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

Revision as of 11:53, 29 January 2021

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 $x1 = x4 = 1/(a1 - a4)$ .

@@ Line 9: / Line 9: @@
 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.
+<cmath>- x1 + x2 + x3 + x4 = 0.</cmath>
 Subtracting the equation for i=4 from that for i=3 and dividing by (a3 - a4) we get:
-      - x1 - x2 - x3 + x4 = 0.
+<cmath>- x1 - x2 - x3 + x4 = 0.</cmath>
 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.
+<cmath>- x1 - x2 + x3 + x4 = 0.</cmath>
-Hence x2 = x3 = 0, and x1 = x4 = 1/(a1 - a4).
+Hence <math>x2 = x3 = 0</math>, and <math>x1 = x4 = 1/(a1 - a4)</math>.
 ==See also==
 {{IMO box|year=1966|num-b=4|num-a=6}}
 [[Category:Olympiad Algebra Problems]]

1966 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
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Difference between revisions of "1966 IMO Problems/Problem 5"

Revision as of 11:53, 29 January 2021

Problem

Solution

See also