<s>[INST] Centroid of a non-polygonal concave shape determined by an equilateral triangle an its incircle I have a diagram here of an equilateral triangle ABC, centre O, where circle centre O has tangents which are all three sides of the triangle ABC. M is the midpoint of AB, and F is the intersection of arc DE and line MC. I know each coordinate of the triangle, and the triangle has edges of length 1.

I need to calculate the centroid of the shape CDE, where the DE vertex is the arc passing through point F. I understand the centroid will be along the line CM, because the shape is symmetrical. I have no idea how to find the exact point though. One thought is that it's the midpoint of line FC, and another thought is
that it's the midpoint of the perpendicular bisector of line DE through C.
Are any of these presumptions right? Or is there no way of working out the centroid of it without actually having it in real life and using a plumbline?
 [/INST] We can do this using calculus. 
Coordinatize by placing the center of the circle, of radius $r$, at the origin. Generalizing slightly, I'll take $\angle COE = \theta$ instead of specifically $\pi/3$; thus, 
$$C = r(0,\sec\theta) \qquad D = r(-\sin\theta,\cos\theta) \qquad E = r(\sin\theta,\cos\theta)$$ 
As OP notes, the centroid of region $CDFE$ lies on $\overline{CM}$, so its $x$-coordinate is $0$. One sees that the $y$-coordinate of that centroid must match that of the half-region $CFE$, which is bounded by $\overleftrightarrow{CE}$ ($f(x) = - x \tan\theta + r \sec\theta$) and the circle ($g(x) = \sqrt{r^2-x^2}$).
By the formula for the centroid of a bounded region,
$$\begin{align}
\bar{y} \cdot (\text{area}\;CFE) &= \frac12\int_{0}^{r\sin\theta}f(x)^2 - g(x)^2 \;dx \tag{1a}\\[4pt]
&= \frac12\int_{0}^{r\sin\theta}( x^2\tan^2\theta - 2 r x\tan\theta\sec\theta + r^2\sec^2\theta) - (r^2-x^2) \;dx \tag{1b}\\[4pt]
&= \frac1{2\cos^2\theta}\int_{0}^{r\sin\theta} x^2\sin^2\theta - 2 r x\sin\theta + r^2 - r^2\cos^2\theta + x^2\cos^2\theta) \;dx \tag{1b}\\[4pt]
&= \frac1{2\cos^2\theta}\int_{0}^{r\sin\theta} x^2 - 2 r x\sin\theta + r^2\sin^2\theta \;dx \tag{1c}\\[4pt]
&= \frac1{2\cos^2\theta}\int_{0}^{r\sin\theta} \left( x - r\sin\theta\right)^2 \;dx \tag{1d}\\[4pt]
&= \left.\frac1{6\cos^2\theta} \left( x - r\sin\theta \right)^3\;\right|_{0}^{r\sin\theta}  \tag{1e}\\[4pt]
&= \frac{r^3\sin^3\theta}{6\cos^2\theta} \tag{1f}
\end{align}$$
(Note: We could get from $(1a)$ to $(1d)$ fairly immediately by observing that $f(x)^2-g(x)^2$ gives the "power", with respect to the circle, of a variable point along $\overline{CE}$. But I digress ...)  Then, since 
$$\begin{align}
\text{area}\;CFE &= \text{area of }\; \triangle COE - \text{area of sector}\;FOE \tag{2a}\\[4pt]
&= \frac12 \cdot r \cdot r\tan\theta - \frac12 r^2 \cdot \theta \tag{2b}\\[4pt]
&= \frac12 r^2 (\tan\theta - \theta) \tag{2c}
\end{align}$$
we have

$$\bar{y} = \frac{r\sin^3\theta}{3\cos\theta(\sin\theta-\theta \cos\theta)} \qquad\stackrel{\theta=\pi/3}{\to}\qquad
\frac{3r\sqrt{3}}{2(3\sqrt{3} -\pi)} = r\cdot 1.26454\ldots
\tag{$\star$}$$


Alternatively, we can use geometric decomposition. 

Writing $\bar{p}$ for the $y$-coordinate of the centroid of $\triangle DCE$ and $\bar{q}$ for the $y$-coordinate of the centroid of sector $DFE$, we have
$$\bar{y} \cdot(\text{area} \;CDFE) = \bar{p}\cdot (\text{area}\; \triangle DCE) - \bar{q}\cdot (\text{area}\; DFE) \tag{3}$$
We "know" that a triangle's centroid is $1/3$ of the way up along a median, and its area is $1/2$-base-times-height, so 
$$\begin{align}
\bar{p} \cdot (\text{area}\;DCE) &= \left( r\cos\theta + \frac13 r ( \sec\theta - \cos\theta ) \right) \cdot \frac12 \cdot 2r\sin\theta \cdot r(\sec\theta - \cos\theta) \tag{4a}\\
&= \frac{r \sin^3\theta}{3 \cos^2\theta} \left( 1 + 2 \cos^2\theta\right) \tag{4b}
\end{align}$$
Consulting a convenient list of centroids, we find
$$\bar{q}\cdot(\text{area}\;DFE) = \frac{4 r \sin^3 \theta}{3(2\theta - \sin 2\theta)}\cdot \frac{r^2}{2}(2\theta-\sin 2\theta) = \frac23 r^3 \sin^3 \theta \tag{5}$$ 
So, the right-hand side of $(3)$ is $(4b)-(5)$, which reduces to twice the value of $(1f)$. Since the area of $CDFE$ is likewise twice the value in $(2c)$, the "twice"s cancel, and $(3)$ yields the result shown in $(\star)$. $\square$
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