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1983 AIME Problems/Problem 5

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Problem

Suppose that the sum of the squares of two complex numbers x and y is 7 and the sum of the cubes is 10. What is the largest real value of x + y can have?

Solution

The best way to solve this problem seems to be by brute force.

x^2+y^2=(x+y)^2-2xy=7 and x^3+y^3=(x+y)(x^2+y^2)-xy(x+y)=(x+y)(7)-xy(x+y)=(7-xy)(x+y)=10

Because we are only left with x+y and xy, substitution won't be too bad. Let w=x+y and z=xy.

We get w^2-2z=7 and w(7-z)=10

Because we want the largest possible w, let's find an expression for z in terms of w.

w^2-7=2z \implies z=\frac{w^2-7}{2}.

Substituting, w^3-21w+20=0. Factored, (w-1)(w+5)(w-4)=0

The largest possible solution is therefore x+y=w=4.

See also

1983 AIME (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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