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Math Help - Similarity of triangles in argand plane

  1. #1
    Member kalyanram's Avatar
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    Similarity of triangles in argand plane

    Given A(\alpha), B(\beta), C(\gamma) and A'(\alpha'), B'(\beta'), C'(\gamma') are the vertices of \triangle ABC and \triangle A'B'C' respectively then show that the triangles are directly similar if
    \displaystyle\sum \alpha{({\beta'}-{\gamma'})} = 0

    Sol:
    I tried out by saying that the \triangle ABC and \triangle A'B'C' are similar if the ratio of their sides are proportional and the \angle A = \angle A' , \angle B = \angle B' and \angle C = \angle C' so we have
    \frac {|\alpha - \beta|}{|\alpha' - \beta'|} = \frac {|\beta - \gamma|}{|\beta' - \gamma'|} = \frac {|\gamma - \alpha|}{|\gamma' - \alpha'|} -- (Eq 1)
    arg(\frac {\alpha - \beta}{\beta - \gamma}) = arg(\frac {\alpha' - \beta'}{\beta' - \gamma'}) , arg(\frac {\beta - \gamma}{\gamma - \alpha}) = arg(\frac {\beta' - \gamma'}{\gamma' - \alpha'}) and arg(\frac {\gamma - \alpha}{\alpha - \beta}) = arg(\frac {\gamma' - \alpha'}{\alpha' - \beta'}) -- (Eq 2)

    From Eq 1 and Eq 2 we can conclude that the complex numbers

    \frac {\alpha - \beta}{\beta - \gamma} = \frac {\alpha' - \beta'}{\beta' - \gamma'} , \frac {\beta - \gamma}{\gamma - \alpha} = \frac {\beta' - \gamma'}{\gamma' - \alpha'} and \frac {\gamma - \alpha}{\alpha - \beta} = \frac {\gamma' - \alpha'}{\alpha' - \beta'} rearranging and solving this I cannot conclude that

    \displaystyle\sum \alpha{({\beta'}-{\gamma'})} = 0

    Can you let me know the mistake with my arguement.

    Thanks and Regards,
    ~Kalyan.
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by kalyanram View Post
    Given A(\alpha), B(\beta), C(\gamma) and A'(\alpha'), B'(\beta'), C'(\gamma') are the vertices of \triangle ABC and \triangle A'B'C' respectively then show that the triangles are directly similar if
    \displaystyle\sum \alpha{({\beta'}-{\gamma'})} = 0

    Sol:
    I tried out by saying that the \triangle ABC and \triangle A'B'C' are similar if the ratio of their sides are proportional and the \angle A = \angle A' , \angle B = \angle B' and \angle C = \angle C' so we have
    \frac {|\alpha - \beta|}{|\alpha' - \beta'|} = \frac {|\beta - \gamma|}{|\beta' - \gamma'|} = \frac {|\gamma - \alpha|}{|\gamma' - \alpha'|} -- (Eq 1)
    \arg(\frac {\alpha - \beta}{\beta - \gamma}) = \arg(\frac {\alpha' - \beta'}{\beta' - \gamma'}) ,\ \arg(\frac {\beta - \gamma}{\gamma - \alpha}) = \arg(\frac {\beta' - \gamma'}{\gamma' - \alpha'}) and \arg(\frac {\gamma - \alpha}{\alpha - \beta}) = \arg(\frac {\gamma' - \alpha'}{\alpha' - \beta'}) -- (Eq 2)

    From Eq 1 and Eq 2 we can conclude that the complex numbers

    \frac {\alpha - \beta}{\beta - \gamma} = \frac {\alpha' - \beta'}{\beta' - \gamma'} ,\ \frac {\beta - \gamma}{\gamma - \alpha} = \frac {\beta' - \gamma'}{\gamma' - \alpha'} and \frac {\gamma - \alpha}{\alpha - \beta} = \frac {\gamma' - \alpha'}{\alpha' - \beta'} rearranging and solving this I cannot conclude that

    \displaystyle\sum \alpha{({\beta'}-{\gamma'})} = 0

    Can you let me know the mistake with my argument.
    What's the problem, you're practically there! If \frac {\alpha - \beta}{\beta - \gamma} = \frac {\alpha' - \beta'}{\beta' - \gamma'} then (multiplying out the fractions) (\alpha - \beta)(\beta' - \gamma') = (\beta - \gamma)(\alpha' - \beta'). Multiply out the brackets, cancel the term -\beta\beta' on both sides, and rearrange to get \alpha(\beta' - \gamma') +\beta(\gamma'-\alpha') + \gamma(\alpha'-\beta'), or in other words \sum \alpha{({\beta'}-{\gamma'})} = 0 .

    In fact, the question asked for the converse result: If \sum \alpha{({\beta'}-{\gamma'})} = 0 then the triangles are similar. But your argument is reversible, so you have proved that result too.
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  3. #3
    Member kalyanram's Avatar
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    Hi Opalg,

    Thanks for the comment. Yeah sometimes I do miss the obvious.

    ~Kalyan.
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