Difference between revisions of "2007 IMO Shortlist Problems/A1"

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== Problem ==
 
== Problem ==
(''New Zealand'') let's solve this problem bois @poco @john
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(''New Zealand'')
You are given a sequence <math>a_1,a_2,\dots ,a_n</math> of numbers. For each <math>i</math> (<math>1\leq 1\leq n</math>) define
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You are given a sequence <math>a_1,a_2,\dots ,a_n</math> of numbers. For each <math>i</math> (<math>1\leq i\leq n</math>) define
  
 
<center><math>d_i=\max\{a_j:1\leq j\leq i\}-\min\{a_j:i\leq j\leq n\}</math></center>
 
<center><math>d_i=\max\{a_j:1\leq j\leq i\}-\min\{a_j:i\leq j\leq n\}</math></center>

Revision as of 21:26, 10 December 2019

Problem

(New Zealand) You are given a sequence $a_1,a_2,\dots ,a_n$ of numbers. For each $i$ ($1\leq i\leq n$) define

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

and let

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

(a) Prove that for arbitrary real numbers $x_1\leq x_2\leq \dots \leq x_n$,

$\max\{|x_i-a_i|:1\leq i\leq n\}\geq \frac{d}{2}$.

(b) Show that there exists a sequence $x_1\leq x_2\leq \dots \leq x_n$ of real numbers such that we have equality in (a).

Solution

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