In this remarkable paper 30 pages are occupied by the proof of the following innocently looking lemma:

Let $K$ be an origin-symmetric convex body in $\mathbb R^3$. There exist three planes through the origin splitting $K$ into $8$ parts of equal volume and such that each two of these planes split the cross-section of $K$ by the third one into $4$ parts of equal area.

I cannot shake off the feeling that there must be a half-page proof of this statement though I don't have one yet. I also know that MO is swarming with good topologists. Anybody up to the challenge?

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    $\begingroup$ Some naive thoughts: We can try and derive this from an intersection theory statement on $(\mathbb{RP}^2)^3$. Points correspond to triples of planes, and the condition that three planes split $K$ into $8$ parts of equal volume will give a codimension $3$ cycle for "generic" $K$. (We get codimension $3$ instead of codimension $7$ because diametrically opposite regions have the same area, by reflection symmetry of $K$). Similarly each of the conditions that two planes split a cross-section into four parts of equal area should lead to a codimension $1$ cycle. $\endgroup$ – dhy Jun 12 '17 at 2:28
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    $\begingroup$ Now one can hope that geometric considerations will show that (in this "generic" situation) the intersection class in $H^6((\mathbb{RP}^2)^3,\mathbb{Z}/2\mathbb{Z})$ will be nontrivial, and that furthermore that there is no contribution from some degenerate case e.g. two planes coinciding. Then it probably still remains necessary to pass to some perturbation of $K$, and to run this argument there instead. So even if this sketch can be made to work, I doubt that a fully rigorous proof would fit in a half-page... $\endgroup$ – dhy Jun 12 '17 at 2:32
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    $\begingroup$ Roman Karasev (private communication) suggests that this essentially follows from the results of [V.V. Makeev. Equipartition of a continuous mass distribution. Journal of Mathematical Sciences, 140:4 (2007), 551--557, mathnet.ru/links/a1722f058ed63ed36cb13d9e67fd255f/znsl299.pdf, theorems 5 and 6] $\endgroup$ – Fedor Petrov Jun 12 '17 at 13:42
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    $\begingroup$ (continued) Makeev partitions two measures which are absolutely continuous w.r.t. Lebesgue measure, but actually this absolute continuity is not essential. $\endgroup$ – Fedor Petrov Jun 12 '17 at 14:14
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    $\begingroup$ @FedorPetrov Except in our case the measures to partition depend on the planes. However the idea that if the solution is unique in some generic position, then there is an odd number of solutions in every generic position is amazing and certainly very promising. It should be rather standard, of course, but the good side of ignorance (mine) is the possibility to get surprised with the facts everybody else considers routine :-). $\endgroup$ – fedja Jun 12 '17 at 15:30

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