Difference between revisions of "1992 OIM Problems/Problem 2"
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And solve for zero: | And solve for zero: | ||
− | <math>\sum_{j\ne i}^{}\left( a_i \prod_{j}^{}\left(x+a_j \right)\right) | + | <math>\sum_{j \ne i}^{}\left( a_i \prod_{j}^{}\left(x+a_j \right)\right)-\prod_{i}^{}\left( x+a_i\right)=0</math> |
* 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. | * 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. |
Revision as of 12:13, 17 December 2023
Problem
Given the collection of positive real numbers
and the function:
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
Since , we can plot
to visualize what we're looking for:
Notice that the intervals will be:
Thus the sum of the intervals will be:
Now we set :
And solve for zero:
- 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.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.