146. There are two cases: 1) the center of the circumscribed sphere coincides with the center of the base, 2) the center of the circumscribed sphere is located at a point on the surface of the inscribed sphere, diametrically opposite to the center of the base. In the first case, the plane angle at the vertex is $\pi / 2$.

Consider the second case. Let $a, b$ and $l$ be the side of the base, the lateral edge, and the apothem of the lateral face, respectively. Then

$$
b^{2}=l^{2}+\frac{a^{2}}{4}
$$

$r$ - the radius of the inscribed sphere is equal to the radius of the circle inscribed in the isosceles triangle with base $a$ and lateral side $l$;

$$
r=\frac{a \sqrt{2 l-a}}{2 \sqrt{2 l+a}}
$$

$R$ - the radius of the circumscribed sphere is equal to the radius of the circle circumscribed around the isosceles triangle with base $a \sqrt{2}$ and lateral side $b$;

$$
R=\frac{b^{2} \sqrt{2}}{2 \sqrt{2 b^{2}-a^{2}}}
$$

The center of the circumscribed circle must be inside the triangle, which means $̈>>$. Since the distance from the center of the circumscribed sphere to the base is $2 r$, then $R^{2}-\frac{a^{2}}{2}=4 r^{2}$. Substituting $R$ and $\boldsymbol{r}$ in this

$$
\frac{\left(b^{2}-a^{2}\right)^{2}}{2\left(2 b^{2}-a^{2}\right)}=\frac{a^{2}(2 l-a)}{2 l+a}
$$

Expressing $b$ in terms of $a$ and $l$ using formula (1), we get

$$
\left(l^{2}-\frac{3 a^{2}}{4}\right)^{2}=a^{2}(2 l-a)^{2} .
$$

Considering that $b>a$ or $l>a \frac{\sqrt{3}}{2}$, we get that $a$ and $l$ satisfy the equation

$$
l^{2}-\frac{3 a^{2}}{4}=a(2 l-a),
$$

from which $\frac{l}{a}=1+\frac{\sqrt{\overline{3}}}{2}$ (for the second root $\left.\frac{l}{a}<\frac{\sqrt{\overline{3}}}{2}\right)$.

Answer: $\frac{\pi}{2}$ or $\frac{\pi}{6}$.
146. In a regular quadrilateral pyramid, the center of the circumscribed sphere lies on the surface of the inscribed sphere. Find the measure of the plane angle at the vertex of the pyramid.