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>. | ||
+ | ==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
is a union of disjoint intervals, the sum of whose length is .
Solution
Consider the graph of . On the values of between and for , the terms of the form for have a finite range. In contrast, the term has an infinite range, from to . Similarly, the term has infinite range from to . Thus, since the two undefined values occur at the distinct endpoints, we can deduce that takes on all values between and for . Thus, by the Intermediate Value Theorem, we are garunteed a such that . Additionally, we have that for , the value of goes from to , since as increases, all the terms go to . Thus, there exists some such that and so for .
So, we have such that . There are obviously no other such since yields a polynomial of degree when combining fractions. Thus, we have that the solution set to the inequality is the union of the intervals (since if for then there would exist another solution to the equation .
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 .
The sum of their lengths is . We have that the equation yields a polynomial with roots . Thus, opposite of the coeficient of divided by the leading coefficient is the sum of the . It is easy to see that the coefficient of is . Thus, since the leading coefficient is we have . Thus, the sum of the lengths of the intervals is 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 |