Elementary Trigonometry problem

1. Jun 28, 2014

GeorgeDirac

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 $AX^2+BX^2=AB^2.$

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

$AX^2+BX^2=AB^2\to X^2(A+B)=AB^2\to X^2=\sqrt{\frac{AB^2}{A+B}}$

This got me nowhere though, so I would appreciate some hints on how to approach the problem.

Last edited: Jun 28, 2014
2. Jun 28, 2014

Simon Bridge

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.

3. Jun 28, 2014

pasmith

I think the OP's problem is "describe the set of points X such that $$\overline{AX}^2 + \overline{BX}^2 = \overline{AB}^2",$$ where $\overline{AB}$ denotes the distance between A and B.

4. Jun 28, 2014

GeorgeDirac

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

5. Jun 28, 2014

Joffan

(and a mind-reading star for pasmith :-) )

6. Jun 28, 2014

Simon Bridge

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.

7. Jul 3, 2014

acegikmoqsuwy

And even if it was $A\times X^2$, then $\displaystyle X^2\ne\sqrt{\frac{AB^2} {A+B}}$