A few days ago
math olympiad question.. try ur best to solve!?
what are all ordered triples of real numbers (x,y,z) that satisfy
(x+y)(x+y+z) =120
(y+z)(x+y+z)=96
(x+z)(x+y+z)=72
show the work.. if u get this question rite.. u r a math geniuss!
Sponsored question was the final question two competers had to face during the math olympiad of 06′. Winner solved it for a grand prize mitsu evo!
Top 1 Answers
A few days ago
Favorite Answer
(x+y)(x+y+z) is
x^2 + y^2 + 2xy + xz + yz
(y+z)(x+y+z) is
y^2 + z^2 + 2yz + xz + yz
(x+z)(x+y+z) is
x^2 + z^2 + 2xz + xy + zy
adding all three you get
2x^2 + 2y^2 + 2z^2 + 4xy + 4yz + 4zx
and simplyfying this you get
2(x+y+z)^2 = 120 + 96+ 72 = 288
or
(x+y+z)^2 = 144
or x+y+z = + or – 12
substitute this in each equation and when i used +12
I got x = 4, y = 6 and z=2
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