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2017 All-Russian Olympiad
8
A nice(Hard) problem from Russia!
A nice(Hard) problem from Russia!
Source: All Russian 2017,Grade 11,day 2,P8
May 1, 2017
geometry
incenter
Angle Chasing
incircle
Problem Statement
Given a convex quadrilateral
A
B
C
D
ABCD
A
BC
D
. We denote
I
A
,
I
B
,
I
C
I_A,I_B, I_C
I
A
,
I
B
,
I
C
and
I
D
I_D
I
D
centers of
ω
A
,
ω
B
,
ω
C
\omega_A, \omega_B,\omega_C
ω
A
,
ω
B
,
ω
C
and
ω
D
\omega_D
ω
D
,inscribed In the triangles
D
A
B
,
A
B
C
,
B
C
D
DAB, ABC, BCD
D
A
B
,
A
BC
,
BC
D
and
C
D
A
CDA
C
D
A
, respectively.It turned out that
∠
B
I
A
A
+
∠
I
C
I
A
I
D
=
18
0
∘
\angle BI_AA + \angle I_CI_AI_D = 180^\circ
∠
B
I
A
A
+
∠
I
C
I
A
I
D
=
18
0
∘
. Prove that
∠
B
I
B
A
+
∠
I
C
I
B
I
D
=
18
0
∘
\angle BI_BA + \angle I_CI_BI_D = 180^{\circ}
∠
B
I
B
A
+
∠
I
C
I
B
I
D
=
18
0
∘
. (A. Kuznetsov)
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