Czech and Slovak Mathematical Olympiad 2003
x 2 − x y + y 2 x 2 y + x y 2 = 7 = − 2
Solve the above system of equations in the set of real numbers. Then, give your answer as the sum of the solution coordinates. That is:
If the system has the solutions ( x 1 , y 1 ) , ( x 2 , y 2 ) , ( x 3 , y 3 ) , … , then determine x 1 + y 1 + x 2 + y 2 + x 3 + y 3 + … .
Give your answer to 3 decimal places.
The answer is 0.
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2 solutions
Nice solution! Argh that I didn't see that coming! XD
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Thanks! After comparing our solutions, however, I see that you have six solutions to my three, since you have two solutions ( x , y ) corresponding to each of the three possible values for x + y . My solution doesn't account for this multiplicity, so I'm wondering if it may be incomplete.
Edit: O.k., I see how to get the multiplicity. Each value of x + y corresponds to two solutions, namely ( a , b ) and ( b , a ) .
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No, your solution IS rigorous and complete! x + y goes very well if you have symmetric solutions as I already found. :)
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@Michael Huang – O.k., great, thanks for the confirmation. I added an edit to my previous comment to indicate how our solutions match up. :)
How did you factor that long polynomial?
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Because the expression can be expressed in terms of x 2 + 2 x y + y 2 , we can include extra x y term in the equation.
I noted quickly that x + y = 1 was one solution and then factored that out of the cubic.
let a = x + y , b = x y ⟹ a 2 = 7 − 3 b ⟹ a b = − 2 ⟹ a = − 2 / b 4 / b 2 = 7 − 3 b 4 = 7 b 2 − 3 b 3 3 b 3 − 7 b 2 + 4 = 0 ( b − 1 ) ( 3 b 2 − 4 b − 4 ) = 0 ( b − 1 ) ( 3 b + 2 ) ( b − 2 ) = 0 ⟹ b = 1 , b = − 2 / 3 , b = 2 ⟹ a = − 2 , a = 3 , a = − 1
we just need to solve for the sum of a's and hence, we will arrive to the answer 0 .
The first of the given equations can be rewritten as ( x + y ) 2 − 3 x y = 7 , and the second as
x y ( x + y ) = − 2 ⟹ x y = − x + y 2 , (as clearly x + y = 0 ). Combining these results gives us
( x + y ) 2 + x + y 6 = 7 ⟹ ( x + y ) 3 − 7 ( x + y ) + 6 = 0
⟹ ( ( x + y ) − 1 ) ( ( x + y ) − 2 ) ( ( x + y ) + 3 ) .
Thus the three possible values for x + y are 1 , 2 and − 3 , which sum to 0 .
Edit: Note that, by symmetry, each of the possible values for x + y corresponds to two solutions ( a , b ) and ( b , a ) , so the complete sum of x k + y k is 1 + 1 + 2 + 2 + ( − 3 ) + ( − 3 ) = 0 .