12.7. In the regular quadrilateral pyramid $V A B C D$, the height has a length of $h$ and is the diameter of a sphere, and $m(\angle A V B)=\varphi$. Determine the length of the curve obtained at the intersection of the sphere with the lateral surface of the pyramid.
## Solution:

![](https://cdn.mathpix.com/cropped/2024_06_06_bf7dcdb792e591bc55feg-2.jpg?height=530&width=1330&top_left_y=83&top_left_x=466)

The curve obtained at the intersection of the sphere with the lateral surface of the pyramid consists of 4 congruent arcs of circles, obtained at the intersection of the lateral faces with the sphere.

We determine the length of the arc $N M$ of the circle with center at point $I$, obtained at the intersection of the sphere with the plane $B V A$.

Let $A B=a$. Then $a=2 V A \sin \frac{\varphi}{2}, A C=a \sqrt{2}, O A=\frac{a \sqrt{2}}{2}$.

In the right triangle $V O A$, we get $h^{2}+\frac{a^{2}}{2}=\frac{a^{2}}{4 \sin ^{2} \frac{\varphi}{2}}$, which implies $a=\frac{2 h \sin \frac{\varphi}{2}}{\sqrt{\cos \varphi}}$ and $V A=\frac{h}{\sqrt{\cos \varphi}}$.

Applying the leg theorem in the triangle $V O A$, we get $V O^{2}=V M \cdot V A$, which implies $V M=h \sqrt{\cos \varphi}$.

Let $M I=R$, then in the triangle $V M N$ we get $2 M I=2 R=\frac{V M}{\sin \left(90^{\circ}-\frac{\varphi}{2}\right)}=\frac{V M}{\cos \frac{\varphi}{2}}=\frac{h \sqrt{\cos \varphi}}{\cos \frac{\varphi}{2}}$, which implies $R=\frac{h \sqrt{\cos \varphi}}{2 \cos \frac{\varphi}{2}}$. Since $m(\angle M I N)=2 \varphi$, we obtain that the length of the arc $M N$ is $l_{M N}=2 \varphi R=\frac{h \varphi \sqrt{\cos \varphi}}{\cos \frac{\varphi}{2}}$, and the length of the curve obtained at the intersection of the sphere with the lateral surface of the pyramid is $l=\frac{4 h \varphi \sqrt{\cos \varphi}}{\cos \frac{\varphi}{2}}$.