Results 1 to 3 of 3

Thread: Similarity of triangles in argand plane

  1. #1
    Member kalyanram's Avatar
    Joined
    Jun 2008
    From
    Uppsala, Sweden
    Posts
    149
    Thanks
    14

    Similarity of triangles in argand plane

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

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

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

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

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

    Can you let me know the mistake with my arguement.

    Thanks and Regards,
    ~Kalyan.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by kalyanram View Post
    Given $\displaystyle A(\alpha), B(\beta), C(\gamma)$ and $\displaystyle A'(\alpha'), B'(\beta'), C'(\gamma')$ are the vertices of $\displaystyle \triangle ABC$ and $\displaystyle \triangle A'B'C'$ respectively then show that the triangles are directly similar if
    $\displaystyle \displaystyle\sum \alpha{({\beta'}-{\gamma'})} = 0 $

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

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

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

    $\displaystyle \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 $\displaystyle \frac {\alpha - \beta}{\beta - \gamma} = \frac {\alpha' - \beta'}{\beta' - \gamma'}$ then (multiplying out the fractions) $\displaystyle (\alpha - \beta)(\beta' - \gamma') = (\beta - \gamma)(\alpha' - \beta')$. Multiply out the brackets, cancel the term $\displaystyle -\beta\beta'$ on both sides, and rearrange to get $\displaystyle \alpha(\beta' - \gamma') +\beta(\gamma'-\alpha') + \gamma(\alpha'-\beta')$, or in other words $\displaystyle \sum \alpha{({\beta'}-{\gamma'})} = 0 $.

    In fact, the question asked for the converse result: If $\displaystyle \sum \alpha{({\beta'}-{\gamma'})} = 0 $ then the triangles are similar. But your argument is reversible, so you have proved that result too.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member kalyanram's Avatar
    Joined
    Jun 2008
    From
    Uppsala, Sweden
    Posts
    149
    Thanks
    14
    Hi Opalg,

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

    ~Kalyan.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Similarity of Triangles
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Sep 18th 2010, 10:20 AM
  2. Similarity of Triangles
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Oct 2nd 2009, 06:39 AM
  3. Straight lines in Argand plane?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Apr 30th 2009, 07:32 AM
  4. Argand Plane?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Apr 29th 2009, 04:55 AM
  5. [SOLVED] Graph of a line in the Argand plane
    Posted in the Advanced Math Topics Forum
    Replies: 3
    Last Post: Feb 15th 2008, 12:23 PM

/mathhelpforum @mathhelpforum