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Old Kyiv MO Geometry
Kyiv City MO 1984-93 - geometry
1993.10.5
cotA +cotB+ cotC=(a^2+b^2+c^2)/4S 1993 Kyiv City MO 10.5
cotA +cotB+ cotC=(a^2+b^2+c^2)/4S 1993 Kyiv City MO 10.5
Source:
July 21, 2021
trigonometry
geometry
Problem Statement
Prove that for the sides
a
,
b
,
c
a, b, c
a
,
b
,
c
, the angles
A
,
B
,
C
A, B, C
A
,
B
,
C
and the area
S
S
S
of the triangle holds
cot
A
+
cot
B
+
cot
C
=
a
2
+
b
2
+
c
2
4
S
.
\cot A+ \cot B + \cot C = \frac{a^2+b^2+c^2}{4S}.
cot
A
+
cot
B
+
cot
C
=
4
S
a
2
+
b
2
+
c
2
.
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