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Elementary Trigonometry problem

  1. Jun 28, 2014 #1
    My problem is from Israel Gelfand's Trigonometry textbook.

    Page 9. Exercise 7: Two points, A and B, are given in the plane. Describe the set of points X such that [itex]AX^2+BX^2=AB^2.[/itex]

    The attempt at a solution
    Since problem asked to describe set of points X such that [itex]AX^2+BX^2=AB^2[/itex], I tried to solve for X, and got

    [itex]AX^2+BX^2=AB^2\to
    X^2(A+B)=AB^2\to
    X^2=\sqrt{\frac{AB^2}{A+B}}[/itex]

    This got me nowhere though, so I would appreciate some hints on how to approach the problem.
     
    Last edited: Jun 28, 2014
  2. jcsd
  3. Jun 28, 2014 #2

    Simon Bridge

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    Well, X is a set of points ... so ##X=\{\vec X_1,\vec X_2,\cdots\}:\vec X_i=(x_{i1},x_{i2},\cdots ) ; i = 1,2,\cdots##?

    Or do you know they mean that ##X \in \mathbb{R}##?

    If the second, then the equation you got gives you a set of curves for different A and B.
    Treat A and B as axes.
     
  4. Jun 28, 2014 #3

    pasmith

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    I think the OP's problem is "describe the set of points X such that [tex]
    \overline{AX}^2 + \overline{BX}^2 = \overline{AB}^2",[/tex] where [itex]\overline{AB}[/itex] denotes the distance between A and B.
     
  5. Jun 28, 2014 #4
    Yes, that is my problem, I am very stupid, I don't know why I thought that AX meant A*X.
     
  6. Jun 28, 2014 #5
    I recommend asking Pythagorus.

    (and a mind-reading star for pasmith :-) )
     
  7. Jun 28, 2014 #6

    Simon Bridge

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    Ah - "points A B and X" ... <sigh>

    More like:
    ##|AX|^2+|BX|^2=|AB|^2##

    ##\overline{AB}## would normally denote the line segment between A and B.

    ... well spotted that individual.
     
  8. Jul 3, 2014 #7
    And even if it was [itex]A\times X^2[/itex], then [itex]\displaystyle X^2\ne\sqrt{\frac{AB^2} {A+B}}[/itex]
     
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