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1987 All Soviet Union Mathematical Olympiad
443
ASU 443 All Soviet Union MO 1987 r. heptagon 1/A_1A_5+1/A_1A_3=1/A_1A_7
ASU 443 All Soviet Union MO 1987 r. heptagon 1/A_1A_5+1/A_1A_3=1/A_1A_7
Source:
August 7, 2019
Heptagon
geometry
Problem Statement
Given a regular heptagon
A
1
.
.
.
A
7
A_1...A_7
A
1
...
A
7
. Prove that
1
∣
A
1
A
5
∣
+
1
∣
A
1
A
3
∣
=
1
∣
A
1
A
7
∣
\frac{1}{|A_1A_5|} + \frac{1}{|A_1A_3| }= \frac{1}{|A_1A_7|}
∣
A
1
A
5
∣
1
+
∣
A
1
A
3
∣
1
=
∣
A
1
A
7
∣
1
.
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