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Czech-Polish-Slovak Match
2021 Czech-Austrian-Polish-Slovak Match
6
6
Part of
2021 Czech-Austrian-Polish-Slovak Match
Problems
(1)
So many circles
Source: 2021 Czech-Polish-Slovak Match, P6
8/3/2021
Let
A
B
C
ABC
A
BC
be an acute triangle and suppose points
A
,
A
b
,
B
a
,
B
,
B
c
,
C
b
,
C
,
C
a
,
A, A_b, B_a, B, B_c, C_b, C, C_a,
A
,
A
b
,
B
a
,
B
,
B
c
,
C
b
,
C
,
C
a
,
and
A
c
A_c
A
c
lie on its perimeter in this order. Let
A
1
≠
A
A_1 \neq A
A
1
=
A
be the second intersection point of the circumcircles of triangles
A
A
b
C
a
AA_bC_a
A
A
b
C
a
and
A
A
c
B
a
AA_cB_a
A
A
c
B
a
. Analogously,
B
1
≠
B
B_1 \neq B
B
1
=
B
is the second intersection point of the circumcircles of triangles
B
B
c
A
b
BB_cA_b
B
B
c
A
b
and
B
B
a
C
b
BB_aC_b
B
B
a
C
b
, and
C
1
≠
C
C_1 \neq C
C
1
=
C
is the second intersection point of the circumcircles of triangles
C
C
a
B
c
CC_aB_c
C
C
a
B
c
and
C
C
b
A
c
CC_bA_c
C
C
b
A
c
. Suppose that the points
A
1
,
B
1
,
A_1, B_1,
A
1
,
B
1
,
and
C
1
C_1
C
1
are all distinct, lie inside the triangle
A
B
C
ABC
A
BC
, and do not lie on a single line. Prove that lines
A
A
1
,
B
B
1
,
C
C
1
,
AA_1, BB_1, CC_1,
A
A
1
,
B
B
1
,
C
C
1
,
and the circumcircle of triangle
A
1
B
1
C
1
A_1B_1C_1
A
1
B
1
C
1
all pass through a common point.Josef Tkadlec (Czech Republic), Patrik Bak (Slovakia)