# One interesting formua

#### ns1954

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

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

-Dan

#### Idea

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 $b,c$

#### ns1954

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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