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Saturday, July 16, 2011

Giza-IQ Test - Solution 2

This is the solution to the second of the two problems I posted on July 11.

1.02    The billiard balls are centred on the vertices of a regular tetrahedron. Their points of contact with each other are in the middle of the edges and, as the radius of the balls is 1, the length of the edges is 2.

To determine the radius of the small ball, we need to subtract 1 (which is the radius of the billiard balls) from the distance between the centre of the tetrahedron (S in the figure) and one of the tetrahedron vertices.

Point A, which is the base of the tetrahedron height, is located at 1/3 of the bottom-face height, while the centre of the tetrahedron is at 1/4 of its height. These two facts are obvious when one considers that A and S are the barycentres respectively of the tetrahedron bottom face and of the whole tetrahedron.

face height = sqrt(22 – 12) = sqrt(3)
=> 2/3 of face height = 2 / sqrt(3)

tetrahedron height = sqrt(22 – (2 / sqrt(3)) 2) = 2 * sqrt(2/3)
=> 3/4 of tetrahedron height = sqrt(3/2)
=> small radius = sqrt(3/2) – 1 = 0.2247..

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