2
$\begingroup$

Question: Let $A$ and $B$ be distinct points of a circle $o$. What is the set of possible centroids of triangles $ABC$ with $C\in o$?

Here is what I have: The angle at $C$ will always be the same as it is always subtended by the same arc as $A$ and $B$ are fixed.

There are 2 cases: Either $C$ lies on the small arc of $AB$ or $C$ lies on the big arc of $AB$.

In the case where $C$ lies on the large arc, by looking at the possible positions of $C$ one can observe that at some point $C$ and $A$ are on the same diameter and at another point $C$ and $B$ are on the same diameter. Also, it is worth mentioning that the midpoint at which $c$ intersects on the chord $AB$ does not depend on $C$ and is thus always the same.

One can observe, by determining various points that satisfy the criteria in a drawing, that the possible points $C$ all seem to lie on a smaller circle contained in the original circle $o$.

One can then guess that the center of this smaller circle is a possible center of homothethy mapping the smaller circle to the larger circle with scale $r_1/r_2$ where $r_1$ is the radius of the larger circle and $r_2$ is the radius of the smaller circle(I am really not sure about this).

I am not sure what to say about the case where $c$ lies on the small arc and am not sure where to continue with the problem.

Any help is appreciated.

$\endgroup$
6
$\begingroup$

[This is just a translation of Carl Schildkraut's answer into synthetic language.]

Note that the centroid of $ABC$ can be constructed by taking the midpoint $M$ of $AB$ and then taking the point $G$ which is $1/3$ of the way along $MC$ (closer to $M$). This means exactly that it is the image of $C$ under the homothety $T$ with center $M$ and scale factor $1/3$. Since homotheties preserve circles and their centers, this homothety maps a circle with center $D$ and radius $r$ to a circle with center $T(D)$ and radius $r/3$.

$\endgroup$
4
$\begingroup$

Here's a moderately obnoxious idea:

If you use complex numbers and set your circle to be the unit circle, then the centroid of the triangle determined by $a,b,c$ is simply $\frac{a+b+c}{3}$. Thus, the locus of possible centroids, as $A$ and $B$ are fixed and $C$ varies, is simply the circle with radius $\frac{1}{3}$ centered at $\frac{a+b}{3}$ (I think you need to get rid of two of the points because $C\neq A,B$, but oh well).

There should be a geometric interpretation of this idea as well, if you essentially try to phrase everything with homotheties.

$\endgroup$
  • $\begingroup$ Wow that is really interesting way of thinking about it, thanks. Unfortunately, I am trying to find a solution using homotheties. $\endgroup$ – rico Dec 11 '18 at 7:15
  • $\begingroup$ In terms of homotheties, the transformation $T:c\mapsto \frac{a+b+c}{3}$ is a homothety with center $\frac{a+b}{2}$ and scale factor $1/3$. So, it maps a circle of radius $r$ and center $d$ to a circle of radius $r/3$ and center $T(d)$. $\endgroup$ – Eric Wofsey Dec 11 '18 at 7:20

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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