3

This question appeared in a 1962 admissions test for an engineering school in Brazil:

Two equal squares with sides of length "m" are placed on parallel planes separated by a distance "h", such that

  • their centers belong to the same perpendicular line to both planes;
  • each side of one square is parallel to one diagonal of the other square.

Now consider the polyhedron whose edges are the sides of these two squares and the line segments that join each vertex of one square to the two vertices closest to it, belonging to the other square. The problem is to express its volume in terms of "m" and "h".

I have tried to start with the squares sides parallel to one another, and after joining the vertices as specified, give a "twist" of 45 degrees to one of them with respect to the other; but to no avail. Or maybe we could embed this polyhedron in a larger solid and then "slice out" some parts, in order to obtain its volume by difference.

Can someone give me any clues?

Eduardo B. T. Leite
  • 425
  • I like your second idea. Try starting with an octagon for the top and bottom faces. – amd May 18 '17 at 23:29

1 Answers1

1

Your polyhedron can be divided into a central prism with octagonal base, plus eight pyramids having as base each a lateral face of the prism (green in the diagram below), plus eight smaller pyramids with triangular base (orange in the diagram).

enter image description here

Every side of the regular octagons has length $(\sqrt2-1)m$, and the volume of the octagonal prism turns out to be $$ V_{prism}=2(\sqrt2-1)m^2h. $$

Each rectangular base pyramid has base area $(\sqrt2-1)mh$ and height $(\sqrt2-1)m/2$, so its volume is $$ V_{repyr}={3-2\sqrt2\over6}m^2h. $$

Each triangular base pyramid has base area $(2-\sqrt2)mh/4$ and height $(\sqrt2-1)m/2$, so its volume is $$ V_{trpyr}={3\sqrt2-4\over24}m^2h. $$

The volume of the polyhedron is then: $$ V=V_{prism}+8V_{repyr}+8V_{trpyr}={2+\sqrt2\over3}m^2h. $$

Intelligenti pauca
  • 50,470
  • 4
  • 42
  • 77