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International Contests
Austrian-Polish
1984 Austrian-Polish Competition
4
4
Part of
1984 Austrian-Polish Competition
Problems
(1)
PA_1+PA_3+PA_5+PA_7 = PA_2+PA_4+PA_6 in regular heptagon
Source: Austrian-Polish 1984
5/25/2019
A regular heptagon
A
1
A
2
.
.
.
A
7
A_1A_2... A_7
A
1
A
2
...
A
7
is inscribed in circle
C
C
C
. Point
P
P
P
is taken on the shorter arc
A
7
A
1
A_7A_1
A
7
A
1
. Prove that
P
A
1
+
P
A
3
+
P
A
5
+
P
A
7
=
P
A
2
+
P
A
4
+
P
A
6
PA_1+PA_3+PA_5+PA_7 = PA_2+PA_4+PA_6
P
A
1
+
P
A
3
+
P
A
5
+
P
A
7
=
P
A
2
+
P
A
4
+
P
A
6
.
geometry
Heptagon
regular polygon