Difference between revisions of "2017 USAMO Problems/Problem 3"
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==Solution== | ==Solution== | ||
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Let <math>X</math> be the point on circle <math>\Omega</math> opposite <math>M \implies \angle MAX = 90^\circ, BC \perp XM.</math> | Let <math>X</math> be the point on circle <math>\Omega</math> opposite <math>M \implies \angle MAX = 90^\circ, BC \perp XM.</math> | ||
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<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> | + | <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 \frac {SN'}{SI'} = \frac {SI}{SN} =2 \implies</ | + | 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. | <math>\angle XN'I' = \angle XSI' = 90^\circ – \angle AXI' = \angle IFX \implies N'XIF</math> is cyclic. | ||
Revision as of 05:38, 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
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.
the points
and
are collinear.
Point is orthocenter
the points
and
are collinear.
is cyclic.
Contact
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