-1
$\begingroup$

Let $a,b,c,x,y,z$ be non-zero numbers such that: $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$ and $\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0$.

Prove that: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$

$\endgroup$
3
  • 3
    $\begingroup$ Is this homework? What have you tried? $\endgroup$ Jan 2, 2014 at 14:39
  • 1
    $\begingroup$ Thanks, Ehsan - I was wondering why it was tagged abstract-algebra... $\endgroup$
    – gt6989b
    Jan 2, 2014 at 14:45
  • $\begingroup$ @gt6989b Lot of people confuse between the term: elementary-algebra and abstract-algebra. $\endgroup$
    – user93957
    Jan 2, 2014 at 15:45

2 Answers 2

5
$\begingroup$

Letting $$\frac xa=s, \frac yb=t, \frac zc=u,$$ we have $$s+t+u=1,\frac 1s+\frac 1t+\frac 1u=0.$$

Letting $A=st+tu+us$, since $$\frac{A}{stu}=0$$ we have $$A=0.$$

So, $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=s^2+t^2+u^2=(s+t+u)^2-2A=1^2-2\cdot 0=1.$$

$\endgroup$
2
$\begingroup$

With

$$u:=\frac{x}{a} , v:=\frac{y}{b} , w:=\frac{z}{c}\ ,$$

we have

u + v + w = 1

and

$\frac{1}{u}+\frac{1}{v}+\frac{1}{w}=0$ ,

multiplied with uvw this gives vw + uw + uv = 0

So we get $$u^2+v^2+w^2 = (u+v+w)^2 - 2uv - 2uw - 2 vw = (u+v+w)^2-2(uv+uw+vw) = 1$$

$\endgroup$
1
  • 1
    $\begingroup$ I assure that I had the same idea and did not see mathlove's answer before writing my own. $\endgroup$
    – Peter
    Jan 2, 2014 at 14:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .