261. Can a regular tetrahedron with an edge of 1 pass through a circular hole with a radius of: a) 0.45; b) 0.44? The thickness of the hole can be neglected.

## § 3. Geometric Loci

## § 3. Geometric Places of Points
261. Let $A B C D$ be a given tetrahedron. Take points $M$ and $N$ on the edges $B C$ and $B D$ and solve the following problem: for what positions of points $M$ and $N$ does the radius of the smallest circle containing the triangle $A M N$ (considering circles located in the plane $A M N$) reach its minimum value? (Obviously, the radius of the smallest hole cannot be less than the value of this radius. For this, it is sufficient to consider the moment in the process of the tetrahedron passing through the hole when two vertices of the tetrahedron are on one side of the plane of the hole, the third is on the other side, and the fourth is in the plane of the hole.)

Suppose that points $M$ and $N$ correspond to the desired triangle. Assume this triangle is acute-angled. Then the smallest circle containing it coincides with the circumscribed circle. Describe a circle around the triangle $A M N$ and consider the body obtained by rotating the arc $A M N$ of this circle around the chord $A N$. The line $B C$ must touch the surface of this body. Otherwise, we could take a point $M_{1}$ on $B C$ such that the radius of the circle circumscribed around $\triangle A M_{1} N$ would be less than the radius of the circle circumscribed around $\triangle A M N$. Moreover, $B C$ must touch the surface of the sphere passing through $A, M$, and $N$, with its center in the plane $A M N$. Similarly, this sphere must touch the line $B D$. Therefore, $|B M|=|B N|$. Let $|B M|=|B N|=x$. Let $K$ be the midpoint of $M N$, and $L$ be the projection of $B$ onto the plane $A M N$ ( $L$ is on the extension of $A K$ ). From the above, it follows that $L M$ and $L N$ are tangents to the circle circumscribed around $\triangle A M N$. $\triangle A M N$ is isosceles, $|A M|=|A N|=\sqrt{x^{2}-x+1},|M N|=x$. If $\widehat{M A N}=\alpha$, then

$$
\begin{gathered}
\cos \alpha=\frac{x^{2}-2 x+2}{2\left(x^{2}-x+1\right)}, \quad \sin \alpha=\frac{x \sqrt{3 x^{2}-4 x+4}}{2\left(x^{2}-x+1\right)} \\
|L K|=|M K| \operatorname{tg} \alpha=\frac{x^{2} \sqrt{3 x^{2}-4 x+4}}{2\left(x^{2}-2 x+2\right)}
\end{gathered}
$$

Consider $\triangle A K B, \widehat{A K B}=\beta>180^{\circ} ; \cos \beta=\frac{3 x-2}{\sqrt{3\left(3 x^{2}-4 x+4\right)}}$, $|L K|=-|K B| \cos \beta=\frac{x(2-3 x)}{2 \sqrt{3 x^{2}-4 x+4}}$. Equating the two expressions for $|L K|$, we get for $x$ after simplification the equation

$$
3 x^{3}-6 x^{2}+7 x-2=0
$$

The radius of the circle circumscribed around $\triangle A M N$ will be

$$
R=\frac{x^{2}-x+1}{\sqrt{3 x^{2}-4 x+4}}
$$

(It can be shown that if $\triangle A M N$ is a right triangle, then its hypotenuse is no less than $\sqrt{15-10 \sqrt{2}}>0.9$.) We will show that the tetrahedron can pass through the hole of the found radius.

Mark points $L$ and $P$ on the edges $C B$ and $C A$ such that $|C L|=$ $=|C P|=|B M|=|B N|=x$, where $x$ satisfies equation (1). Place the tetrahedron on the plane containing the given hole so that $M$ and $N$ are on the boundary of the hole. Rotate the tetrahedron around the line $M N$ until the edge $A B$, passing through the hole, becomes parallel to our plane. Then, keeping $A B$ parallel to this plane, move the tetrahedron $A B C D$ so that points $P$ and $L$ fall on the boundary of the hole. Finally, rotate the tetrahedron around $P L$ until the edge $D C$ exits the hole. (The tetrahedron will end up on the other side of our plane, with the face $A B C$ lying in this plane.)

Answer: the radius of the smallest hole $R=\frac{x^{2}-x+1}{\sqrt{3 x^{2}-4 x+4}}$, where $x$ is the root of the equation $3 x^{3}-6 x^{2}+7 x-2=0$. The corresponding calculations yield the approximate values $x \approx 0.3913, R \approx 0.4478$ with an error not exceeding 0.00005.