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Brazil geometry
Problem
An equilateral triangle has side . A square is constructed on the outside of each side of the triangle. A right regular pyramid with sloping side is placed on each square. These pyramids are rotated about the sides of the triangle so that the apex of each pyramid comes to a common point above the triangle. Show that when this has been done the other vertices of the bases of the pyramids (apart from the vertices of the triangle) form a regular hexagon.
Solution
The key observation is that the midpoints of the edges of a regular tetrahedron form the vertices of a regular octahedron. So we can place a tetrahedron (side ) with its base on a plane next to an octahedron (side ) with its base on the same plane and with a face of each coinciding. Now the pyramid is just half an octahedron, so evidently if we rotate it from its initial position so that its apex coincides with the apex of a regular tetrahedron (whose base is ), then one face of the pyramid is parallel to the base of the tetrahedron. Then the result follows immediately.
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