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Let $A$ be a compact subset of $\mathbb{R}^n$. Consider the set of hyperplanes in $\mathbb{R}^n$ that separate $A$ into two pieces with equal measure (Lebesgue). Now take the intersection over all such planes. Call a point $x$ a centroid of $A$ if it lies in this intersection. It seems like this definition is equivalent to the usual definitions for centroid, but if not please let me know. My main question is: Is the resulting intersection always non-empty?

I'm mostly interested in the cases $n=2$ and $n=3$, but it seems like the answer will probably be the same for all $n > 1$.

For example, consider the sphere $S^2 \subset \mathbb{R}^3$. Any plane through the origin will cut $S^2$ into two sets of equal measure, and no other plane will, so there is a unique centroid.

EDIT: This post may be related, but I don't think it answers my question. https://mathoverflow.net/questions/248206/a-question-about-the-centroids-of-compact-subsets-of-euclidean-spaces

EDIT 2: I realized that I'm asking the wrong question, since the answer to the above is clearly no, as demonstrated in the comments. My revised question is: Given a compact set $A$, does there exists a point $x$ such that any hyperplane through $x$ separates $A$ into two sets of equal measure? In the example of a union of two disjoint disks given below, the centroid of the set would be such a point. Does such a point always exist?

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    I don't think this is a good definition of centroid. Consider two disjoint disks in the plane, with the same radius. The centroid should be the midpoint of their centers. However, *any* line that has one circle on either side will cut the figure into two pieces of equal measure, and such a line need not go through the midpoint. – 2017-01-25
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    No. If you consider a triangle on a plane, then there are three medians which separate it into equal areas, but any other line which goes through the medians intersection breaks the triangle into parts with different areas. – 2017-01-25
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    @SergeiGolovan Also, there exist lines which may not pass through it's centroid but nevertheless, may divide the triangle into 2 equal measures. – 2017-01-25
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    Sergei's example provides the counter example to your revised question. – 2017-01-25
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    @PragnyaJha Sergei's example demonstrates that the intersection of the three medians is not a point satisfying what I want. But as far as I can tell, this does not demonstrate that no such point exists for the triangle. – 2017-01-25
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    True, but symmetry would inform us that no point other than the intersection of the medians will be a suitable candidate for your revised question. Consider the special case of an equilateral triangle and assume that there exists a point inside the triangle through which every line passing divides the triangle into 2 equal areas. Now consider three lines passing through the point which are parallel to three of the triangle's sides. It can be shown that they can't at the same time divide the triangle into two portions of equal areas and hence a contradiction would ensue. – 2017-01-25
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    @PragnyaJha Thank you! – 2017-01-25

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Not every line through the centroid of a triangle splits the triangle in pieces with the same area: enter image description here and not every line splitting the triangle in halves goes through a common point: $GH$ is a "splitting segment" iff $CG\cdot CH=\frac{1}{2}CB\cdot CA$, but here enter image description here the red splitting segments are not concurrent since $\frac{1}{\sqrt{2}}\neq\frac{2}{3}$.