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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.

geometric diagram

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?

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  • $\begingroup$ You can't make random guesses of where the centroid lies. $\endgroup$ – Yves Daoust Jun 4 '18 at 8:06
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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.

enter image description here

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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  • $\begingroup$ Looks great - really appreciate the time and effort you put into this. I'm going through it now and I'm confused how you deduced the equation of the line through C and E, f(x). I feel like it'll be obvious when you explain it but I'm having a mental-block right now! $\endgroup$ – Adam Bromiley Jun 4 '18 at 12:23
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    $\begingroup$ Actually don't worry - after manipulating the unit circle around I figured why the equation is that :) $\endgroup$ – Adam Bromiley Jun 4 '18 at 12:37
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enter image description here

The centroid is $.3650417045$ from point O. I figured it out by determining the centroid $C_k$ of the kite CDOE and the centroid $C_s$ of sector ODFE and by subtracting moments using areas was able to determine the centroid of CDFE.

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Hint:

Your shape is a triangle from which a circular segment has been removed.

The area and centroid of the triangle are well known.

From the length of the chord and the radius, you can determine the aperture angle of the arc. Then use the formulas for the area and centroid, as given here. (Caution, the centroid is given wrt the circle center.)

Now you deduce the ordinate of the shape using the weighted sum

$$y_{shape}=\frac{A_{\text{triangle}}y_{\text{triangle}}-A_{\text{segment}}y_{\text{segment}}}{A_{\text{triangle}}-A_{\text{segment}}}.$$

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