3. Let $AB$ be the diameter of the circle $K$ with radius $R$ lying in the plane $\alpha$, and let $S$ be a point on the normal to the plane $\alpha$ at point $B$. A plane $\beta$ is drawn through point $S$, which is perpendicular to the plane $ABS$ and forms an angle of $\frac{\pi}{3}$ with the plane $\alpha$.

a) Determine the boundary values for $x = BS$ such that $\beta$ intersects the circle $K$ at two points $C$ and $D$.

b) Prove that the angles $SCA$ and $SDA$ are right angles.

c) Determine the sum $y$ of the squares of the edges of the tetrahedron $SDAC$ as a function of $x$ and $R$.

d) Determine the geometric locus of the extremum of the function $y(x)$ as $R$ varies.
Solution. a) Let $E$ be the intersection of the plane $\beta$ and the line $A B$. The triangle $S E B$ is a right triangle, where $\measuredangle S B E=\frac{\pi}{2}, \measuredangle S E B=\frac{\pi}{3}$, so $B E=\frac{x}{\sqrt{3}}$. The point $E$ is an internal point of the segment $A B$ (i.e., the circle $K$ and the plane $\beta$ have two common points) if and only if $0<x<2 R \sqrt{3}$.

b) Let $0<x<2 R \sqrt{3}$ and let $O$ be the center of the circle $K$. Then $O E=|B E-B O|=\left|\frac{x}{\sqrt{3}}-R\right|$. Note that

![](https://cdn.mathpix.com/cropped/2024_06_05_17a2a9aba963f6a4e511g-4.jpg?height=419&width=296&top_left_y=860&top_left_x=1169)
the triangles $C E O, C E A, S B E, S B A, S E C$ are right triangles with right angles at the vertices $E, E, B, B, E$ (diagram on the right). Now, by the Pythagorean theorem, we have:

$$
\begin{aligned}
& C E^{2}=C O^{2}-O E^{2}=R^{2}-\left(\frac{x}{\sqrt{3}}-R\right)^{2}=\frac{2 R x}{\sqrt{3}}-\frac{x^{2}}{3} \\
& A C^{2}=A E^{2}+C E^{2}=\left(2 R-\frac{x}{\sqrt{3}}\right)^{2}+\frac{2 R x}{\sqrt{3}}-\frac{x^{2}}{3}=4 R^{2}-\frac{2 R x}{\sqrt{3}} \\
& S C^{2}=S E^{2}+C E^{2}=S B^{2}+B E^{2}+C E^{2}=x^{2}+\frac{2 R x}{\sqrt{3}} \\
& S C^{2}+A C^{2}=x^{2}+4 R^{2}=B S^{2}+A B^{2}=S A^{2}
\end{aligned}
$$

Accordingly, the triangle $S C A$ is a right triangle with a right angle at vertex $C$. Similarly, it is proven that the angle $S D A$ is a right angle.

c) We have:

$$
\left.y=S A^{2}+S D^{2}+A D^{2}+S C^{2}+A C^{2}+C D^{2}=3\right\rangle S A^{2}+C D^{2}=\frac{5}{3} x^{2}+\frac{8 R}{\sqrt{3}} x+12 R^{2}
$$

d) The coordinates of the vertex of the parabola are

$$
x=-\frac{4 \sqrt{3}}{5} R, y=\frac{44}{5} R^{2}
$$

from which, by eliminating the parameter $R$, we obtain the geometric locus of the vertex

$$
y=\frac{55}{12} x^{2}, \quad x<0
$$

(The quadratic function obtained is considered as a function defined on the set of all real numbers. The function $y(x), 0<x<2 R \sqrt{3}$ has no extrema.)