Difference between revisions of "2023 USAMO Problems/Problem 6"
m |
m (→Solution 1) |
||
Line 138: | Line 138: | ||
</asy> | </asy> | ||
− | We claim that <math>\angle{}I_BE'I_C+\angle{}I_BDI_C=180^\circ</math>. Construct <math>D'</math> to be the intersection of line <math>\overline{AE}</math> and the circumcircle of <math>\triangle{} | + | We claim that <math>\angle{}I_BE'I_C+\angle{}I_BDI_C=180^\circ</math>. Construct <math>D'</math> to be the intersection of line <math>\overline{AE'}</math> and the circumcircle of <math>\triangle{}E'I_BI_C</math> and let <math>B'</math> and <math>C'</math> be the intersections of lines <math>\overline{AC}</math> and <math>\overline{AB}</math> with the circumcircle of <math>\triangle{}BI_BI_C</math>. Since <math>B'</math> and <math>C'</math> are the reflections of <math>B</math> and <math>C</math> over <math>\overline{I_BI_C}</math>, it is sufficient to prove that <math>A,B',C',D'</math> are concyclic. Since <math>\overline{B'C},\overline{D'E'},</math> and <math>\overline{I_BI_C}</math> concur and <math>D',E',I_B,I_C</math> and <math>I_B,I_C,B',C</math> are concyclic, we have that <math>B',C,D',E'</math> are concyclic, so <math>\angle{}B'D'A=\angle{}ACE'=\angle{}AC'B'</math>, so <math>A,B',C',D'</math> are concyclic, proving the claim. We can similarly get that <math>\angle{}IE'I_A=\angle{}IDI_A</math>. |
<asy> | <asy> |
Revision as of 17:31, 29 October 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
See Also
2023 USAMO (Problems • Resources) | ||
Preceded by Problem 5 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.