4.34. All edges (including the base sides) of a triangular pyramid are equal. Find the ratio of the radius of the sphere inscribed in the pyramid to its height.
4.34. Let $a$ be the edge of the pyramid $SABC$, $OM=r$ be the radius of the inscribed sphere, and $SH=\boldsymbol{h}$ be the height of the pyramid (Fig. 4.30). In $\triangle SAF$, we have $SF=\frac{a \sqrt{3}}{2}$, $HF=\frac{1}{3} AF=\frac{a \sqrt{3}}{6}$; therefore,

$$
h=SH=\sqrt{SF^{2}-HF^{2}}=\sqrt{\frac{3 a^{2}}{4}-\frac{3 a^{2}}{36}}=\frac{a \sqrt{6}}{3}
$$

Since $\triangle SOM \sim \Delta SFH$, we have $\frac{OM}{OS}=\frac{HF}{SF}$, or $\frac{r}{h-r}=\frac{\frac{a \sqrt{3}}{6}}{\frac{a \sqrt{3}}{2}}$, i.e., $\frac{r}{h-r}=\frac{1}{3}$. Thus, $\frac{r}{h}=\frac{1}{4}$.

Answer: $1: 4$.