Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1965 IMO Problems/Problem 2"

(I'm pretty sure this solution is good.)
 
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
 
Consider the system of equations
 
Consider the system of equations
 
<cmath>a_{11}x_1 + a_{12}x_2 + a_{13}x_3 = 0</cmath>
 
<cmath>a_{11}x_1 + a_{12}x_2 + a_{13}x_3 = 0</cmath>
Line 13: Line 14:
  
 
Prove that the given system has only the solution <math>x_1 = x_2 = x_3 = 0</math>.
 
Prove that the given system has only the solution <math>x_1 = x_2 = x_3 = 0</math>.
 +
  
 
== Solution ==
 
== Solution ==
 +
 
Clearly if the <math>x_i</math> are all equal, then they are equal to 0. Now let's assume WLOG that <math>x_1=0</math>. If <math>x_2</math> or <math>x_3</math> is 0, then the other is clearly zero, so let's consider the case where neither are 0. <math>a_{12}</math> and <math>a_{21}</math> are negative, so exactly one of <math>x_2</math> or <math>x_3</math> is positive. Unfortunately this means that one of <math>a_{22}x_2 + a_{23}x_3</math> or <math>a_{32}x_2 + a_{33}x_3 = 0</math> is positive and the other is negative, so the equation couldn't possibly be satisfied if <math>x_2</math> or <math>x_3</math> isn't 0. We have covered the case where one of the <math>x_i</math> is 0, now let's assume that none of them are 0.
 
Clearly if the <math>x_i</math> are all equal, then they are equal to 0. Now let's assume WLOG that <math>x_1=0</math>. If <math>x_2</math> or <math>x_3</math> is 0, then the other is clearly zero, so let's consider the case where neither are 0. <math>a_{12}</math> and <math>a_{21}</math> are negative, so exactly one of <math>x_2</math> or <math>x_3</math> is positive. Unfortunately this means that one of <math>a_{22}x_2 + a_{23}x_3</math> or <math>a_{32}x_2 + a_{33}x_3 = 0</math> is positive and the other is negative, so the equation couldn't possibly be satisfied if <math>x_2</math> or <math>x_3</math> isn't 0. We have covered the case where one of the <math>x_i</math> is 0, now let's assume that none of them are 0.
  
Line 21: Line 24:
 
Case 1: The <math>x_i</math> are all positive. WLOG <math>x_1\leq x_2\leq x_3</math>. Now consider the third equation, <math>a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0</math>. Therefore <math>x_2(a_{31} +a_{32}+a_{33})+ a_{31}(x_1-x_2)+a_{33}(x_3-x_2)= 0</math>, but all of the terms on the LHS are non-negative and the first one is positive, so this is impossible.
 
Case 1: The <math>x_i</math> are all positive. WLOG <math>x_1\leq x_2\leq x_3</math>. Now consider the third equation, <math>a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0</math>. Therefore <math>x_2(a_{31} +a_{32}+a_{33})+ a_{31}(x_1-x_2)+a_{33}(x_3-x_2)= 0</math>, but all of the terms on the LHS are non-negative and the first one is positive, so this is impossible.
  
−
Case 2: The <math>x_i</math> are all negative. WLOG <math>x_1\leq x_2\leq x_3</math>. Consider the third equation, <math>a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0</math>. Therefore <math>x_3(a_{31}+a_{32}+a_{33})+a_{31}(x_1-x_3)+a_{32}(x_2-x_3)=0</math>, but all of the terms on the LHS are non-positive and the first one is negative, so this is impossible.
+
Case 2: The <math>x_i</math> are all negative. WLOG <math>x_1\geq x_2\geq x_3</math>. Consider the third equation, <math>a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0</math>. Therefore <math>x_3(a_{31}+a_{32}+a_{33})+a_{31}(x_1-x_3)+a_{32}(x_2-x_3)=0</math>, but all of the terms on the LHS are non-positive and the first one is negative, so this is impossible.
  
 
Therefore at least one of the <math>x_i</math> is 0, which implies all of them are 0.
 
Therefore at least one of the <math>x_i</math> is 0, which implies all of them are 0.
 +
 +
 +
== Solution 2 ==
 +
 +
We will prove that the matrix
 +
 +
<math>a_{11}\ \ \ \ a_{12}\ \ \ \ a_{13}</math>
 +
 +
<math>a_{21}\ \ \ \ a_{22}\ \ \ \ a_{23}</math>
 +
 +
<math>a_{31}\ \ \ \ a_{32}\ \ \ \ a_{33}</math>
 +
 +
has its determinant <math>> 0.</math>
 +
 +
The determinant is
 +
 +
<math>a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{32}a_{21} -
 +
a_{13}a_{22}a_{31} - a_{11}a_{23}a_{32} - a_{33}a_{12}a_{21}.</math>
 +
 +
After a little algebraic manipulation we can rewrite this as
 +
 +
<math>[a_{33}(a_{11} + a_{13}) - a_{13}(a_{31} + a_{33})](a_{21} + a_{22} + a_{23}) -
 +
(a_{21}a_{33} - a_{23}a_{31})(a_{11} + a_{12} + a_{13}) -
 +
(a_{11}a_{23} - a_{13}a_{21})(a_{31} + a_{32} + a_{33})</math>
 +
 +
(or we can just verify that this is true).
 +
 +
Note that <math>a_{11} + a_{12} + a_{13} > 0</math> implies <math>a_{11} + a_{12} > 0</math>
 +
and <math>a_{11} + a_{13} > 0.</math>  The expression above is clearly
 +
<math>> 0</math>.  To show this in a simple way, I will just write out
 +
the sign of each factor:
 +
 +
<math>[(+)(+) - (-)(+)]\ (+) - ((-)(+) - (-)(-))\ (+) - ((+)(-) - (-)(-))\ (+)</math>
 +
 +
so now we can see that the end expression is <math>(+).</math>
 +
 +
[Solution by pf02, November 2024]
 +
 +
 +
{{IMO box|year=1965|num-b=1|num-a=3}}

Latest revision as of 17:58, 10 November 2024

Problem

Consider the system of equations \[a_{11}x_1 + a_{12}x_2 + a_{13}x_3 = 0\] \[a_{21}x_1 + a_{22}x_2 + a_{23}x_3 = 0\] \[a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0\] with unknowns $x_1$, $x_2$, $x_3$. The coefficients satisfy the conditions:

(a) $a_{11}$, $a_{22}$, $a_{33}$ are positive numbers;

(b) the remaining coefficients are negative numbers;

(c) in each equation, the sum of the coefficients is positive.

Prove that the given system has only the solution $x_1 = x_2 = x_3 = 0$.


Solution

Clearly if the $x_i$ are all equal, then they are equal to 0. Now let's assume WLOG that $x_1=0$. If $x_2$ or $x_3$ is 0, then the other is clearly zero, so let's consider the case where neither are 0. $a_{12}$ and $a_{21}$ are negative, so exactly one of $x_2$ or $x_3$ is positive. Unfortunately this means that one of $a_{22}x_2 + a_{23}x_3$ or $a_{32}x_2 + a_{33}x_3 = 0$ is positive and the other is negative, so the equation couldn't possibly be satisfied if $x_2$ or $x_3$ isn't 0. We have covered the case where one of the $x_i$ is 0, now let's assume that none of them are 0.

If two are positive and one is negative, then when the negative $x_i$ is paired with one of the positive $a_i$, the corresponding equation is negative. This is bad. If two are negative and one is positive, then when the positive $x_i$ is paired with one of the positive $a_i$, the corresponding equation is positive. This is also bad. Therefore the $x_i$ all have the same sign.

Case 1: The $x_i$ are all positive. WLOG $x_1\leq x_2\leq x_3$. Now consider the third equation, $a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0$. Therefore $x_2(a_{31} +a_{32}+a_{33})+ a_{31}(x_1-x_2)+a_{33}(x_3-x_2)= 0$, but all of the terms on the LHS are non-negative and the first one is positive, so this is impossible.

Case 2: The $x_i$ are all negative. WLOG $x_1\geq x_2\geq x_3$. Consider the third equation, $a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0$. Therefore $x_3(a_{31}+a_{32}+a_{33})+a_{31}(x_1-x_3)+a_{32}(x_2-x_3)=0$, but all of the terms on the LHS are non-positive and the first one is negative, so this is impossible.

Therefore at least one of the $x_i$ is 0, which implies all of them are 0.


Solution 2

We will prove that the matrix

$a_{11}\ \ \ \ a_{12}\ \ \ \ a_{13}$

$a_{21}\ \ \ \ a_{22}\ \ \ \ a_{23}$

$a_{31}\ \ \ \ a_{32}\ \ \ \ a_{33}$

has its determinant $> 0.$

The determinant is

$a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{32}a_{21} - a_{13}a_{22}a_{31} - a_{11}a_{23}a_{32} - a_{33}a_{12}a_{21}.$

After a little algebraic manipulation we can rewrite this as

$[a_{33}(a_{11} + a_{13}) - a_{13}(a_{31} + a_{33})](a_{21} + a_{22} + a_{23}) - (a_{21}a_{33} - a_{23}a_{31})(a_{11} + a_{12} + a_{13}) - (a_{11}a_{23} - a_{13}a_{21})(a_{31} + a_{32} + a_{33})$

(or we can just verify that this is true).

Note that $a_{11} + a_{12} + a_{13} > 0$ implies $a_{11} + a_{12} > 0$ and $a_{11} + a_{13} > 0.$ The expression above is clearly $> 0$. To show this in a simple way, I will just write out the sign of each factor:

$[(+)(+) - (-)(+)]\ (+) - ((-)(+) - (-)(-))\ (+) - ((+)(-) - (-)(-))\ (+)$

so now we can see that the end expression is $(+).$

[Solution by pf02, November 2024]


1965 IMO (Problems) • Resources
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1965_IMO_Problems/Problem_2&oldid=233111"