1971 IMO Problems/Problem 2

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]

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

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