2023 USAMO Problems/Problem 6
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
[asy]
size(500);
pair A,B,C,D,E,F,G,H,I,J,K,IA,IB,IC,P,Q;
B=(0,0);
C=(8,0);
A=intersectionpoint(Circle(B,6),Circle(C,9));
I=incenter(A,B,C);
path c=circumcircle(A,B,C);
J=intersectionpoint(I--(4*I-3*A),c);
IA=2*J-I;
IB=2*intersectionpoint(I--(4*I-3*B),c)-I;
IC=2*intersectionpoint(I--(4*I-3*C),c)-I;
K=intersectionpoint(IB--IC,c);
D=intersectionpoint(I--(I+(10,-12)),c);
path c1=circumcircle(D,I,IA),c2=circumcircle(D,IB,IC);
F=intersectionpoints(c1,c2)[1];
E=extension(B,C,D,F);
G=intersectionpoint(c1,c);
H=intersectionpoint(c2,c);
P=extension(A,I,B,C);
Q=extension(IB,IC,B,C);
draw(A--B--C--A);
draw(c);
draw(A--J);
draw(circumcircle(D,I,IA));
draw(circumcircle(D,IB,IC));
draw(D--F);
draw(B--Q--IB);
draw(G--J,dashed);
draw(H--K,dashed);
dot("",A,dir(A-circumcenter(A,B,C)));
dot("
",B,1/2*dir(B-dir(circumcenter(A,B,C))*dir(90)+dir(B-C)));
dot("
",C,dir(C-circumcenter(A,B,C))*dir(15));
dot("
",D,dir(dir(90)*dir(circumcenter(D,I,IA)-D)+dir(90)*dir(D-circumcenter(D,IB,IC))));
dot("
",E,dir(dir(H-K)+dir(B-C)));
dot("
",F,dir(dir(90)*dir(F-circumcenter(D,I,IA))+dir(90)*dir(circumcenter(D,IB,IC)-F)));
dot("
",G,dir(dir(90)*dir(G-circumcenter(D,I,IA))+dir(90)*dir(circumcenter(A,B,C)-G)));
dot("
",H,dir(dir(90)*dir(circumcenter(D,IB,IC)-H)+dir(90)*dir(H-circumcenter(A,B,C))));
dot("
",I,dir(dir(90)*dir(circumcenter(D,I,IA)-I)+dir(A-I)));
dot("
",J,dir(J-circumcenter(A,B,C)));
dot("
",K,dir(K-circumcenter(A,B,C)));
dot("
",IA,dir(IA-circumcenter(D,I,IA)));
dot("
",IB,dir(dir(IB-IC)+dir(IB-IA)));
dot("
",IC,dir(dir(90)*dir(circumcenter(D,IB,IC)-IC)+dir(IC-IB)));
dot("
",P,dir(dir(A-I)+dir(C-B)));
dot("
",Q,dir(dir(IC-IB)+dir(B-C)));
[/asy]
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
.
[asy]
size(500);
pair A,B,C,D,E,F,G,H,I,J,K,IA,IB,IC,P,Q,GP;
B=(0,0);
C=(8,0);
A=intersectionpoint(Circle(B,6),Circle(C,9));
I=incenter(A,B,C);
path c=circumcircle(A,B,C);
J=intersectionpoint(I--(4*I-3*A),c);
IA=2*J-I;
IB=2*intersectionpoint(I--(4*I-3*B),c)-I;
IC=2*intersectionpoint(I--(4*I-3*C),c)-I;
K=intersectionpoint(IB--IC,c);
D=intersectionpoint(I--(I+(10,-12)),c);
path c1=circumcircle(D,I,IA),c2=circumcircle(D,IB,IC);
F=intersectionpoints(c1,c2)[1];
E=extension(B,C,D,F);
G=intersectionpoint(c1,c);
H=intersectionpoint(c2,c);
P=extension(A,I,B,C);
Q=extension(IB,IC,B,C);
GP=extension(A,2*foot(G,A,I)-G,B,C);
draw(A--B--C--A);
draw(c);
draw(A--J--G--D);
draw(C--GP);
draw(circumcircle(A,E,J),dashed);
dot("",A,dir(A-circumcenter(A,B,C)));
dot("
",B,dir(B-circumcenter(A,B,C)));
dot("
",C,dir(dir(90)*dir(circumcenter(A,B,C)-C)+dir(C-B)));
dot("
",D,dir(dir(90)*dir(circumcenter(D,I,IA)-D)+dir(90)*dir(D-circumcenter(D,IB,IC))));
dot("
",E,dir(dir(J-G)+dir(B-C)));
dot("
",G,dir(dir(90)*dir(G-circumcenter(D,I,IA))+dir(90)*dir(circumcenter(A,B,C)-G)));
dot("
",I,dir(dir(90)*dir(circumcenter(D,I,IA)-I)+dir(A-I)));
dot("
",J,dir(J-circumcenter(A,B,C)));
dot("
",P,dir(dir(A-I)+dir(C-B)));
dot("
",GP,dir(GP-B));
[/asy]
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
.
[asy]
size(500);
pair A,B,C,D,E,F,G,H,I,J,K,IA,IB,IC,P,Q,GP,BP,CP,DP;
B=(0,0);
C=(8,0);
A=intersectionpoint(Circle(B,6),Circle(C,9));
I=incenter(A,B,C);
path c=circumcircle(A,B,C);
J=intersectionpoint(I--(4*I-3*A),c);
IA=2*J-I;
IB=2*intersectionpoint(I--(4*I-3*B),c)-I;
IC=2*intersectionpoint(I--(4*I-3*C),c)-I;
K=intersectionpoint(IB--IC,c);
D=intersectionpoint(I--(I+(10,-12)),c);
path c1=circumcircle(D,I,IA),c2=circumcircle(D,IB,IC);
F=intersectionpoints(c1,c2)[1];
E=extension(B,C,D,F);
G=intersectionpoint(c1,c);
H=intersectionpoint(c2,c);
P=extension(A,I,B,C);
Q=extension(IB,IC,B,C);
BP=2*foot(B,IB,IC)-B;
CP=2*foot(C,IB,IC)-C;
DP=2*foot(D,IB,IC)-D;
draw(A--B--C--A);
draw(E--DP);
draw(BP--A--CP);
draw(IB--IC);
draw(c);
draw(circumcircle(B,IB,IC));
draw(circumcircle(E,IB,IC));
dot("",A,2*dir(dir(IB-A)+dir(C-A)));
dot("
",B,dir(B-circumcenter(A,B,C)));
dot("
",C,dir(dir(90)*dir(circumcenter(A,B,C)-C)+dir(C-B)));
dot("
",D,dir(dir(90)*dir(circumcenter(D,I,IA)-D)+dir(90)*dir(D-circumcenter(D,IB,IC))));
dot("
",E,dir(B-C)*dir(90));
dot("
",IB,dir(dir(IB-IC)+dir(IB-IA)));
dot("
",IC,dir(dir(90)*dir(circumcenter(D,IB,IC)-IC)+dir(IC-IB)));
dot("
",BP,dir(BP-circumcenter(B,IB,IC)));
dot("
",CP,dir(CP-circumcenter(B,IB,IC)));
dot("
",DP,dir(DP-E));
[/asy]
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
.
[asy]
size(500);
pair A,B,C,D,E,F,G,H,I,J,K,IA,IB,IC,P,Q,JP,KP;
B=(0,0);
C=(8,0);
A=intersectionpoint(Circle(B,6),Circle(C,9));
I=incenter(A,B,C);
path c=circumcircle(A,B,C);
J=intersectionpoint(I--(4*I-3*A),c);
IA=2*J-I;
IB=2*intersectionpoint(I--(4*I-3*B),c)-I;
IC=2*intersectionpoint(I--(4*I-3*C),c)-I;
K=intersectionpoint(IB--IC,c);
D=intersectionpoint(I--(I+(10,-12)),c);
path c1=circumcircle(D,I,IA),c2=circumcircle(D,IB,IC);
F=intersectionpoints(c1,c2)[1];
E=extension(B,C,D,F);
G=intersectionpoint(c1,c);
H=intersectionpoint(c2,c);
P=extension(A,I,B,C);
Q=extension(IB,IC,B,C);
JP=2*J-E;
KP=2*K-E;
draw(A--B--C--A);
draw(c);
draw(A--J);
draw(circumcircle(D,I,IA));
draw(circumcircle(D,IB,IC));
draw(D--F,dashed);
draw(B--Q--IB);
draw(G--JP);
draw(H--KP);
dot("",A,dir(A-circumcenter(A,B,C)));
dot("
",B,1/2*dir(B-dir(circumcenter(A,B,C))*dir(90)+dir(B-C)));
dot("
",C,dir(C-circumcenter(A,B,C))*dir(15));
dot("
",D,dir(dir(90)*dir(circumcenter(D,I,IA)-D)+dir(90)*dir(D-circumcenter(D,IB,IC))));
dot("
",E,dir(dir(H-K)+dir(B-C)));
dot("
",F,dir(dir(90)*dir(F-circumcenter(D,I,IA))+dir(90)*dir(circumcenter(D,IB,IC)-F)));
dot("
",G,dir(dir(90)*dir(G-circumcenter(D,I,IA))+dir(90)*dir(circumcenter(A,B,C)-G)));
dot("
",H,dir(dir(90)*dir(circumcenter(D,IB,IC)-H)+dir(90)*dir(H-circumcenter(A,B,C))));
dot("
",I,dir(dir(90)*dir(circumcenter(D,I,IA)-I)+dir(A-I)));
dot("
",J,dir(dir(circumcenter(A,B,C)-J)*dir(90)+dir(J-G)));
dot("
",K,dir(dir(K-circumcenter(A,B,C))*dir(90)+dir(K-H)));
dot("
",IA,dir(IA-circumcenter(D,I,IA)));
dot("
",IB,dir(dir(IB-IC)+dir(IB-IA)));
dot("
",IC,dir(dir(90)*dir(circumcenter(D,IB,IC)-IC)+dir(IC-IB)));
dot("
",P,dir(dir(A-I)+dir(C-B)));
dot("
",Q,dir(dir(IC-IB)+dir(B-C)));
dot("
",JP,dir(JP-circumcenter(D,I,IA)));
dot("
",KP,dir(KP-circumcenter(D,IB,IC)));
[/asy]
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
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.