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It is well-known that the centroid of a triangle is the intersection point of its three medians. The medians happen to be area bisectors, but it seems that most (all?) other lines through the centroid are not area bisectors. With other polygons there are area bisectors which pass through the centroid but not through a vertex (for instance any line which passes through the centroid of a square), but based on the examples that I am able to compute it seems that any line which passes through a vertex and the centroid is an area bisector. This leads me to pose the following general question:

Let $C$ be a convex body in the plane and let $L$ be a line which passes through the centroid of $C$ and an extreme point of $C$. Is $L$ necessarily an area bisector?

I couldn't find any relevant tools in my usual convex geometry references, but I apologize if I missed something obvious.

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    $\begingroup$ Take a line which goes through the centroid and is far from being an area bisector. You can deform C infinitesimally so that the intersection of L with the boundary is now a vertex. Now the line connecting the new vertex with the centroid is still far from being an area bisector. $\endgroup$
    – Yoav Kallus
    Commented Feb 17, 2014 at 14:08
  • $\begingroup$ Is it clear that you can deform $C$ as you describe while preserving convexity? $\endgroup$
    – Paul Siegel
    Commented Feb 17, 2014 at 14:17
  • $\begingroup$ Yes, I guess it is: e.g. replace the edge of a triangle with the arc of a circle with a very large radius. $\endgroup$
    – Paul Siegel
    Commented Feb 17, 2014 at 14:18
  • $\begingroup$ Please use top-level tags. $\endgroup$
    – user9072
    Commented Feb 17, 2014 at 14:25

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No, a line connecting a vertex to the centroid is not necessarily an area bisector. This follows easily from your observation that not every line through the centroid bisects area: take a line which goes through the centroid and is far from being an area bisector. You can deform C infinitesimally so that the intersection of L with the boundary is now a vertex (e.g. take the convex hull of C together with a point along L just outside C). Now the line connecting the new vertex with the new centroid is still far from being an area bisector.

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The statement is obviously wrong. Take a square with side $a$ and triangle of hight $a\sqrt3$ on one of its sides. This side passes through the centroid but is not an area bisector.

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  • $\begingroup$ Sorry, what do you mean by the "triangle of height $a \sqrt{3}$ on one of its sides"? $\endgroup$
    – Paul Siegel
    Commented Feb 17, 2014 at 14:16
  • $\begingroup$ ??? Triangle whose base is the side and height equals $a\sqrt3$. Actually, Yoav's comment answers this even better. $\endgroup$
    – Alex Degtyarev
    Commented Feb 17, 2014 at 14:21

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