Difference between revisions of "2023 USAMO Problems/Problem 6"
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Consider points <math>G,H,J,K,P,</math> and <math>Q</math> such that the intersections of the circumcircle of <math>\triangle{}ABC</math> with the circumcircle of <math>\triangle{}DII_A</math> are <math>D</math> and <math>G</math>, the intersections of the circumcircle of <math>\triangle{}ABC</math> with the circumcircle of <math>\triangle{}DI_BI_C</math> are <math>D</math> and <math>H</math>, the intersections of the circumcircle of <math>\triangle{}ABC</math> with line <math>\overline{II_A}</math> are <math>A</math> and <math>J</math>, the intersections of the circumcircle of <math>\triangle{}ABC</math> with line <math>\overline{I_BI_C}</math> are <math>A</math> and <math>K</math>, the intersection of lines <math>\overline{II_A}</math> and <math>\overline{BC}</math> is <math>P</math>, and the intersection of lines <math>\overline{I_BI_C}</math> and <math>\overline{BC}</math> is <math>Q</math>. | Consider points <math>G,H,J,K,P,</math> and <math>Q</math> such that the intersections of the circumcircle of <math>\triangle{}ABC</math> with the circumcircle of <math>\triangle{}DII_A</math> are <math>D</math> and <math>G</math>, the intersections of the circumcircle of <math>\triangle{}ABC</math> with the circumcircle of <math>\triangle{}DI_BI_C</math> are <math>D</math> and <math>H</math>, the intersections of the circumcircle of <math>\triangle{}ABC</math> with line <math>\overline{II_A}</math> are <math>A</math> and <math>J</math>, the intersections of the circumcircle of <math>\triangle{}ABC</math> with line <math>\overline{I_BI_C}</math> are <math>A</math> and <math>K</math>, the intersection of lines <math>\overline{II_A}</math> and <math>\overline{BC}</math> is <math>P</math>, and the intersection of lines <math>\overline{I_BI_C}</math> and <math>\overline{BC}</math> is <math>Q</math>. | ||
− | Since <math>IBI_AC</math> is cyclic, the pairwise radical axes of the circumcircles of <math>\triangle{}DII_A,\triangle{}ABC,</math> and <math>IBI_AC</math> concur. The pairwise radical axes of these circles are <math>\overline{GD},\overline{II_A},</math> and <math>\overline{BC}</math>, so <math>G,P,</math> and <math>D</math> are collinear. Similarly, since <math>BCI_BI_C</math> is cyclic, the pairwise radical axes of the cirucmcircles of <math>\triangle{}DI_BI_C,\triangle{}ABC,</math> and <math>BCI_BI_C</math> concur. The pairwise radical axes of these circles are <math>\overline{HD},\overline{I_BI_C},</math> and <math>\overline{BC}</math>, so <math>H,Q,</math> and <math>D</math> are collinear. This means that <math>-1=(Q,P;B,C)\stackrel{D}{=}(H,G;B,C)</math>, so the tangents to the circumcircle of <math>\triangle{}ABC</math> at <math>G</math> and <math>H</math> intersect on <math>\overline{BC}</math>. Let this intersection be <math>X</math>. Also, let the intersection of the tangents to the circumcircle of <math>\triangle{}ABC</math> at <math>K</math> and <math>J</math> be a point at infinity on <math>\overline{BC}</math> called <math>Y</math> and let the intersection of lines <math>\overline{KG}</math> and <math>\overline{}HJ</math> be <math>Z</math>. Then, let the intersection of lines <math>\overline{GJ}</math> and <math>\overline{HK}</math> be <math>E'</math>. By Pascal's Theorem on <math>GGJHHK</math> and <math>GJJHKK</math>, we get that <math>X,E',</math> and <math>Z</math> are collinear and that <math>E',Y,</math> and <math>Z</math> are collinear, so <math>E',X,</math> and <math>Y</math> are collinear, meaning that <math>E</math> lies on <math>\overline{BC}</math> since both <math>X</math> and <math>Y</math> lie on <math>\overline{BC}</math>. | + | Since <math>IBI_AC</math> is cyclic, the pairwise radical axes of the circumcircles of <math>\triangle{}DII_A,\triangle{}ABC,</math> and <math>IBI_AC</math> concur. The pairwise radical axes of these circles are <math>\overline{GD},\overline{II_A},</math> and <math>\overline{BC}</math>, so <math>G,P,</math> and <math>D</math> are collinear. Similarly, since <math>BCI_BI_C</math> is cyclic, the pairwise radical axes of the cirucmcircles of <math>\triangle{}DI_BI_C,\triangle{}ABC,</math> and <math>BCI_BI_C</math> concur. The pairwise radical axes of these circles are <math>\overline{HD},\overline{I_BI_C},</math> and <math>\overline{BC}</math>, so <math>H,Q,</math> and <math>D</math> are collinear. This means that <math>-1=(Q,P;B,C)\stackrel{D}{=}(H,G;B,C)</math>, so the tangents to the circumcircle of <math>\triangle{}ABC</math> at <math>G</math> and <math>H</math> intersect on <math>\overline{BC}</math>. Let this intersection be <math>X</math>. Also, let the intersection of the tangents to the circumcircle of <math>\triangle{}ABC</math> at <math>K</math> and <math>J</math> be a point at infinity on <math>\overline{BC}</math> called <math>Y</math> and let the intersection of lines <math>\overline{KG}</math> and <math>\overline{}HJ</math> be <math>Z</math>. Then, let the intersection of lines <math>\overline{GJ}</math> and <math>\overline{HK}</math> be <math>E'</math>. By Pascal's Theorem on <math>GGJHHK</math> and <math>GJJHKK</math>, we get that <math>X,E',</math> and <math>Z</math> are collinear and that <math>E',Y,</math> and <math>Z</math> are collinear, so <math>E',X,</math> and <math>Y</math> are collinear, meaning that <math>E'</math> lies on <math>\overline{BC}</math> since both <math>X</math> and <math>Y</math> lie on <math>\overline{BC}</math>. |
<asy> | <asy> |
Revision as of 18:37, 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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