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International Mathematics Competition
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2023

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IMC2023: Day 2, Problem 9

Problem 9. We say that a real number V is good if there exist two closed convex subsets X, Y of the unit cube in R3, with volume V each, such that for each of the three coordinate planes (that is, the planes spanned by any two of the three coordinate axes), the projections of X and Y onto that plane are disjoint.

Find sup.

Josef Tkadlec and Arseniy Akopyan

Solution. We prove that \displaystyle \sup\{V\mid V \text{ is good}\} = 1/4.

We will use the unit cube \displaystyle U=[-1/2,1/2]^3.

For \displaystyle \varepsilon\to 0, the axis-parallel boxes \displaystyle X=[-1/2,-\varepsilon]\times [-1/2,-\varepsilon]\times [-1/2,1/2] and \displaystyle Y=[\varepsilon,1/2]\times[\varepsilon,1/2]\times[-1/2,1/2] show that \displaystyle \sup\{V\}\ge 1/4.

To prove the other bound, consider two admissible convex bodies \displaystyle X\displaystyle Y. For any point \displaystyle P=[x,y,z]\in U with \displaystyle xyz\ne 0, let \displaystyle \overline{P}=\{[\pm x,\pm y, \pm z]\} be the set consisting of 8 points (the original \displaystyle P and its 7 ``symmetric'' points). If for each such \displaystyle P we have \displaystyle |\overline{P} \cap (X\cup Y)|\le 4, then the conclusion follows by integrating. Suppose otherwise and let \displaystyle P be a point with \displaystyle |\overline{P} \cap (X\cup Y)|\ge 5. Below we will complete the proof by arguing that:

  1. [(1)] we can replace one of the two bodies (the ``thick'' one) with the reflection of the other body about the origin, and
  2. [(2)] for such symmetric pairs of bodies we in fact have \displaystyle |\overline{P} \cap (X\cup Y)|\le 4, for all \displaystyle P.

To prove Claim (1), we say that a convex body is thick if each of its three projections contains the origin. We claim that one of the two bodies \displaystyle X, \displaystyle Y is thick. This is a short casework on the 8 points of \displaystyle \overline{P}. Since \displaystyle |\overline{P} \cap (X\cup Y)|\ge 5, by pigeonhole principle, we find a pair of points in \displaystyle \overline{P} \cap (X\cup Y) symmetric about the origin. If both points belong to one body (say to \displaystyle X), then by convexity of \displaystyle X the origin belongs to \displaystyle X, thus \displaystyle X is thick. Otherwise, label \displaystyle \overline{P} as \displaystyle ABCDA'B'C'D'. Wlog \displaystyle A\in X, \displaystyle C'\in Y is the pair of points in \displaystyle \overline{P} symmetric about the origin. Wlog at least 3 points of \displaystyle \overline{P} belong to \displaystyle X. Since \displaystyle X, \displaystyle Y have disjoint projections, we have \displaystyle C,B',D'\not\in X, so wlog \displaystyle B,D\in X. Then \displaystyle Y can contain no other point of \displaystyle \overline{P} (apart from \displaystyle C'), so \displaystyle X must contain at least 4 points of \displaystyle \overline{P} and thus \displaystyle A'\in X. But then each projection of \displaystyle X contains the origin, so \displaystyle X is indeed thick.

Note that if \displaystyle X is thick then none of the three projections of \displaystyle Y contains the origin. Consider the reflection \displaystyle Y'=-Y of \displaystyle Y about the origin. Then \displaystyle (Y,Y') is an admissible pair with the same volume as \displaystyle (X,Y): the two bodies \displaystyle Y and \displaystyle Y' clearly have equal volumes \displaystyle V and they have disjoint projections (by convexity, since the projections of \displaystyle Y miss the origin). This proves Claim (1).

Claim (2) follows from a similar small casework on the 8-tuple \displaystyle \overline{P}: For contradiction, suppose \displaystyle |\overline{P}\cap Y'|=|\overline{P}\cap Y|\ge 3. Wlog \displaystyle A\in Y'. Then \displaystyle C'\in Y, so \displaystyle C,B',D'\not\in Y', so wlog \displaystyle B,D\in Y'. Then \displaystyle B',D'\in Y, a contradiction with \displaystyle (Y,Y') being admissible.

There are more examples with \displaystyle V\to1/4, e.g. \displaystyle X a union of two triangular pyramids with base \displaystyle ACD' – one with apex \displaystyle D, one with apex at the origin (and \displaystyle Y symmetric with \displaystyle X about the origin).

The word ``convex'' matters. E.g., in a \displaystyle 3\times 3\times 3 cube, one can set \displaystyle X to be a \displaystyle 2\times 2\times 2 sub-cube, and \displaystyle Y to be the (non-convex) 3D L-shape consisting of 7 unit cubes. This shows that without convexity we have \displaystyle V\ge 7/27>1/4.

IMC
2023

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