Difference between revisions of "1992 OIM Problems/Problem 2"
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~translated into English by Tomas Diaz. ~orders@tomasdiaz.com | ~translated into English by Tomas Diaz. ~orders@tomasdiaz.com | ||
| − | * Note. I actually competed at this event in Venezuela when I was in High School representing Puerto Rico. I got a ZERO on this one because I didn't even know what was I supposed to do, nor did I know what the sum of the lengths of the intervals, disjoint two by two meant. A decade ago I finally solved it but now I don't remember. | + | == Solution == |
| + | * Note. I actually competed at this event in Venezuela when I was in High School representing Puerto Rico. I got a ZERO on this one because I didn't even know what was I supposed to do, nor did I know what the sum of the lengths of the intervals, disjoint two by two meant. A decade ago I finally solved it but now I don't remember how. I will attempt to solve this one later. | ||
| − | |||
{{solution}} | {{solution}} | ||
== See also == | == See also == | ||
https://www.oma.org.ar/enunciados/ibe7.htm | https://www.oma.org.ar/enunciados/ibe7.htm | ||
Revision as of 18:21, 14 December 2023
Problem
Given the collection of 
positive real numbers 
and the function:
![\[f(x) = \frac{a_1}{x+a_1}+\frac{a_2}{x+a_2}+\cdots +\frac{a_n}{x+a_n}\]](http://latex.artofproblemsolving.com/b/d/b/bdb3b28e005214b8b860c79ba8f7995604c9b6d3.png)
Determine the sum of the lengths of the intervals, disjoint two by two, formed by all 
.
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
- Note. I actually competed at this event in Venezuela when I was in High School representing Puerto Rico. I got a ZERO on this one because I didn't even know what was I supposed to do, nor did I know what the sum of the lengths of the intervals, disjoint two by two meant. A decade ago I finally solved it but now I don't remember how. I will attempt to solve this one later.
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