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ABC-is-a-triangle-whose-one-vertex-is-moving-along-a-circular-path-Discuss-the-nature-of-the-path-along-which-the-centroid-of-the-triangle-is-moving-




Question Number 6640 by Rasheed Soomro last updated on 07/Jul/16
ABC is a triangle,whose one vertex   is moving along a circular path.   Discuss the nature of the  path, along  which the centroid of the triangle  is moving.
$$\mathrm{ABC}\:{is}\:{a}\:{triangle},{whose}\:{one}\:{vertex}\: \\ $$$${is}\:{moving}\:{along}\:{a}\:{circular}\:{path}.\: \\ $$$${Discuss}\:{the}\:{nature}\:{of}\:{the}\:\:{path},\:{along} \\ $$$${which}\:{the}\:\boldsymbol{{centroid}}\:{of}\:{the}\:{triangle} \\ $$$${is}\:{moving}. \\ $$
Commented by Rasheed Soomro last updated on 08/Jul/16
If  A  in a △ABC  is a moving point, then   BC^(−)   and objects tied with it will be static.
$${If}\:\:\mathrm{A}\:\:{in}\:{a}\:\bigtriangleup\mathrm{ABC}\:\:{is}\:{a}\:{moving}\:{point},\:{then}\: \\ $$$$\overline {\mathrm{BC}}\:\:{and}\:{objects}\:{tied}\:{with}\:{it}\:{will}\:{be}\:\boldsymbol{{static}}. \\ $$
Commented by Yozzii last updated on 07/Jul/16
Centroid O moves in a circle either   outside, on or inside the circle along  which one of A,B or C moves.  Let AO be the line joining the vertex  A and centroid O of △ABC. Let N be  the centre of the circle of radius r  along which A moves. Suppose r>∣OA∣.  Let θ=∠OAN where 0≤θ≤π. Let θ=π.  Then, O,A,N are collinear and the point O  moves in a circle of radius r+∣AO∣, centre N.  Generally, N,A and O form a triangle  where ∣NO∣ is the radius of the circular  path of O since the entire triangle ABC  rotates about N; all properties of △ABC  are conserved under the motion of A.
$${Centroid}\:{O}\:{moves}\:{in}\:{a}\:{circle}\:{either}\: \\ $$$${outside},\:{on}\:{or}\:{inside}\:{the}\:{circle}\:{along} \\ $$$${which}\:{one}\:{of}\:{A},{B}\:{or}\:{C}\:{moves}. \\ $$$${Let}\:{AO}\:{be}\:{the}\:{line}\:{joining}\:{the}\:{vertex} \\ $$$${A}\:{and}\:{centroid}\:{O}\:{of}\:\bigtriangleup{ABC}.\:{Let}\:{N}\:{be} \\ $$$${the}\:{centre}\:{of}\:{the}\:{circle}\:{of}\:{radius}\:{r} \\ $$$${along}\:{which}\:{A}\:{moves}.\:{Suppose}\:{r}>\mid{OA}\mid. \\ $$$${Let}\:\theta=\angle{OAN}\:{where}\:\mathrm{0}\leqslant\theta\leqslant\pi.\:{Let}\:\theta=\pi. \\ $$$${Then},\:{O},{A},{N}\:{are}\:{collinear}\:{and}\:{the}\:{point}\:{O} \\ $$$${moves}\:{in}\:{a}\:{circle}\:{of}\:{radius}\:{r}+\mid{AO}\mid,\:{centre}\:{N}. \\ $$$${Generally},\:{N},{A}\:{and}\:{O}\:{form}\:{a}\:{triangle} \\ $$$${where}\:\mid{NO}\mid\:{is}\:{the}\:{radius}\:{of}\:{the}\:{circular} \\ $$$${path}\:{of}\:{O}\:{since}\:{the}\:{entire}\:{triangle}\:{ABC} \\ $$$${rotates}\:{about}\:{N};\:{all}\:{properties}\:{of}\:\bigtriangleup{ABC} \\ $$$${are}\:{conserved}\:{under}\:{the}\:{motion}\:{of}\:{A}.\: \\ $$
Commented by Yozzii last updated on 07/Jul/16
So, ∣NO∣=(√(r^2 +∣AO∣^2 −2r∣AO∣cosθ)).  If O lies on the circle just as A moves   along it, ∣NO∣=r ⇒cosθ=((∣AO∣)/(2r)).   θ=cos^(−1) ((∣AO∣)/(2r))  Since ∣cosθ∣≤1 ∀θ∈[0,π], ⇒∣AO∣≤2r or r≥(1/2)∣AO∣  So, if the triangle is to be positioned  so that both A and the centroid O move  on the same circle of radius r it is necessary that  r≥0.5∣AO∣ and θ=cos^(−1) ((∣AO∣)/(2r)).  If θ=0⇒∣NO∣=r−∣OA∣ so O moves  along a circle inside the circle for A.
$${So},\:\mid{NO}\mid=\sqrt{{r}^{\mathrm{2}} +\mid{AO}\mid^{\mathrm{2}} −\mathrm{2}{r}\mid{AO}\mid{cos}\theta}. \\ $$$${If}\:{O}\:{lies}\:{on}\:{the}\:{circle}\:{just}\:{as}\:{A}\:{moves}\: \\ $$$${along}\:{it},\:\mid{NO}\mid={r}\:\Rightarrow{cos}\theta=\frac{\mid{AO}\mid}{\mathrm{2}{r}}.\: \\ $$$$\theta={cos}^{−\mathrm{1}} \frac{\mid{AO}\mid}{\mathrm{2}{r}} \\ $$$${Since}\:\mid{cos}\theta\mid\leqslant\mathrm{1}\:\forall\theta\in\left[\mathrm{0},\pi\right],\:\Rightarrow\mid{AO}\mid\leqslant\mathrm{2}{r}\:{or}\:{r}\geqslant\frac{\mathrm{1}}{\mathrm{2}}\mid{AO}\mid \\ $$$${So},\:{if}\:{the}\:{triangle}\:{is}\:{to}\:{be}\:{positioned} \\ $$$${so}\:{that}\:{both}\:{A}\:{and}\:{the}\:{centroid}\:{O}\:{move} \\ $$$${on}\:{the}\:{same}\:{circle}\:{of}\:{radius}\:{r}\:{it}\:{is}\:{necessary}\:{that} \\ $$$${r}\geqslant\mathrm{0}.\mathrm{5}\mid{AO}\mid\:{and}\:\theta={cos}^{−\mathrm{1}} \frac{\mid{AO}\mid}{\mathrm{2}{r}}. \\ $$$${If}\:\theta=\mathrm{0}\Rightarrow\mid{NO}\mid={r}−\mid{OA}\mid\:{so}\:{O}\:{moves} \\ $$$${along}\:{a}\:{circle}\:{inside}\:{the}\:{circle}\:{for}\:{A}. \\ $$
Commented by Rasheed Soomro last updated on 08/Jul/16
Nice!  Actually, I think,  every point tied with triangle,  (which is  not on the opposite side of moving vertex)  is  moving in circular path(whether the points  are on the triangle or inside the triangle or  outside the triangle.
$${Nice}! \\ $$$${Actually},\:{I}\:{think},\:\:{every}\:{point}\:\boldsymbol{{tied}}\:\boldsymbol{{with}}\:\boldsymbol{{triangle}}, \\ $$$$\left({which}\:{is}\:\:{not}\:{on}\:{the}\:{opposite}\:{side}\:{of}\:{moving}\:{vertex}\right) \\ $$$${is}\:\:{moving}\:{in}\:{circular}\:{path}\left({whether}\:{the}\:{points}\right. \\ $$$${are}\:{on}\:{the}\:{triangle}\:{or}\:{inside}\:{the}\:{triangle}\:{or} \\ $$$${outside}\:{the}\:{triangle}. \\ $$

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