Difference between revisions of "2023 USAMO Problems/Problem 6"
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~Zhaom | ~Zhaom | ||
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+ | ==Solution 2== | ||
+ | |||
+ | Set <math>\triangle ABC</math> as the reference triangle, and let <math>D=(x_0,y_0,z_0)</math> with homogenized coordinates. To find the equation of circle <math>DII_A</math>, we note that <math>I=(a:b:c)</math> (not homogenized) and <math>I_A=(-a:b:c)</math>. Thus, for this circle with equation <math>a^2yz+b^2xz+c^2xy=(x+y+z)(u_1x+v_1y+w_1z)</math>, we compute that <cmath>u_1=bc,</cmath> <cmath>v_1=\frac{bc^2x_0}{bz_0-cy_0},</cmath> <cmath>w_1=\frac{b^2cx_0}{cy_0-bz_0}.</cmath> | ||
+ | For circle <math>DI_BI_C</math> with equation <math>a^2yz+b^2xz+c^2xy=(x+y+z)(u_2x+v_2y+w_2z)</math>,, we find that <cmath>u_2=-bc,</cmath> <cmath>v_2=\frac{bc^2x_0}{bz_0+cy_0},</cmath> <cmath>w_2=\frac{b^2cx_0}{cy_0+bz_0}.</cmath> | ||
+ | The equation of the radical axis is <math>ux+vy+wz=0</math> with <math>u=u_1-u_2</math>, <math>v=v_1-v_2</math>, and <math>w=w_1-w_2</math>. We want to consider the intersection of this line with line <math>\overline{BC}</math>, so set <math>x=0</math>. The equation reduces to <math>vy+wz=0</math>. We see that <math>v=\frac{2bc^3x_0y_0}{b^2z_0^2-c^2y_0^2}</math> and <math>w=\frac{2b^3cx_0z_0}{c^2y_0^2-b^2z_0^2}</math>, so <cmath>\frac{v}{w}=\frac{b^2z_0}{c^2y_0},</cmath> which is the required condition for <math>\overline{AD}</math> and <math>\overline{AE}</math> to be isogonal. | ||
==See Also== | ==See Also== | ||
{{USAMO newbox|year=2023|num-b=5|after=Last Problem}} | {{USAMO newbox|year=2023|num-b=5|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:36, 10 March 2024
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
.
Video Solution by mop 2024
https://youtube.com/watch?v=LAuyU2OuVzE
~r00tsOfUnity
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
Solution 2
Set as the reference triangle, and let
with homogenized coordinates. To find the equation of circle
, we note that
(not homogenized) and
. Thus, for this circle with equation
, we compute that
For circle
with equation
,, we find that
The equation of the radical axis is
with
,
, and
. We want to consider the intersection of this line with line
, so set
. The equation reduces to
. We see that
and
, so
which is the required condition for
and
to be isogonal.
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.