Centroid locus

As I did this numerically, this is just a sketch solution. First of all, here's a picture of the locus:

Some notation: $O(0,0)$ is the circumcentre; $I$ is the incentre; $d=OI$ is the distance between them; $G$ is the centroid; $R=4$ and $r=1$ are the circumradius and inradius. Let $A(R,0)$ be the fixed vertex.

By Euler's theorem, $OI^2=R^2-2Rr$ so $d$ is fixed and $I$ lies on a circle of radius $d$ centred at $O$ . So we can parameterise the point $I$ (and hence the triangles) by saying $I(d \cos(q),d \sin(q))$ .

The aim is to calculate the coordinates of $B$ and $C$ in terms of $q$ , so that we can find $G$ .

Since we can now work out $AI$ , we also get $\angle BAC=2\sin^{-1} \frac{r}{AI}$ .

This is in fact enough; we know the gradient of the line $AB$ , and that it goes through point $A$ ; since $OB=R$ we can work out the coordinates of $B$ (and likewise $C$ ) from this.

By tabulating the position of $G$ for a high resolution of $q$ values, we can work out the path length to be approximately $\boxed{15.55673\ldots}$ .

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This is messy, and (to me) quite unsatisfactory. For a deeper dive into this, we can use some other triangle formulas to find $G$ .

We have $GO^2 = \frac19 \left(R^2-a^2-b^2-c^2 \right)$ and $GA^2 = \frac19 \left(2b^2+2c^2-a^2 \right)$ , where $a$ is as usual the length of the side opposite $A$ , etc. We'll also use $A$ to denote the angle $\angle BAC$ . Note that these are symmetric in lengths $b$ and $c$ - this is to be expected from the way they're defined.

By the extended sine rule, $a=2R \sin A$ .

Now consider the triangle $BIC$ . By definition of the incentre, the altitude from $I$ has length $r$ . Also it's easy to show that $\angle BIC = \frac{A+\pi}{2}$ . Using the fact the incentre lies on the angle bisectors, $\angle IBC=\frac{B}{2}$ .

Combining all of this, $BC=a=2R\sin A=r\left(\cot \frac{B}{2}+\cot \frac{C}{2}\right)=r\left(\cot \frac{B}{2}+\tan \frac{A+B}{2}\right)$

This can be written as a quadratic in $\cot\frac{B}{2}$ .

Ultimately, we can get expressions for $b+c$ and $bc$ , and from these work out $b^2+c^2$ to get $GO$ and $GA$ . This gives a slightly neater way to work out the position of $G$ ; but the expressions don't reduce very nicely. I'll post what I've found if anyone's interested, but I'd be really keen to see a less numerical solution.