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1996 Moldova Team Selection Test
3
Prove that $\frac{\sin(B+C)}{\sin B \cdot \sin C} \geq \frac{2}{3}.$
Prove that $\frac{\sin(B+C)}{\sin B \cdot \sin C} \geq \frac{2}{3}.$
Source: Moldova TST 1996
August 8, 2023
trigonometry
geometry
Problem Statement
In triangle
A
B
C
ABC
A
BC
medians from
B
B
B
and
C
C
C
are perpendicular. Prove that
sin
(
B
+
C
)
sin
B
⋅
sin
C
≥
2
3
.
\frac{\sin(B+C)}{\sin B \cdot \sin C} \geq \frac{2}{3}.
s
i
n
B
⋅
s
i
n
C
s
i
n
(
B
+
C
)
≥
3
2
.
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