https://artofproblemsolving.com/wiki/index.php?title=1988_IMO_Problems/Problem_4&feed=atom&action=history 1988 IMO Problems/Problem 4 - Revision history 2021-01-28T05:19:21Z Revision history for this page on the wiki MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1988_IMO_Problems/Problem_4&diff=27631&oldid=prev Cosinator: New page: 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 $1988$. Con... 2008-08-30T15:56:22Z <p>New page: Show that the solution set of the inequality &lt;math&gt;\sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4}&lt;/math&gt; is a union of disjoint intervals, the sum of whose length is &lt;math&gt;1988&lt;/math&gt;. Con...</p> <p><b>New page</b></p><div>Show that the solution set of the inequality <br /> <br /> &lt;math&gt;\sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4}&lt;/math&gt;<br /> <br /> is a union of disjoint intervals, the sum of whose length is &lt;math&gt;1988&lt;/math&gt;.<br /> <br /> Consider the graph of &lt;math&gt;f(x)=\sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4}&lt;/math&gt;. On the values of &lt;math&gt;x&lt;/math&gt; between &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; for &lt;math&gt;n\in\mathbb{N}&lt;/math&gt; &lt;math&gt;1\le n\le 69&lt;/math&gt;, the terms of the form &lt;math&gt;\frac{k}{x-k}&lt;/math&gt; for &lt;math&gt;k\ne n,n+1&lt;/math&gt; have a finite range. In contrast, the term &lt;math&gt;\frac{n}{x-n}&lt;/math&gt; has an infinite range, from &lt;math&gt;+\infty&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt;. Similarly, the term &lt;math&gt;\frac{n+1}{x-n-1}&lt;/math&gt; has infinite range from &lt;math&gt;-n-1&lt;/math&gt; to &lt;math&gt;-\infty&lt;/math&gt;. Thus, since the two undefined values occur at the distinct endpoints, we can deduce that &lt;math&gt;f(x)&lt;/math&gt; takes on all values between &lt;math&gt;+\infty&lt;/math&gt; and &lt;math&gt;-\infty&lt;/math&gt; for &lt;math&gt;x\in(n,n+1)&lt;/math&gt;. Thus, by the Intermediate Value Theorem, we are garunteed a &lt;math&gt;n&lt;r_n&lt;n+1&lt;/math&gt; such that &lt;math&gt;f(r_n)=\frac{5}{4}&lt;/math&gt;. Additionally, we have that for &lt;math&gt;x&gt;70&lt;/math&gt;, the value of &lt;math&gt;f(x)&lt;/math&gt; goes from &lt;math&gt;+\infty&lt;/math&gt; to &lt;math&gt;0&lt;/math&gt;, since as &lt;math&gt;x&lt;/math&gt; increases, all the terms go to &lt;math&gt;0&lt;/math&gt;. Thus, there exists some &lt;math&gt;r_{70}&gt;70&lt;/math&gt; such that &lt;math&gt;f(r_{70})=\frac{5}{4}&lt;/math&gt; and so &lt;math&gt;f(x)\ge\frac{5}{4}&lt;/math&gt; for &lt;math&gt;x\in(70,r_{70})&lt;/math&gt;. <br /> <br /> So, we have &lt;math&gt;70&lt;/math&gt; &lt;math&gt;r_i&lt;/math&gt; such that &lt;math&gt;f(r_i)=\frac{5}{4}&lt;/math&gt;. There are obviously no other such &lt;math&gt;r_i&lt;/math&gt; since &lt;math&gt;f(x)=\frac{5}{4}&lt;/math&gt; yields a polynomial of degree &lt;math&gt;70&lt;/math&gt; when combining fractions. Thus, we have that the solution set to the inequality &lt;math&gt;f(x)\ge\frac{5}{4}&lt;/math&gt; is the union of the intervals &lt;math&gt;(n,r_n]&lt;/math&gt; (since if &lt;math&gt;f(x)&lt;\frac{5}{4}&lt;/math&gt; for &lt;math&gt;x\in(n,r_n)&lt;/math&gt; then there would exist another solution to the equation &lt;math&gt;f(x)=\frac{5}{4}&lt;/math&gt;. <br /> <br /> 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 &lt;math&gt;1988&lt;/math&gt;.<br /> <br /> The sum of their lengths is &lt;math&gt;r_1+r_2+\cdots+r_{70}-(1+2+\cdots+70)=r_1+r_2+\cdots+r_{70}-35\cdot71&lt;/math&gt;. We have that the equation &lt;math&gt;f(x)=\frac{5}{4}&lt;/math&gt; yields a polynomial with roots &lt;math&gt;r_i&lt;/math&gt;. Thus, opposite of the coeficient of &lt;math&gt;x^{69}&lt;/math&gt; divided by the leading coefficient is the sum of the &lt;math&gt;r_i&lt;/math&gt;. It is easy to see that the coefficient of &lt;math&gt;x^{69}&lt;/math&gt; is &lt;math&gt;-5(1+2+\cdots+70)-4(1+2+\cdots+70)=-9\cdot35\cdot 71&lt;/math&gt;. Thus, since the leading coefficient is &lt;math&gt;5&lt;/math&gt; we have &lt;math&gt;r_1+r_2+\cdots+r_{70}=9\cdot7\cdot71&lt;/math&gt;. Thus, the sum of the lengths of the intervals is &lt;math&gt;63\cdot71-35\cdot71=28\cdot71=1988&lt;/math&gt; as desired.</div> Cosinator