Difference between revisions of "1966 IMO Problems/Problem 5"

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Solve the system of equations
 
Solve the system of equations
  
<math>|a_1 - a_2| x_2 +|a_1 - a_3| x_3 +|a_1 - a_4| x_4 = 1\\ |a_2 - a_1| x_1 +|a_2 - a_3| x_3 +|a_2 - a_4| x_4 = 1\\ |a_3 - a_1| x_1 +|a_3 - a_2| x_2 +|a_3-a_4|x_4= 1\\ |a_4 - a_1| x_1 +|a_4 - a_2| x_2 +|a_4 - a_3| x_3 = 1</math>
+
<math>\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_1 - a_2| x_2 + |a_1 - a_3| x_3 + |a_1 - a_4| x_4 = 1 \\
 +
|a_2 - a_1| x_1 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_2 - a_3| x_3 + |a_2 - a_4| x_4 = 1 \\
 +
|a_3 - a_1| x_1 + |a_3 - a_2| x_2 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_3-a_4| x_4 = 1 \\
 +
|a_4 - a_1| x_1 + |a_4 - a_2| x_2 + |a_4 - a_3| x_3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1</math>
  
 
where <math>a_1, a_2, a_3, a_4</math> are four different real numbers.
 
where <math>a_1, a_2, a_3, a_4</math> are four different real numbers.

Revision as of 20:55, 22 September 2024

Problem

Solve the system of equations

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_1 - a_2| x_2 + |a_1 - a_3| x_3 + |a_1 - a_4| x_4 = 1 \\ |a_2 - a_1| x_1 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_2 - a_3| x_3 + |a_2 - a_4| x_4 = 1 \\ |a_3 - a_1| x_1 + |a_3 - a_2| x_2 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_3-a_4| x_4 = 1 \\ |a_4 - a_1| x_1 + |a_4 - a_2| x_2 + |a_4 - a_3| x_3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1$

where $a_1, a_2, a_3, a_4$ are four different real numbers.

Solution

Take a1 > a2 > a3 > a4. Subtracting the equation for i=2 from that for i=1 and dividing by (a1 - a2) we get:

\[- x1 + x2 + x3 + x4 = 0.\]

Subtracting the equation for i=4 from that for i=3 and dividing by (a3 - a4) we get:

\[- x1 - x2 - x3 + x4 = 0.\]

Hence x1 = x4. Subtracting the equation for i=3 from that for i=2 and dividing by (a2 - a3) we get:

\[- x1 - x2 + x3 + x4 = 0.\]

Hence $x2 = x3 = 0$, and $x1 = x4 = 1/(a1 - a4)$.

See also

1966 IMO (Problems) • Resources
Preceded by
Problem 4
1 2 3 4 5 6 Followed by
Problem 6
All IMO Problems and Solutions