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 | + | <cmath>E \in \Omega.</cmath> |
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==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.
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