One interesting formua

Dec 2009
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0
If $t_a,\ t_b,\ t_c$ are medians of triangle, $\alpha,\ \beta,\ \gamma$ interior angles, $P$ area of triangle, prove formulas:
$${t_a}^2=\frac{1}{4}a^2+2P\cot\alpha$$
$${t_b}^2=\frac{1}{4}b^2+2P\cot\beta$$
$${t_c}^2=\frac{1}{4}c^2+2P\cot\gamma$$
 

topsquark

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Jan 2006
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We aren't going to do your homework for you, but we can help. How far have you got on these?

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Jun 2013
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Lebanon
we know the following

\(\displaystyle 2\left(b^2+c^2\right)=\left(2t_a\right){}^2+a^2\)

\(\displaystyle a^2=b^2+c^2-2 b c \cos \alpha\)

\(\displaystyle P=\frac{1}{2}b c \sin \alpha\)

Eliminate $b,c$
 
Dec 2009
54
0
Yes, this is solution. You forgot to mention formulas $abc=4PR$ and $\frac{a}{\sin\alpha}=2R$, where $P$ is area of triangle, and $R$ is radius of circumscribed circle.
 
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