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Difference between revisions of "1992 OIM Problems/Problem 2"
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== Solution ==
 
== Solution ==
  
−
Since <math>a_1 < a_2 < a_3 < \cdots < a_n</math>, we can plot <math>f(x)</math> to visualize what we're looking for:
+
Since <math>0<a_1 < a_2 < a_3 < \cdots < a_n</math>, we can plot <math>f(x)</math> to visualize what we're looking for:
−
[[File:1992_OIM_P2a.png|center|600px]]
+
[[File:1992_OIM_P2b.png|center|800px]]
  
 
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>
Line 17: Line 17:
 
Thus the sum of the intervals will be: <math>\sum_{i}^{}\left( r_i+a_i \right)</math>
 
Thus the sum of the intervals will be: <math>\sum_{i}^{}\left( r_i+a_i \right)</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.
+
Now we set <math>f(x)=1</math>:
 +
 
 +
<math>f(x)=\frac{\sum_{j\ne i}^{}\left( a_i \prod_{j}^{}\left(x+a_j  \right)\right)}{\prod_{i}^{}\left(  x+a_i\right)}=1</math>
 +
 
 +
And solve for zero:
 +
 
 +
<math>\prod_{i}^{}\left(  x+a_i\right)-\sum_{j \ne i}^{}\left( a_i \prod_{j}^{}\left(x+a_j  \right)\right)=0</math>
 +
 
 +
<math>\left( x^n+\sum_{i}^{}a_ix^{n-1}+K_{n-2}x^{n-2}+\cdots+K_1x+K_0\right)-\left( \sum_{i}^{}a_ix^{n-1}+L_{n-2}x^{n-2}+\cdots+L_1x+L_0\right)=0</math>
 +
 
 +
<math> x^n+\left( \sum_{i}^{}a_i-\sum_{i}^{}a_i \right)x^{n-1}+\left( K_{n-2}-L_{n-2} \right)x^{n-2}+\cdots+\left( K_{1}-L_{1} \right)x+\left( K_{0}-L_{0} \right)=0</math>
 +
 
 +
Where <math>K_i</math> and <math>L_i</math> are coefficients of the respective polynomials for each <math>x^i</math>
 +
 
 +
<math> x^n+(0)x^{n-1}+\left( K_{n-2}-L_{n-2} \right)x^{n-2}+\cdots+\left( K_{1}-L_{1} \right)x+\left( K_{0}-L_{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>
 +
 
 +
Therefore, <math>\sum_{i}^{}r_i=-0/1=0</math>
 +
 
 +
and, <math>\sum_{i}^{}\left( r_i+a_i \right)=\sum_{i}^{}r_i+\sum_{i}^{}a_i=\sum_{i}^{}a_i</math>
 +
 
 +
Thus the sum of the intervals is <math>\sum_{i}^{}a_i</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.  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}}
 
{{alternate solutions}}
  
 
== See also ==
 
== See also ==
 +
[[OIM Problems and Solutions]]
 +
 
https://www.oma.org.ar/enunciados/ibe7.htm
 
https://www.oma.org.ar/enunciados/ibe7.htm

Latest revision as of 08:42, 23 December 2023

Problem

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

Solution

Since $0<a_1 < a_2 < a_3 < \cdots < a_n$, we can plot $f(x)$ to visualize what we're looking for:

1992 OIM P2b.png

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: $\sum_{i}^{}\left( r_i+a_i \right)$

Now we set $f(x)=1$:

$f(x)=\frac{\sum_{j\ne i}^{}\left( a_i \prod_{j}^{}\left(x+a_j \right)\right)}{\prod_{i}^{}\left( x+a_i\right)}=1$

And solve for zero:

$\prod_{i}^{}\left( x+a_i\right)-\sum_{j \ne i}^{}\left( a_i \prod_{j}^{}\left(x+a_j \right)\right)=0$

$\left( x^n+\sum_{i}^{}a_ix^{n-1}+K_{n-2}x^{n-2}+\cdots+K_1x+K_0\right)-\left( \sum_{i}^{}a_ix^{n-1}+L_{n-2}x^{n-2}+\cdots+L_1x+L_0\right)=0$

$x^n+\left( \sum_{i}^{}a_i-\sum_{i}^{}a_i \right)x^{n-1}+\left( K_{n-2}-L_{n-2} \right)x^{n-2}+\cdots+\left( K_{1}-L_{1} \right)x+\left( K_{0}-L_{0} \right)=0$

Where $K_i$ and $L_i$ are coefficients of the respective polynomials for each $x^i$

$x^n+(0)x^{n-1}+\left( K_{n-2}-L_{n-2} \right)x^{n-2}+\cdots+\left( K_{1}-L_{1} \right)x+\left( K_{0}-L_{0} \right)=0$

From properties of polynomials, we know that the sum of the roots of a polynomial of degree n is $-b/a$ where $b$ is the coefficient of $x^{n-1}$ and $a$ is the coefficient of $x^n$

Therefore, $\sum_{i}^{}r_i=-0/1=0$

and, $\sum_{i}^{}\left( r_i+a_i \right)=\sum_{i}^{}r_i+\sum_{i}^{}a_i=\sum_{i}^{}a_i$

Thus the sum of the intervals is $\sum_{i}^{}a_i$

  • 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.

See also

OIM Problems and Solutions

https://www.oma.org.ar/enunciados/ibe7.htm

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