4. Given the two sides $\mathrm{a}$ and $\mathrm{b}$ of a triangle, and it is known that the medians on these two sides intersect at a right angle, find the third side of the triangle. Under what conditions does this triangle exist?
$\mathrm{BC}=\mathrm{a}$. Now find $\mathrm{AB}$, denoted by $\mathrm{c}$.
Let $O D=x, O E=y$. Then, in $\triangle A O E$, $\triangle A O B$, and $\triangle B O D$, we can obtain:
$$
4 x^{2}+y^{2}=\frac{b^{2}}{4}, 4 x^{2}+4 y^{2}=c^{2}, 4 x^{2}+16 y^{2}=a^{2} \text {. }
$$

Eliminating $\mathbf{x}$ and $y$, we get
$$
\mathrm{c}^{2}=\frac{\mathrm{a}^{2}+\mathrm{b}^{2}}{5} .
$$

Thus, the existence condition for a triangle with sides $a$, $b$, and $c$ is:
$$
5(a+b)^{2}>a^{2}+b^{2}, 5(a-b)^{2}<a^{2}+b^{2} \text {. }
$$

Obviously, the first inequality is satisfied for any $a$ and $b$. The second inequality can be transformed into:
$$
\mathrm{a}^{2}-\frac{5}{2} \mathrm{ab}+\mathrm{b}^{2}<0 .
$$

Solving this inequality for $\frac{a}{b}$, we finally get:
$$
\frac{1}{2}<\frac{a}{b}<2 .
$$
4. In $\triangle \triangle B G$ (Fig. 4), with $O$ as the midpoint,
$\mathrm{F}$ point, $\mathrm{AC}=\mathrm{b}$,