Difference between revisions of "1988 IMO Problems/Problem 4"

(New page: Show that the solution set of the inequality <math>\sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4}</math> is a union of disjoint intervals, the sum of whose length is <math>1988</math>. Con...)
 
 
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==Problem==
 
Show that the solution set of the inequality  
 
Show that the solution set of the inequality  
  
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is a union of disjoint intervals, the sum of whose length is <math>1988</math>.
 
is a union of disjoint intervals, the sum of whose length is <math>1988</math>.
  
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==Solution==
 
Consider the graph of <math>f(x)=\sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4}</math>.  On the values of <math>x</math> between <math>n</math> and <math>n+1</math> for <math>n\in\mathbb{N}</math> <math>1\le n\le 69</math>, the terms of the form <math>\frac{k}{x-k}</math> for <math>k\ne n,n+1</math> have a finite range.  In contrast, the term <math>\frac{n}{x-n}</math> has an infinite range, from <math>+\infty</math> to <math>n</math>.  Similarly, the term <math>\frac{n+1}{x-n-1}</math> has infinite range from <math>-n-1</math> to <math>-\infty</math>.  Thus, since the two undefined values occur at the distinct endpoints, we can deduce that <math>f(x)</math> takes on all values between <math>+\infty</math> and <math>-\infty</math> for <math>x\in(n,n+1)</math>.  Thus, by the Intermediate Value Theorem, we are garunteed a <math>n<r_n<n+1</math> such that <math>f(r_n)=\frac{5}{4}</math>.  Additionally, we have that for <math>x>70</math>, the value of <math>f(x)</math> goes from <math>+\infty</math> to <math>0</math>, since as <math>x</math> increases, all the terms go to <math>0</math>.  Thus, there exists some <math>r_{70}>70</math> such that <math>f(r_{70})=\frac{5}{4}</math> and so <math>f(x)\ge\frac{5}{4}</math> for <math>x\in(70,r_{70})</math>.   
 
Consider the graph of <math>f(x)=\sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4}</math>.  On the values of <math>x</math> between <math>n</math> and <math>n+1</math> for <math>n\in\mathbb{N}</math> <math>1\le n\le 69</math>, the terms of the form <math>\frac{k}{x-k}</math> for <math>k\ne n,n+1</math> have a finite range.  In contrast, the term <math>\frac{n}{x-n}</math> has an infinite range, from <math>+\infty</math> to <math>n</math>.  Similarly, the term <math>\frac{n+1}{x-n-1}</math> has infinite range from <math>-n-1</math> to <math>-\infty</math>.  Thus, since the two undefined values occur at the distinct endpoints, we can deduce that <math>f(x)</math> takes on all values between <math>+\infty</math> and <math>-\infty</math> for <math>x\in(n,n+1)</math>.  Thus, by the Intermediate Value Theorem, we are garunteed a <math>n<r_n<n+1</math> such that <math>f(r_n)=\frac{5}{4}</math>.  Additionally, we have that for <math>x>70</math>, the value of <math>f(x)</math> goes from <math>+\infty</math> to <math>0</math>, since as <math>x</math> increases, all the terms go to <math>0</math>.  Thus, there exists some <math>r_{70}>70</math> such that <math>f(r_{70})=\frac{5}{4}</math> and so <math>f(x)\ge\frac{5}{4}</math> for <math>x\in(70,r_{70})</math>.   
  
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The sum of their lengths is <math>r_1+r_2+\cdots+r_{70}-(1+2+\cdots+70)=r_1+r_2+\cdots+r_{70}-35\cdot71</math>.  We have that the equation <math>f(x)=\frac{5}{4}</math> yields a polynomial with roots <math>r_i</math>.  Thus, opposite of the coeficient of <math>x^{69}</math> divided by the leading coefficient is the sum of the <math>r_i</math>.  It is easy to see that the coefficient of <math>x^{69}</math> is <math>-5(1+2+\cdots+70)-4(1+2+\cdots+70)=-9\cdot35\cdot 71</math>.  Thus, since the leading coefficient is <math>5</math> we have <math>r_1+r_2+\cdots+r_{70}=9\cdot7\cdot71</math>.  Thus, the sum of the lengths of the intervals is <math>63\cdot71-35\cdot71=28\cdot71=1988</math> as desired.
 
The sum of their lengths is <math>r_1+r_2+\cdots+r_{70}-(1+2+\cdots+70)=r_1+r_2+\cdots+r_{70}-35\cdot71</math>.  We have that the equation <math>f(x)=\frac{5}{4}</math> yields a polynomial with roots <math>r_i</math>.  Thus, opposite of the coeficient of <math>x^{69}</math> divided by the leading coefficient is the sum of the <math>r_i</math>.  It is easy to see that the coefficient of <math>x^{69}</math> is <math>-5(1+2+\cdots+70)-4(1+2+\cdots+70)=-9\cdot35\cdot 71</math>.  Thus, since the leading coefficient is <math>5</math> we have <math>r_1+r_2+\cdots+r_{70}=9\cdot7\cdot71</math>.  Thus, the sum of the lengths of the intervals is <math>63\cdot71-35\cdot71=28\cdot71=1988</math> as desired.
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== See Also == {{IMO box|year=1988|num-b=3|num-a=5}}

Latest revision as of 10:36, 30 January 2021

Problem

Show that the solution set of the inequality

$\sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4}$

is a union of disjoint intervals, the sum of whose length is $1988$.

Solution

Consider the graph of $f(x)=\sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4}$. On the values of $x$ between $n$ and $n+1$ for $n\in\mathbb{N}$ $1\le n\le 69$, the terms of the form $\frac{k}{x-k}$ for $k\ne n,n+1$ have a finite range. In contrast, the term $\frac{n}{x-n}$ has an infinite range, from $+\infty$ to $n$. Similarly, the term $\frac{n+1}{x-n-1}$ has infinite range from $-n-1$ to $-\infty$. Thus, since the two undefined values occur at the distinct endpoints, we can deduce that $f(x)$ takes on all values between $+\infty$ and $-\infty$ for $x\in(n,n+1)$. Thus, by the Intermediate Value Theorem, we are garunteed a $n<r_n<n+1$ such that $f(r_n)=\frac{5}{4}$. Additionally, we have that for $x>70$, the value of $f(x)$ goes from $+\infty$ to $0$, since as $x$ increases, all the terms go to $0$. Thus, there exists some $r_{70}>70$ such that $f(r_{70})=\frac{5}{4}$ and so $f(x)\ge\frac{5}{4}$ for $x\in(70,r_{70})$.

So, we have $70$ $r_i$ such that $f(r_i)=\frac{5}{4}$. There are obviously no other such $r_i$ since $f(x)=\frac{5}{4}$ yields a polynomial of degree $70$ when combining fractions. Thus, we have that the solution set to the inequality $f(x)\ge\frac{5}{4}$ is the union of the intervals $(n,r_n]$ (since if $f(x)<\frac{5}{4}$ for $x\in(n,r_n)$ then there would exist another solution to the equation $f(x)=\frac{5}{4}$.

Thus we have proven that the solution set is the union of disjoint intervals. Now we are to prove that the sum of their lengths is $1988$.

The sum of their lengths is $r_1+r_2+\cdots+r_{70}-(1+2+\cdots+70)=r_1+r_2+\cdots+r_{70}-35\cdot71$. We have that the equation $f(x)=\frac{5}{4}$ yields a polynomial with roots $r_i$. Thus, opposite of the coeficient of $x^{69}$ divided by the leading coefficient is the sum of the $r_i$. It is easy to see that the coefficient of $x^{69}$ is $-5(1+2+\cdots+70)-4(1+2+\cdots+70)=-9\cdot35\cdot 71$. Thus, since the leading coefficient is $5$ we have $r_1+r_2+\cdots+r_{70}=9\cdot7\cdot71$. Thus, the sum of the lengths of the intervals is $63\cdot71-35\cdot71=28\cdot71=1988$ as desired.

See Also

1988 IMO (Problems) • Resources
Preceded by
Problem 3
1 2 3 4 5 6 Followed by
Problem 5
All IMO Problems and Solutions
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