of 04:11, 21 September 2022
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
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
is cyclic
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
In
is orthocenter of
is midpoint
is midpoint
is orthocenter of
is root of height
circle
is the nine-point circle of
lies on circle
Let
is cyclic.
the points
and
are collinear.
Point is orthocenter
the points
and
are collinear.
is cyclic.
Contact
Contact v_Enhance at https://www.facebook.com/v.Enhance.