Menu Close

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-

Tinku Tara 0 Comments
Pin




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.
ABCisatriangle,whoseonevertexismovingalongacircularpath.Discussthenatureofthepath,alongwhichthecentroidofthetriangleismoving.
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.
IfAina△ABCisamovingpoint,thenBC―andobjectstiedwithitwillbestatic.
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.
CentroidOmovesinacircleeitheroutside,onorinsidethecirclealongwhichoneofA,BorCmoves.LetAObethelinejoiningthevertexAandcentroidOof△ABC.LetNbethecentreofthecircleofradiusralongwhichAmoves.Supposer>∣OA∣.Letθ=∠OANwhere0⩽θ⩽π.Letθ=π.Then,O,A,NarecollinearandthepointOmovesinacircleofradiusr+∣AO∣,centreN.Generally,N,AandOformatrianglewhere∣NO∣istheradiusofthecircularpathofOsincetheentiretriangleABCrotatesaboutN;allpropertiesof△ABCareconservedunderthemotionofA.
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,∣NO∣=r2+∣AO∣2−2r∣AO∣cosθ.IfOliesonthecirclejustasAmovesalongit,∣NO∣=r⇒cosθ=∣AO∣2r.θ=cos−1∣AO∣2rSince∣cosθ∣⩽1∀θ∈[0,π],⇒∣AO∣⩽2rorr⩾12∣AO∣So,ifthetriangleistobepositionedsothatbothAandthecentroidOmoveonthesamecircleofradiusritisnecessarythatr⩾0.5∣AO∣andθ=cos−1∣AO∣2r.Ifθ=0⇒∣NO∣=r−∣OA∣soOmovesalongacircleinsidethecircleforA.
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,Ithink,everypointtiedwithtriangle,(whichisnotontheoppositesideofmovingvertex)ismovingincircularpath(whetherthepointsareonthetriangleorinsidethetriangleoroutsidethetriangle.

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *