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Sunday, July 24, 2011

Giza-IQ Test - Solution 8

If you lie the octahedron on its ADE face, its top view is as follows:

The area of the smallest enclosing rectangle is given by BC, which is 1, multiplied by the projection of DF onto the horizontal plane, which is twice the distance between the centre of a face and one of its vertices = 2 * 2/3 * sqrt(3)/ 2 = 2/3 * sqrt(3)

The height of our box coincides with the distance between the faces ADE and BCF, which is twice the radius of the inscribed sphere.

Using the coordinate axes shown in the figure below, I write the equation of the sphere centred in the origin that has on its surface the centre of one of the faces. If you find a better method for calculating the radius, I’d be happy to hear from you. I just use the first method that comes to mind.

Considering that CDEF is a square of side 1, the distance of each vertex from the centre of the octahedron is sqrt(2)/2. Therefore, the coordinates of the vertices B, C, and F are:
B: (0, 0, sqrt(2)/2)
C: (0, sqrt(2)/2, 0)
F: (sqrt(2)/2, 0, 0)

To calculate the coordinates of the centre of the BCF face, I just average the coordinates of its vertices:

centre of BCF: (sqrt(2)/6, sqrt(2)/6, sqrt(2)/6)

The equation of a sphere centred in the origin is simply x2 + y2 + z2 = r2, and r turns out to be sqrt(6)/6.

We can finally calculate the volume of our box as:

V = area of base * height = [2/3 * sqrt(3)] * [sqrt(6) / 3] = 2/3 * sqrt(2)

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