1
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ 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 Feb 17 '14 at 14:08
  • $\begingroup$ Is it clear that you can deform $C$ as you describe while preserving convexity? $\endgroup$ – Paul Siegel Feb 17 '14 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 Feb 17 '14 at 14:18
  • $\begingroup$ Please use top-level tags. $\endgroup$ – user9072 Feb 17 '14 at 14:25
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Sorry, what do you mean by the "triangle of height $a \sqrt{3}$ on one of its sides"? $\endgroup$ – Paul Siegel Feb 17 '14 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 Feb 17 '14 at 14:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.