Difference between revisions of "2023 USAMO Problems/Problem 6"
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− | Consider the transformation which is the composition of an inversion centered at <math>A</math> and a reflection over the angle bisector of <math>\angle{}CAB</math> that sends <math>B</math> to <math>C</math> and <math>C</math> to <math>B</math>. We claim that this sends <math>D</math> to <math>E'</math> and <math>E'</math> to <math>D</math>. It is sufficient to prove that if the transformation sends <math>G</math> to <math>G'</math>, then <math>AE'JG'</math> is cyclic. Notice that <math>\triangle{}AGB\sim\triangle{}ACG'</math> since <math>\angle{}GAB=\angle{}G'AC</math> and <math>\tfrac{AG'}{AC}=\tfrac{\frac{AB\cdot{}AC}{AG}}{ | + | Consider the transformation which is the composition of an inversion centered at <math>A</math> and a reflection over the angle bisector of <math>\angle{}CAB</math> that sends <math>B</math> to <math>C</math> and <math>C</math> to <math>B</math>. We claim that this sends <math>D</math> to <math>E'</math> and <math>E'</math> to <math>D</math>. It is sufficient to prove that if the transformation sends <math>G</math> to <math>G'</math>, then <math>AE'JG'</math> is cyclic. Notice that <math>\triangle{}AGB\sim\triangle{}ACG'</math> since <math>\angle{}GAB=\angle{}G'AC</math> and <math>\tfrac{AG'}{AC}=\tfrac{\frac{AB\cdot{}AC}{AG}}{AC}=\tfrac{AB}{AG}</math>. Therefore, we get that <math>\angle{}AG'E'=\angle{}ABG=\angle{}AJE'</math>, so <math>AE'JG'</math> is cyclic, proving the claim. This means that <math>\angle{}BAE'=\angle{}CAD</math>. |
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Revision as of 18:41, 22 September 2023
Problem
Let ABC be a triangle with incenter and excenters
,
,
opposite
,
, and
, respectively. Given an arbitrary point
on the circumcircle of
that does not lie on any of the lines
,
, or
, suppose the circumcircles of
and
intersect at two distinct points
and
. If
is the intersection of lines
and
, prove that
.
Solution 1
Consider points and
such that the intersections of the circumcircle of
with the circumcircle of
are
and
, the intersections of the circumcircle of
with the circumcircle of
are
and
, the intersections of the circumcircle of
with line
are
and
, the intersections of the circumcircle of
with line
are
and
, the intersection of lines
and
is
, and the intersection of lines
and
is
.
Since is cyclic, the pairwise radical axes of the circumcircles of
and
concur. The pairwise radical axes of these circles are
and
, so
and
are collinear. Similarly, since
is cyclic, the pairwise radical axes of the cirucmcircles of
and
concur. The pairwise radical axes of these circles are
and
, so
and
are collinear. This means that
, so the tangents to the circumcircle of
at
and
intersect on
. Let this intersection be
. Also, let the intersection of the tangents to the circumcircle of
at
and
be a point at infinity on
called
and let the intersection of lines
and
be
. Then, let the intersection of lines
and
be
. By Pascal's Theorem on
and
, we get that
and
are collinear and that
and
are collinear, so
and
are collinear, meaning that
lies on
since both
and
lie on
.
Consider the transformation which is the composition of an inversion centered at and a reflection over the angle bisector of
that sends
to
and
to
. We claim that this sends
to
and
to
. It is sufficient to prove that if the transformation sends
to
, then
is cyclic. Notice that
since
and
. Therefore, we get that
, so
is cyclic, proving the claim. This means that
.
We claim that . Construct
to be the intersection of line
and the circumcircle of
and let
and
be the intersections of lines
and
with the circumcircle of
. Since
and
are the reflections of
and
over
, it is sufficient to prove that
are concyclic. Since
and
concur and
and
are concyclic, we have that
are concyclic, so
, so
are concyclic, proving the claim. We can similarly get that
.
Let line intersect the circumcircle of
at
and
. Notice that
is the midpoint of
and
, so
is a parallelogram with center
, so
. Similarly, we get that if line
intersects the circumcircle of
at
and
, we have that
, so
, so
, so
are concyclic. Then, the pairwise radical axes of the circumcircles of
and
are
and
, so
and
concur, so
and
concur, so
. We are then done since
.
~Zhaom
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