Elementary Trigonometry problem

GeorgeDirac
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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<br /> X^2(A+B)=AB^2\to<br /> 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.
 
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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.
 
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.
 
pasmith said:
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.

Yes, that is my problem, I am very stupid, I don't know why I thought that AX meant A*X.
 
I recommend asking Pythagorus.

(and a mind-reading star for pasmith :-) )
 
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.
 
GeorgeDirac said:
[itex]AX^2+BX^2=AB^2\to<br /> X^2(A+B)=AB^2\to<br /> X^2=\sqrt{\frac{AB^2}{A+B}}[/itex]

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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