11.20. 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.)

## Problems for Independent Solving
11.20. In the process of a tetrahedron passing through an opening, there will inevitably be a moment when vertex \( B \) is on one side of the plane of the opening, vertex \( A \) is in the plane of the opening, and vertices \( C \) and \( D \) are on the other side of the plane of the opening (or in the plane of the opening). Let at this moment the plane of the opening intersect the edges \( BC \) and \( BD \) at points \( M \) and \( N \); the circle of the opening contains the triangle \( AMN \). We will now determine in which position of points \( M \) and \( N \) the radius of the smallest circle containing the triangle \( AMN \) will be the smallest.

First, let's assume that the triangle \( AMN \) is acute-angled. Then the smallest circle containing it is its circumcircle (see Prasolov, II, 15.127). If the sphere, whose equator is the circumcircle of the triangle \( AMN \), does not touch, for example, the edge \( BC \), then inside this sphere on the edge \( BC \) we can take a point \( M' \) near \( M \) such that the triangle \( AM'N \) is acute-angled and the radius of its circumcircle is less than the radius of the circumcircle of the triangle \( AMN \). Therefore, in the position where the radius of the circumcircle of the triangle \( AMN \) is minimal, the considered sphere touches the edges \( BC \) and \( BD \), and thus \( BM = BN = x \). The triangle \( AMN \) is isosceles, and in it \( MN = x \), \( AM = AN = \sqrt{x^2 - x + 1} \). Let \( K \) be the midpoint of \( MN \), and \( L \) be the projection of \( B \) onto the plane \( AMN \). Since the center of the sphere lies in this plane, and the lines \( BM \) and \( BN \) are tangent to the given sphere, \( LN \) and \( LM \) are tangents to the circumcircle of the triangle \( AMN \). If \( \angle MAN = \alpha \), then \( LK = MK \lg \alpha = x^2 \sqrt{3x^2 - 4x + 4 / 2} (x^2 - 2x + 2) \). In the triangle \( A \), the angle \( AKB = \beta \) is obtuse and \( \cos \beta = (3x - 2) / \sqrt{3(3x^2 - 4x + 4)} \). Therefore, \( LK = -KB \cos \beta = x(2 - 3x) / 2 \sqrt{3x^2 - 4x + 4} \). Equating the two expressions for \( LK \), we get the equation for \( x \):

\[
3x^3 - 6x^2 + 7x - 2 = 0
\]

The radius \( R \) of the circumcircle of the triangle \( AMN \) is \( (x^2 - x + 1) / \sqrt{3x^2 - 4x + 4} \). Approximate calculations of the roots of the equation give values \( x \approx 0.3913 \), \( R \approx 0.4478 \) (with an error not exceeding 0.00005).

Now let's assume that the triangle \( AMN \) is not acute-angled. Let \( BM = x \), \( BN = y \). Then \( AM^2 = 1 - x + x^2 \), \( AN^2 = 1 - y + y^2 \), and \( MN^2 = x^2 + y^2 - xy \). Since \( \angle MAN \) is acute, \( AM^2 + AN^2 > MN^2 \). Suppose for definiteness that the angle \( \angle ANM \) is not acute, i.e., \( 1 - x + x^2 \geq (x^2 + y^2 - xy) + (1 - y + y^2) \). Then \( 0 \leq x \leq y(1 - 2y) / (1 - y) \), so \( y \leq 0.5 \), and thus \( x \leq 2y(1 - 2y) \leq 1/4 \). On the segment [0, 1/2], the quadratic polynomial \( 1 - x + x^2 \) decreases, so \( AM^2 \geq 1 - 1/4 + 1/16 = 13/16 > (0.9)^2 \), i.e., in the case of an obtuse-angled triangle \( AMN \), the radius of the smallest circle containing it is greater than in the case of an acute-angled triangle.

We will now prove that the tetrahedron can pass through the opening of the found radius \( R \). We will mark segments of length \( x \), where \( x \) is the root of equation (1), on the edges of the tetrahedron as shown in Fig. 84, and perform the following sequence of movements:

a) position the tetrahedron so that the circle of the opening is the circumcircle of the triangle \( AMN \), and rotate the tetrahedron around the line \( MN \) until point \( V \) reaches the plane of the opening;

b) move the tetrahedron so that the plane \( VMN \) remains parallel to its original position, and points \( P \) and \( Q \) reach the boundary of the opening;

c) rotate the tetrahedron around the line \( PQ \) until vertex \( D \) reaches the plane of the opening.

We will prove that all these operations are feasible. When rotating the tetrahedron around the line \( MN \), the plane of the opening intersects it along a trapezoid, the diagonal of which decreases from \( NA \) to \( NV \), and the acute angle at the larger base increases to \( 90^\circ \). Therefore, the radius of the circumcircle of the trapezoid decreases. Hence, operation a) and the analogous operation c) are feasible.

Take a point \( T \) on the edge \( DC \). The section of the tetrahedron \( ABCD \), parallel to \( VMN \) and passing through point \( T \), is a rectangle with a diagonal \( \sqrt{t^2 + (1 - t)^2} = \sqrt{2(t - 0.5)^2 + 0.5^2} \), where \( t = BT \). This implies the feasibility of operation b).

Answer: the tetrahedron can pass through an opening of radius 0.45, but it cannot pass through an opening of radius 0.4.