Mock AIME II 2012 Problems/Problem 13
Regular octahedron (such that points , , , and are coplanar and form the vertices of a square) is divided along plane , parallel to line , into two polyhedra of equal volume. The cosine of the acute angle plane makes with plane is . Given that , find the area of the cross section made by plane with octahedron .
Let be the center of the octahedron. The plane must pass through in order to bisect the area of the octahedron. We see that the cross-section will be a hexagon as it passes through six of the eight faces. By symmetry, the area of this hexagon is twice that of the trapezoid contained within square pyramid .
We proceed by finding the height of the square pyramid, which is . The altitude from to of is . The distance from to the midpoint of is half the side length of the square , so its . The altitude of , the segment joining and the midpoint of , and make a right triangle. Then by the Pythagorean Theorem, so . Now consider of this right triangle. We have . This will be important later.
Back to the trapezoid. One of its bases is on and the other base must be on or (it cannot be on the other two faces as it is parallel to ). WLoG let the base be on . Let be the midpoint of this base and consider . We have that is the complement of the angle between the plane and the square . Then . Now consider . This is the same angle as the we had before. Then .
Now by the Law of Sines so . By the Law of Cosines, . This solves to and . We reject the solution as it is too large; must be largest side in as is the largest angle. We know this as the other angles have sines less that and so have values less than .
Now on we have that the second base is parallel to . Let be the length of the second base. Then by the two similar triangles, where we have compared the side length of the triangles to their heights. As is a given, this solves to .
Then the area of the trapezoid is . The area of the whole hexagon is twice this, so the final answer is