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.
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 $\frac{\pi}{6}$.