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Ukraine National Mathematical Olympiad
2023 Ukraine National Mathematical Olympiad
9.6
Angle condition again
Angle condition again
Source: Ukraine MO 2023 9.6
April 5, 2023
geometry
Problem Statement
A point
O
O
O
lies inside
△
A
B
C
\triangle ABC
△
A
BC
so that
∠
B
O
C
=
90
−
∠
B
A
C
\angle BOC=90-\angle BAC
∠
BOC
=
90
−
∠
B
A
C
. Let
B
O
,
C
O
BO, CO
BO
,
CO
meet
A
C
,
A
B
AC, AB
A
C
,
A
B
at
K
,
L
K, L
K
,
L
. Points
K
1
,
L
1
K_1, L_1
K
1
,
L
1
lie on the segments
C
L
,
B
K
CL, BK
C
L
,
B
K
so that
K
1
B
=
K
1
K
K_1B=K_1K
K
1
B
=
K
1
K
and
L
1
C
=
L
1
L
L_1C=L_1L
L
1
C
=
L
1
L
. If
M
M
M
is the midpoint of
B
C
BC
BC
, then prove that
∠
K
1
M
L
1
=
9
0
o
\angle K_1ML_1=90^{o}
∠
K
1
M
L
1
=
9
0
o
.Proposed by Anton Trygub
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