Difference between revisions of "1992 OIM Problems/Problem 2"
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<math> x^n+(0)x^{n+1}+\left( K_{n-2}-P_{n-2} \right)x^{n-2}+\cdots+\left( K_{1}-P_{1} \right)x+\left( K_{0}-P_{0} \right)=0</math> | <math> x^n+(0)x^{n+1}+\left( K_{n-2}-P_{n-2} \right)x^{n-2}+\cdots+\left( K_{1}-P_{1} \right)x+\left( K_{0}-P_{0} \right)=0</math> | ||
+ | |||
+ | From properties of polynomials, we know that the sum of the roots of a polynomial of degree n is <math>-b/a</math> where <math>b</math> is the coefficient of <math>x^{n-1}</math> and <math>a</math> is the coefficient of <math>x^n</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 12:33, 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:
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
- 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.