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Poland Contests
Poland - Second Round
2020 Poland - Second Round
4.
Easy Geometry - Hexagon
Easy Geometry - Hexagon
Source: Poland - Second Round P4
February 8, 2020
geometry
Problem Statement
Let
A
B
C
D
E
F
ABCDEF
A
BC
D
EF
be a such convex hexagon that
A
B
=
C
D
=
E
F
and
B
C
=
D
E
=
.
F
A
AB=CD=EF\; \text{and} \; BC=DE=.FA
A
B
=
C
D
=
EF
and
BC
=
D
E
=
.
F
A
Prove that if
∢
F
A
B
+
∢
A
B
C
=
∢
F
A
B
+
∢
E
F
A
=
24
0
∘
\sphericalangle FAB + \sphericalangle ABC=\sphericalangle FAB + \sphericalangle EFA = 240^{\circ}
∢
F
A
B
+
∢
A
BC
=
∢
F
A
B
+
∢
EF
A
=
24
0
∘
, then
∢
F
A
B
+
∢
C
D
E
=
24
0
∘
\sphericalangle FAB+\sphericalangle CDE=240^{\circ}
∢
F
A
B
+
∢
C
D
E
=
24
0
∘
.
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