Dolbilin $H$.

On a plane, there are three pipes (circular cylinders of the same size with a circumference of 4 m). Two of them lie parallel and, touching each other along a common generatrix, form a tunnel above the plane. The third, perpendicular to the first two, cuts out a chamber in the tunnel. Find the area of the boundary of this chamber.
Horizontal sections of the chamber are rectangles with perimeters equal to twice the diameter of the pipe. For each such rectangle, the angle between its plane and the tangent to the surface of the chamber is the same at all points. The midpoints of the sides of these rectangles, when the section is moved, describe quarters of the pipe's circumference. Therefore, the surface area of the chamber is equal to the surface area of a tetrahedron, the faces of which are equal isosceles triangles with a base equal to the diameter of the pipe, and a height equal to a quarter of its circumference.

## Answer

$8 / \pi \mathrm{M}^{2}$.

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