## Task B-3.5.

A regular right quadrilateral pyramid with apex $V$ and base $ABCD$ is inscribed in a sphere of radius $R$. Let $\cos \angle AVB = \frac{3}{4}$. What is the height of the pyramid expressed in terms of $R$?
## First Solution.

Let $ABCD$ be the base of the pyramid, and $N$ the foot of the perpendicular from the vertex $V$.

The lateral face of the pyramid is an isosceles triangle $ABV$. By applying the cosine rule, we get

$|AB|^2 = 2|AV|^2 - 2|AV|^2 \cos \angle AVB = 2|AV|^2 - \frac{3}{2}|AV|^2 = \frac{1}{2}|AV|^2$, so $|AV|^2 = 2|AB|^2$.

Furthermore, $|AN| = \frac{\sqrt{2}}{2}|AB|$.

In the cross-section of the pyramid, consider triangles $ANV$ and $ANS$, where point $S$ is the center, and $R$ is the radius of the sphere.

![](https://cdn.mathpix.com/cropped/2024_05_30_ea3ef29009412d1e0d2bg-16.jpg?height=542&width=394&top_left_y=760&top_left_x=820)

From the right triangle $ANV$, express the height of the pyramid $VN$:

$|VN|^2 = |AV|^2 - |AN|^2 = 2|AB|^2 - \frac{1}{2}|AB|^2 = \frac{3}{2}|AB|^2$, hence $|VN| = \frac{\sqrt{6}}{2}|AB|$.

Apply the Pythagorean theorem in the right triangle $ANS$. We get

$|AN|^2 + |NS|^2 = |AS|^2$

$\frac{1}{2}|AB|^2 + \left(\frac{\sqrt{6}}{2}|AB| - R\right)^2 = R^2$

$\frac{1}{2}|AB|^2 + \frac{3}{2}|AB|^2 - \sqrt{6}|AB| \cdot R + R^2 = R^2$

By simplifying the expression, we get $|AB| = \frac{\sqrt{6}}{2} R$.

Finally, $|VN| = \frac{\sqrt{6}}{2}|AB| = \frac{3}{2} R$.