Simple But Uncommon Equation from 1968 IMO

Here is a shortlisted problem submitted by Hungary at the 1968 International Mathematical Olympiad [Compendium, p. 50]:

If $a_i$ $(i=1,2,\ldots,n)$ are distinct non-zero real numbers, prove that the equation

$\displaystyle\frac{a_1}{a_1-x}+\frac{a_2}{a_2-x}+\cdots\frac{a_n}{a_n-x}=n$

has at least $n-1$ real roots. State a condition that insures the existence of $n$ real roots.

Solution

References

  1. D. Djukic et al, The IMO Compendium, Springer, 2011 (Second edition)

|Contact| |Front page| |Contents|

Copyright © 1996-2017 Alexander Bogomolny

If $a_i$ $(i=1,2,\ldots,n)$ are distinct non-zero real numbers, prove that the equation

$\displaystyle\frac{a_1}{a_1-x}+\frac{a_2}{a_2-x}+\cdots\frac{a_n}{a_n-x}=n$

has at least $n-1$ real roots. State a condition that insures the existence of $n$ real roots.

Solution

Let $\displaystyle f(x)=\frac{a_1}{a_1-x}+\frac{a_2}{a_2-x}+\cdots\frac{a_n}{a_n-x}-n.$

The distinct coefficients $a_i$ can be assumed to be ordered: $a_1\lt a_2\lt\ldots\lt a_n.$ Note that only one of the intervals $(a_k,a_{k+1}),$ $k=1,2,\ldots,n-1,$ may contain $0.$ For all other intervals, $a_k$ and $a_{k+1}$ are of the same sign. It follows that, $\displaystyle\lim_{x\rightarrow a_k^{+}}f(x)$ and $\displaystyle\lim_{x\rightarrow a_{k+1}^{-}}f(x)$ have different signs so that there is certainly a root of $f(x)=0$ on that interval. Observe that $f(0)=0,$ implying that $x=0$ is always a root of the equation. If all the coefficients $a_i$ have the same sign, then $x=0$ lies outside of all $n-1$ intervals $(a_k,a_{k+1}),$ $k=1,2,\ldots,n-1,$ thus becoming the root number $n.$ The sufficient condition, therefore, for there being $n$ roots is for all the coefficients $a_i$ to be of the same sign.

Of course, for $n=2,$ two real roots pop up automatically. The equation $\displaystyle\frac{a}{a-x}+\frac{b}{b-x}=2$ is equivalent to

$\displaystyle\frac{2ab-ax-bx-2(a-x)(b-x)}{(a-x)(b-x)}=\frac{x(2x-(a+b))}{(a-x)(b-x)}=0,$

from which $x=0$ or $x=\frac{1}{2}(a+b).$ For $n=3,$ there are also $3$ real roots. Besides $x=0,$ there are

$\displaystyle x_{2,3}=\frac{(a+b+c)\pm\sqrt{\frac{1}{2}[(a-b)^2+(b-c)^2+(c-a)^2]}}{3}.$

|Contact| |Front page| |Contents|

Copyright © 1996-2017 Alexander Bogomolny

 62609152

Search by google: