Difference between revisions of "1992 OIM Problems/Problem 2"

Revision as of 11:02, 17 December 2023

Given the collection of $n$ positive real numbers $a_1 < a_2 < a_3 < \cdots < a_n$ and the function:

$f(x) = \frac{a_1}{x+a_1}+\frac{a_2}{x+a_2}+\cdots +\frac{a_n}{x+a_n}$

Determine the sum of the lengths of the intervals, disjoint two by two, formed by all $f(x) = 1$ .

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Since $a_1 < a_2 < a_3 < \cdots < a_n$ , we can plot $f(x)$ to visualize what we're looking for:

Notice that the intervals will be: $I_1=r_1-(-a_1), I_2=r_2-(-a_2), \cdots , I_n=r_n-(-a_n)$

Thus the sum of the intervals will be: $\sum_{i}^{}\left( r_i+a_i \right)$

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.

@@ Line 13: / Line 13: @@
 [[File:1992_OIM_P2a.png|center|600px]]
-Notice that the intervals will be: <math>I_1=r_1-(-a_1), I_2=r_2-(-a_2_, \cdots , I_n=r_n-(-a_n)</math>
+Notice that the intervals will be: <math>I_1=r_1-(-a_1), I_2=r_2-(-a_2), \cdots , I_n=r_n-(-a_n)</math>
 Thus the sum of the intervals will be: <math>\sum_{i}^{}\left( r_i+a_i \right)</math>