1971 IMO Problems/Problem 2

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Problem

Consider a convex polyhedron $P_1$ with nine vertices $A_1, A_2, \cdots, A_9;$ let $P_i$ be the polyhedron obtained from $P_1$ by a translation that moves vertex $A_1$ to $A_i(i=2,3,\cdots, 9).$ Prove that at least two of the polyhedra $P_1, P_2,\cdots, P_9$ have an interior point in common.

Solution

WLOG let $A_1$ be the origin $0$. Take any point $A_i$, then $P_i=A_i+P_1$, lies in $2 P_1$, the polyhedron $P_1$ stretched by the factor $2$ on $P_1=0$. More general: take any $p,q$ in any convex shape $S$. Then $p+q \in 2S$. Prove: since $S$ is convex, $\frac{p+q}{2} \in S$, thus $p+q \in 2S$.

Now all these nine polyhedrons lie inside $2 P_1$. Let $V$ be the volume of $P_1$. Then some polyhedrons with total sum of volumes $9V$ lie in a shape of volume $8V$, thus they must overlap, meaning that they have an interior point in common.

The above solution was posted by ZetaX. The original thread for this problem can be found here: [1]

See Also

1971 IMO (Problems) • Resources
Preceded by
Problem 1
1 2 3 4 5 6 Followed by
Problem 3
All IMO Problems and Solutions
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