a) Estimation. Symmetry either swaps polyhedra $A$ and $B$, or leaves each in place. In the first case, it swaps their centers of mass, so the plane of symmetry is perpendicular to the segment between the centers of mass and passes through its midpoint. In the second case, the plane of symmetry of the figure is a plane of symmetry for each of the polyhedra $A$ and $B$. Hence the estimate $1+2012=2013$.

Example. Let $A$ be a regular 2012-sided pyramid. On its axis of symmetry (obviously, the only one), choose a point outside $A$ and draw a plane $P$ perpendicular to the axis through this point. Let $B$ be the result of reflecting $A$ across $P$. All conditions of the problem are satisfied, and $P$ and the 2012 planes of symmetry of the pyramid $A$ are planes of symmetry of the resulting figure.

b) Estimation. Since polyhedra $A$ and $B$ have a different number of planes of symmetry, they are not equal and cannot be transformed into each other by the symmetry of the entire figure. Therefore, this symmetry leaves each of them in place and, in particular, is a symmetry of the polyhedron $A$, but it has only 2012 planes of symmetry by condition.

Example. Let $A$, as in part a), be a regular 2012-sided pyramid. Choose a point on its axis outside $A$, draw a plane perpendicular to the axis through this point, and reflect the base of the pyramid across this plane. Above this base, construct a straight prism that does not intersect $A$, and this will be $B$. Clearly, $B$ has 2013 planes of symmetry: one is parallel to the planes of the bases of the prism and is located midway between them, and the other 2012 pass through the axis of the prism and two opposite vertices of the base. They are also planes of symmetry for $A$ and for the entire figure.

c) Estimation. Since polyhedra $A$ and $B$ have a different number of axes of symmetry, they are not equal and cannot be transformed into each other by the symmetry of the entire figure. Therefore, this symmetry leaves each of them in place. It leaves the center of mass of each polyhedron in place. These centers do not coincide since the polyhedra are convex (this is significant!). Thus, the line connecting them has two fixed points. Therefore, it is the axis of symmetry.

Example. Let $A$ be a straight prism with bases that are regular 2011-sided polygons. The planes of the bases are considered horizontal. Then the prism has one vertical axis of symmetry, and 2011 horizontal axes of symmetry pass through the midpoint of the segment between the bases. Next, let $B$ be a straight prism with horizontal bases that are regular 2012-sided polygons, and the vertical axis is the same as that of $A$, with $A$ and $B$ having no common points. Then $A$ has 2012 axes of symmetry, $B$ has 2013, and the figure composed of them has a vertical axis of symmetry.

## Answer

a) 2013, b) 2012, c) 1.

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Frankinn B.P.
Convex polyhedra $A$ and $B$ have no common points. Polyhedron $A$ has exactly 2012 planes of symmetry. What is the maximum possible number of planes of symmetry of the figure consisting of $A$ and $B$, if $B$ has
a) 2012,
b) 2013 planes of symmetry?
c) What would the answer in part b) be if planes of symmetry are replaced by axes of symmetry?