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For the variable Triangle $\Delta ABC$ with fixed vertex at $C(1,2)$ and $A,\,B$ having co-ordinates $(\cos t, \sin t)$, $(\sin t, -\cos t)$, find the locus of its centroid.

Hosein Rahnama
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Ger Wyn
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    Possible duplicate of Equation of Locus of a point – Alex M. Jan 04 '16 at 18:05
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    Please do not ask the same question several times. If necessary, use the previous question to ask for details or clarifications. – Alex M. Jan 04 '16 at 18:06
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    It appears that you accepted the hints provided in the answer. This suggests that you understood the direction in which you need to go to attain an answer. Then you should post your work so that others might proof read it. – Em. Jan 04 '16 at 19:26
  • This is a good question, please add your first ideas or what you have tried before seeing the answers so that your post can be reopened. :) – Hosein Rahnama Jan 07 '16 at 14:01

2 Answers2

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The parametric equation of the locus curve for centroid is

$$\begin{align} \vec G &= \frac{1}{3} (\vec A + \vec B + \vec C)\\ &= \frac{1}{3}[(1+\cos t+\sin t){\bf{i}}+(2+\sin t -\cos t){\bf{j}}] \\ &\equiv x_G {\bf{i}} + y_G {\bf{j}} \end{align}$$

In fact, this locus is a circle and the details of it is given in the answer by marwalix. It's Cartesian equation is

$$(x_G-{1\over 3})^2+(y_G-{2\over 3})^2={2\over 9}$$

So it is a circle centered at $(\frac{1}{3},\frac{2}{3})$ with the radius $R=\frac{\sqrt{2}}{3}$.

This animation helps to visualize the locus better. As you can see, if one of the points $A$ or $B$ lies at the intersection of the two circles then all the points lie on a line and the centroid is $A$ or $B$ itself!

enter image description here

Hosein Rahnama
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The. Centroid $G$ of ${A,B,C}$ is such that

$$\vec{GA}+\vec{GB}+\vec{GC}=\vec{0}$$

Using Chasles identity with the origin $O$ one gets

$$\vec{GO}+\vec{OA}+\vec{GO}+\vec{OB}+\vec{GO}+\vec{OB}=\vec{0}$$

Using $\vec{GO}=-\vec{OG}$ one gets

$$\vec{OG}={\vec{OA}+\vec{OB}+\vec{OC}\over 3}$$

And this translates in coordinates

$$\vec{OG}={\left(1+\cos{t}+\sin{t}\right)\over 3}\vec{i}+{\left(2+\sin{t}-\cos{t}\right)\over 3}\vec{j}$$

One can check that

$$(x_G-{1\over 3})^2+(y_G-{2\over 3})^2={\cos^2{t}+2\cos{t}\sin{t}+\sin^2{t}+\sin^2{t}-2\sin{t}\cos{t}+\cos^2{t}\over 9}$$

And this simplifies to

$$(x_G-{1\over 3})^2+(y_G-{2\over 3})^2={2\over 9}$$

And this is the equation of a circle centred in $(1/3,2/3)$ with radius ${\sqrt{2}\over 3}$

marwalix
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