In triangle $ABC$ given $BC=a$ and the altitude $BH=h_{b}=x$ and median $AM=m_{a}=y$
Then find the sides of triangle $ABC$ mean find $AB,AC$ ?
Also how we can represented ?
My try but not complete :
I know that : $m_{a}^{2}=\frac{2(b^{2}+c^{2})-a^{2})}{4}$
And I know that : $h_{b}=\frac{2S}{b}$ I have many relations but I can't use it ? I have already see your solution ?
The following two equations for the altitude $x$ and the median $y$ can be established,
$$b^2+c^2=\frac{a^2}2+2y^2$$ $$\sqrt{c^2-x^2}=b-\sqrt{a^2-x^2}$$
Square the second equation and eliminate $c$ to get,
$$b^2-b\sqrt{a^2-x^2}+\frac{a^2}4-y^2=0$$
Solve for the side $b$ in terms of known $a$, $x$ and $y$,
$$b= \frac12 \sqrt{a^2-x^2} +\frac12\sqrt{4y^2-x^2}$$
The side $c$ can then be obtained from $b$,
$$c= \sqrt{\frac{a^2}4+2y^2-b^2}$$ $$=\left[\frac14\left(\sqrt{a^2-x^2}-\sqrt{4y^2-x^2}\right)^2+x^2\right]^{1/2}$$