Let A and B be two different points in the plane. Find the Locus of all the points C so ∠ A C B = 2 π 3 .

Question

Let A and B be two different points in the plane. Find the Locus of all the points C so   ∠  A  C  B  =            2      π        3  .

elverku7 · Accepted Answer

Step 1We need to fix constant values to A(m,n); B(r,s)so that we can use cosine ruleOtherwiseStep 2If the gradient of CA, BC are m, n respectively   tan  ⁡                    2        π            3        =  ±                    m        −        n                    1        +        m        n            .

muilasqk · Accepted Answer

Step 1Let the coordinates of the points   A  (      x    a    ,      y    a    ) and   B  (      x    b    ,      y    b    ). Then, the corresponding midpoint of AB is   M  (      x    m    ,      y    m    )  =  M  (                    x        a            +              x        b              2    ,                    y        a            +              y        b              2    ) and the directional vector perpendicular to the chord AB is   (      y    b    −      y    a    ,  −      x    b    +      x    a    ). The center of the locus circle lies on the line passing through the midpoint M and perpendicular to AB, which can be parametrized as,   (  x  ,  y  )  =  (      x    m    ,      y    m    )  +  t  (      y    b    −      y    a    ,  −      x    b    +      x    a    )Knowing that the triangle AOM is an 30-60-90 right triangle, we have   O  M  =      1    2    A  B    cot  ⁡  60  =      1          2              3              A  B where   O      M    2    =      t    2    [  (      x    b    −      x    a        )    2    +  (      y    b    −      y    a        )    2    ]  A      B    2    =  (      x    b    −      x    a        )    2    +  (      y    b    −      y    a        )    2  Step 2Substitute OM and AB into above equation to get the center parameter   t  =  ±      1          2              3            . Thus, the center of the circle is                    (1)                    (                  x          0                ,                  y          0                )        =        (                                            x              a                        +                          x              b                                2                ±                                            y              b                        −                          y              a                                            2                          3                                      ,                                                    y              a                        +                          y              b                                2                ∓                                            x              b                        −                          x              a                                            2                          3                                      )             and its radius is   R  =      1    2    A  B  csc  ⁡  60, or                     (2)                              R          2                =                  1          3                [        (                  x          b                −                  x          a                          )          2                +        (                  y          b                −                  y          a                          )          2                ]            As a result, the equation of the locus, expressed in terms of known coordinates   (      x    a    ,      y    a    ) and   (      x    b    ,      y    b    ), is   (  x  −      x    0        )    2    +  (  y  −      y    0        )    2    =      R    2   where the center   (      x    0    ,      y    0    ) and the radius R are given by (1) and (2), respectively. Note that there are two circles as shown by the two centers in (1).

Let A and B be two different points in the plane. Find the Locus of all the points C so angle ACB=(2pi)/3.

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