Difference between revisions of "1966 IMO Problems/Problem 5"
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Solve the system of equations | Solve the system of equations | ||
− | <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. | ||
==Solution== | ==Solution== | ||
− | + | Take a1 > a2 > a3 > a4. Subtracting the equation for i=2 from that for i=1 and dividing by (a1 - a2) we get: | |
+ | |||
+ | <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: | ||
+ | |||
+ | <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: | ||
+ | |||
+ | <cmath>- x1 - x2 + x3 + x4 = 0.</cmath> | ||
+ | |||
+ | Hence <math>x2 = x3 = 0</math>, and <math>x1 = x4 = 1/(a1 - a4)</math>. | ||
==See also== | ==See also== | ||
{{IMO box|year=1966|num-b=4|num-a=6}} | {{IMO box|year=1966|num-b=4|num-a=6}} | ||
[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] |
Revision as of 12:53, 29 January 2021
Problem
Solve the system of equations
where 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:
Subtracting the equation for i=4 from that for i=3 and dividing by (a3 - a4) we get:
Hence x1 = x4. Subtracting the equation for i=3 from that for i=2 and dividing by (a2 - a3) we get:
Hence , and .
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 |