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Solve the equations:-

$\frac{x}{a+l}+\frac{y}{b+l}+\frac{z}{c+l}=1$

$\frac{x}{a+u}+\frac{y}{b+u}+\frac{z}{c+u}=1$

$\frac{x}{a+v}+\frac{y}{b+v}+\frac{z}{c+v}=1$

My friend had asked me this question and then gave me a solution which i dont understand . Consider a fourth equation

$\frac{x}{a+\theta }+\frac{y}{b+\theta }+\frac{z}{c+\theta }=1-\frac{\left(\theta -l\right)\left(\theta -u\right)\left(\theta -v\right)}{\left(a+\theta \right)\left(b+\theta \right)\left(c+\theta \right)}$

then multiply the fourth equation by $a+\theta $ and then put $a+\theta =0$

thus $x=\frac{\left(a+l\right)\left(a+u\right)\left(a+v\right)}{\left(a-b\right)\left(a-c\right)}$

using symmetry we can write y and z also

i don't understand how can someone assume a fourth equation and then use it to get the solution .I could have even assumed $\frac{x}{a+\theta }+\frac{y}{b+\theta }+\frac{z}{c+\theta }=1-\frac{1}{2}\frac{\left(\theta -l\right)\left(\theta -u\right)\left(\theta -v\right)}{\left(a+\theta \right)\left(b+\theta \right)\left(c+\theta \right)}$ but that would give me wrong answer.x y and z would be less by a factor of two in that case*

Adesh
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1 Answers1

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One way to resolve this mystery is to view $l, u, v$ as roots of the rational function $$P(\theta) = 1 - \frac{x}{a+\theta} - \frac{y}{b+\theta} - \frac{z}{c+\theta}$$ Combine them into a fraction with a common denominator, we can rewrite $P(\theta)$ into the form $$P(\theta) = \frac{Q(\theta)}{(a+\theta)(b+\theta)(c+\theta)}$$ where $Q(\theta)$ is a cubic polynomial in $\theta$.

Now $l, u, v$ are roots of $P(\theta)$. If they are distinct, then $Q(\theta)$ must factor as $\lambda (\theta - l)(\theta - u)(\theta - v)$ for some suitably chosen constant $\lambda$. To fix $\lambda$, we can send $\theta \to \infty$ and get

$$\lambda = \lim_{\theta\to\infty}\frac{Q(\theta)}{(a+\theta)(b+\theta)(c+\theta)} = \lim_{\theta\to\infty} P(\theta) = 1$$

This leads to the mysterious fourth equation:

$$\frac{x}{a+\theta} + \frac{y}{b+\theta} + \frac{z}{c+\theta} = 1 - P(\theta) = 1 - \frac{(\theta-l)(\theta-u)(\theta-v)}{(a+\theta)(b+\theta)(c+\theta)}$$

achille hui
  • 122,701