Is there an easier way to solve a "Find the locus" problem?Note: I am not concerned with the accuracy of my solution so you don't need to redo any of my calculations.Original question: Suppose ABC is an equilateral triangular lamina of side length unity, resting in two-dimensions. If A and B were constrained to move on the x- and y-axis respectively, then what is the locus of the centre of the triangle?My solution: Let the vertices of Δ A B C be A = ( x 0 , 0 ), B = ( 0 , y 0 ) and C = ( x , y ). Let the centre of the triangle be D = ( X , Y ). We wish to find the locus of D under the constraints: x 0 2 + y 0 2 = 0 ( x − x 0 ) 2 + y 2 = 0 x 2 + ( y − y 0 ) 2 = 0 Since D is the centre of the triangle, we know X = x + x 0 3 and Y = y + y 0 3 Parametrizing: Let θ be the angle that the edge AB makes with the x-axis. We use (theta) as our parameter of choice to write, x 0 = cos ⁡ θ y 0 = sin ⁡ θ for θ ∈ [ 0 , 2 π )This implies, ( x − cos ⁡ θ ) 2 + y 2 = 0We can parametrize this with another parameter ϕ ∈ [ 0 , 2 π ) as x = cos ⁡ θ + cos ⁡ ϕ y = sin ⁡ ϕ Plugging these back into third equation gives ( cos ⁡ θ + cos ⁡ ϕ ) 2 + ( sin ⁡ ϕ − sin ⁡ θ ) 2 = 1After lots of algebraic manipulations... X = 1 2 cos ⁡ θ ± 1 2 3 sin ⁡ θ Y = 1 2 sin ⁡ θ ± 1 2 3 cos ⁡ θ

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Is there an easier way to solve a &quot;Find the locus&quot; problem?Note: I am not concerned with the accuracy of my solution so you don&#039;t need to redo any of my calculations.Original question: Suppose ABC is an equilateral triangular lamina of side length unity, resting in two-dimensions. If A and B were constrained to move on the x- and y-axis respectively, then what is the locus of the centre of the triangle?My solution: Let the vertices of   Δ  A  B  C be   A  =  (      x    0    ,  0  ),   B  =  (  0  ,      y    0    ) and   C  =  (  x  ,  y  ). Let the centre of the triangle be   D  =  (  X  ,  Y  ). We wish to find the locus of D under the constraints:                              x          0          2                +                  y          0          2                                    =        0                            (        x        −                  x          0                          )          2                +                  y          2                                    =        0                                      x          2                +        (        y        −                  y          0                          )          2                                    =        0            Since D is the centre of the triangle, we know  X  =            x      +              x        0              3      and    Y  =            y      +              y        0              3  Parametrizing: Let   θ be the angle that the edge AB makes with the x-axis. We use (theta) as our parameter of choice to write,                               x          0                                    =        cos        ⁡        θ                                      y          0                                    =        sin        ⁡        θ            for   θ  ∈  [  0  ,  2  π  )This implies,   (  x  −  cos  ⁡  θ      )    2    +      y    2    =  0We can parametrize this with another parameter   ϕ  ∈  [  0  ,  2  π  ) as                    x                            =        cos        ⁡        θ        +        cos        ⁡        ϕ                            y                            =        sin        ⁡        ϕ            Plugging these back into third equation gives  (  cos  ⁡  θ  +  cos  ⁡  ϕ      )    2    +  (  sin  ⁡  ϕ  −  sin  ⁡  θ      )    2    =  1After lots of algebraic manipulations...                    X                            =                  1          2                cos        ⁡        θ        ±                  1                      2                          3                                      sin        ⁡        θ                    Y                            =                  1          2                sin        ⁡        θ        ±                  1                      2                          3                                      cos        ⁡        θ

Jaylene Tyler · Accepted Answer

Step 1It is probably easier to embed the construction in the complex plane and set   A  =  sin  ⁡  θ  ,    B  =  i  cos  ⁡  θ to have   C  =  (  A  −  B  )      e          ±      i      π              /            3        +  B  =      e          −      i      θ      ±      i      π              /            3        +  Bso that:   3  D  =  A  +  B  +  C  =      e          i      θ        +      e          −      i      θ      ±      i      π              /            3        +  i  cos  ⁡  θ  ..Step 2That leads to the same equations for the ellipses.If three vertices of a tetrahedron lie on the x-, y- and z- axis we may assume they are A(a,0,0), B(0,b,0), C(0,0,c) and       a    2    +      b    2    =      a    2    +      c    2    =      b    2    +      c    2   must hold, so       |    a      |    =      |    b      |    =      |    c      |   and there are just 16 possible locations for D and the centroid G of ABCD.

Suppose ABC is an equilateral triangular lamina of side length unity, resting in two-dimensions. If A and B were constrained to move on the x- and y-axis respectively, then what is the locus of the centre of the triangle?

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