Difference between revisions of "2007 IMO Problems/Problem 1"

Latest revision as of 22:15, 31 March 2010

Problem

Real numbers $a_1, a_2, \dots , a_n$ are given. For each $i$ ( $1\le i\le n$ ) define

$d_i=\max\{a_j:1\le j\le i\}-\min\{a_j:i\le j\le n\}$

and let

$d=\max\{d_i:1\le i\le n\}$ .

(a) Prove that, for any real numbers $x_1\le x_2\le \cdots\le x_n$ ,

$\max\{|x_i-a_i|:1\le i\le n\}\ge \dfrac{d}{2} (*)$

(b) Show that there are real numbers $x_1\le x_2\le x_n$ such that equality holds in (*)

Solution

Since $d_i=\max\{a_j:1\le j\le i\}-\min\{a_j:i\le j\le n\}$ , all $d_i$ can be expressed as $a_p-a_q$ , where $1\le p\le i\le q \le n$ .

Thus, $d$ can be expressed as $a_p-a_q$ for some $p$ and $q$ , $1\le p\le q\le n$

Lemma) $d\ge 0$

Assume for contradiction that $d<0$ , then for all $i$ , $a_i \le \max\{a_j:1\le j\le i\}\le \min\{a_j:i\le j\le n\}\le a_{i+1}$

$a_i\le a_{i+1}$

Then, ${a_i}$ is a non-decreasing function, which means, $\max\{a_j:1\le j\le i\}=a_i$ , and $\min\{a_j:i\le j\le n\}\le a_{i+1}=a_i$ , which means, ${d_i}={0}$ .

Then, $d=0$ and contradiction.

a)

Case 1) $d=0$

If $d=0$ , $\max\{|x_i-a_i|:1\le i\le n\}$ is the maximum of a set of non-negative number, which must be at least $0$ .

Case 2) $d>0$ (We can ignore $d<0$ because of lemma)

Using the fact that $d$ can be expressed as $a_p-a_q$ for some $p$ and $q$ , $1\le p\le q\le n$ . $x_p\le x_q$

Assume for contradiction that $\max\{|x_i-a_i|:1\le i\le n\}<\dfrac{d}{2}$ .

Then, $\forall x_i$ , $|x_i-a_i|<\dfrac{d}{2}$ .

$|x_p-a_p|<\dfrac{d}{2}$ , and $|x_q-a_q|<\dfrac{d}{2}$

Thus, $x_p>a_p-\dfrac{d}{2}$ and $x_q<a_q+\dfrac{d}{2}$ .

Subtracting the two inequality, we will obtain:

$x_p-x_q>a_p-a_q-d=a_p-a_q-a_p+a_q=0$

$x_p>x_q$ --- contradiction ( $p\le q \rightarrow x_p\le x_q$ ).

Thus, $\max\{|x_i-a_i|:1\le i\le n\}\ge\dfrac{d}{2}$

(b)

A set of ${x_i}$ where the equality in (*) holds is:

$x_i=\max\{a_j:1\le j\le i\}-\frac{d}{2}$

Since $\max\{a_j:1\le j\le i\}$ is a non-decreasing function, $x_i$ is non-decreasing.

$\forall x_i$ :

Let $a_m=\max\{a_j:1\le j\le i\}$ , $a_m-a_i<a_m-\min\{a_j:i\le j\le n\}=d_i$ .

Thus, $0\le a_m-a_i \le d$ ( $0\le a_m-a_i$ because $a_m$ is the max of a set including $a_i$ )

$|x_i-a_i|=\left|a_m-\dfrac{d}{2}-a_i\right|=\left|(a_m-a_i)\dfrac{d}{2}\right|$

$0\le a_m-a_i\le d$

$-\dfrac{d}{2} \le (a_m-a_i)\dfrac{d}{2} \le \dfrac{d}{2}$

$\left|(a_m-a_i)-\dfrac{d}{2}\right|=|x_i-a_i|\le \frac{d}{2}$

Since $\max\{|x_i-a_i|:1\le i\le n\}\ge\dfrac{d}{2}$ and $|x_i-a_i|\le \frac{d}{2}$ $\forall x_i$ , $\max\{|x_i-a_i|:1\le i\le n\}=\dfrac{d}{2}$

This is written by Mo Lam--- who is a horrible proof writer, so please fix the proof for me. Thank you. O, also the formatting.

--> Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

2007 IMO (Problems) • Resources
Preceded by First question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
All IMO Problems and Solutions

<url>viewtopic.php?p=894656#p894656 Discussion on AoPS/MathLinks</url>

@@ Line 64: / Line 64: @@
 <cmath>x_p-x_q>a_p-a_q-d=a_p-a_q-a_p+a_q=0</cmath>
-<math>x_p>x_q</math> --- contradiction.
+<math>x_p>x_q</math> --- contradiction (<math>p\le q \rightarrow x_p\le x_q</math>).
 <br/>
@@ Line 96: / Line 96: @@
 <math>-\dfrac{d}{2} \le (a_m-a_i)\dfrac{d}{2} \le \dfrac{d}{2}</math>
-<math>\left|(a_m-a_i)\dfrac{d}{2}\right|=|x_i-a_i|\le \frac{d}{2}</math>
+<math>\left|(a_m-a_i)-\dfrac{d}{2}\right|=|x_i-a_i|\le \frac{d}{2}</math>
 <br/>

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Difference between revisions of "2007 IMO Problems/Problem 1"

Latest revision as of 22:15, 31 March 2010

Problem

Solution

Resources