Difference between revisions of "1992 OIM Problems/Problem 2"
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https://www.oma.org.ar/enunciados/ibe7.htm | https://www.oma.org.ar/enunciados/ibe7.htm |
Latest revision as of 09:42, 23 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:
Where and
are coefficients of the respective polynomials for each
From properties of polynomials, we know that the sum of the roots of a polynomial of degree n is where
is the coefficient of
and
is the coefficient of
Therefore,
and,
Thus the sum of the intervals is
- 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. I didn't even get points for drawing the function. haha. Several decades ago I was able to finally solve it. But even now, I'm still unsure about the "disjunct two by two" wording...
~Tomas Diaz. ~orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.