Maths Olympiad Prep

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Geometry Difficulty 6.1 National olympiad Prove it Argentina

A cube with edge 1010 is cut into 2727 parallelepipeds by three pairs of planes parallel to its faces. The edges of the interior parallelepiped have lengths 11, 22 and 33. Find the sum of the volumes of the 88 corner parallelepipeds.

Solution

Imagine the eight corner parts yellow and the rest of the cube white. The two horizontal cuts produce three parallelepipeds. The middle one is white and has the same vertical dimension as the central piece. Assume the latter dimension to be 11 and remove the middle part. A 10×10×910 \times 10 \times 9 parallelepiped is obtained with 44 yellow parts instead of 88. This is because the initial 88 yellow parts come into 44 pairs with equal horizontal dimensions (the two of them) in every pair. So after removing the middle white part every pair becomes a single yellow piece. Repeat the same with the two cuts in direction left-right. They also produce three parts; the middle one is white. Its width equals the front-back dimension of the central piece which we assume to be 22. So removing the middle part yields a 10×8×910 \times 8 \times 9 parallelepiped in which the yellow portion consists of two parallelepipeds separated by a white parallelepiped 3×8×93 \times 8 \times 9. Remove this white piece; the entire yellow part remains in the shape of a parallelepiped 7×8×97 \times 8 \times 9, hence its volume is 789=5047 \cdot 8 \cdot 9 = 504.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.