Difference between revisions of "2017 USAMO Problems/Problem 3"
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− | Let <math>X</math> be the point on circle <math>\Omega</math> opposite <math>M | + | 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. | ||
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<math>AXFL</math> is circle <math>\implies AI \cdot IF = IL \cdot XI\implies</math> | <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 = | + | <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> | <cmath>E \in \Omega.</cmath> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
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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. | ||
+ | |||
+ | {{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.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.