Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.8 National Olympiad Prove it Italy

Problem:

Let a rectangular parallelepiped $ABCD A'B'C'D'$ be given, where $ABCD$ is the bottom face with the letters assigned in clockwise order, and $A, B, C$ , and $D$ are below $A', B', C'$ , and $D'$ respectively. The parallelepiped is divided into eight pieces by three planes mutually orthogonal and parallel to the faces of the parallelepiped. For each vertex $P$ of the parallelepiped let $V_P$ denote the volume of the piece of the parallelepiped that contains $P$ . Given that $V_A=40$ , $V_C=300$ , $V_{B'}=360$ and $V_{C'}=90$ , what is the volume of the parallelepiped $ABCD A'B'C'D'$ ?

Solution

Solution:

Let us call $x, y, z$ the distances of $A$ from the three cutting planes, and $x', y', z'$ the distances of $C'$ from the same planes, so that $x + x' = AB$ , $y + y' = AD$ , $z + z' = AA'$ .
When two pieces share a common face, their volumes are proportional to their respective heights, and thus we obtain the following equalities:
$\frac{y}{y'} = 4 = \frac{V_{B'}}{V_{C'}} = \frac{V_B}{V_C}, \quad \frac{z}{z'} = \frac{10}{3} = \frac{V_C}{V_{C'}}, \quad \frac{x}{x'} = \frac{V_A}{V_B}$
From the first equality we obtain that $V_B = 4 V_C = 1200$ , and substituting into the last one we obtain that $\frac{x}{x'} = \frac{1}{30}$ . If we call $V$ the volume of the parallelepiped, we have:
$\frac{V}{V_{C'}} = \frac{(x + x')(y + y')(z + z')}{x' y' z'} = \left(\frac{x}{x'} + 1\right)\left(\frac{y}{y'} + 1\right)\left(\frac{z}{z'} + 1\right) = \frac{403}{18}$
from which $V = \frac{403}{18} V_{C'} = \frac{403}{18} \cdot 90 = 2015$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.