AIME 1983 problem

mun · **Joined:** Fri, 28 Dec 2007 12:01:53 UTC **Posts:** 1261

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 that x + y can have?
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outermeasure · **Posted:** Fri, 1 Jul 2011 18:49:56 UTC

mun wrote:

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 that x + y can have?

There are 6 points of intersection of the plane quadratic x^2+y^2=7

and the plane cubic x^3+y^3=7

, and it is not that difficult to find what they are:
$(x,y)=\left(2\pm\dfrac{1}{\sqrt{2}}i,2\mp\dfrac{1}{\sqrt{2}}i\right), \left(\dfrac{-5\pm\sqrt{11}i}{2},\dfrac{-5\mp\sqrt{11}i}{2}\right), \left(\dfrac{1\pm\sqrt{13}}{2},\dfrac{1\mp\sqrt{13}}{2}\right).$
so the answer is...

Shadow · **Posted:** Fri, 1 Jul 2011 22:07:05 UTC

This belongs in Miscellaneous. Topic moved.

Soroban · **Posted:** Sat, 2 Jul 2011 13:38:27 UTC

Hello, mun!

I think I've solved it . . .

Quote:

Suppose that the sum of the squares of two complex numbers

and

is 7,
and the sum of the cubes is 10. .What is the largest real value that x + y

can have?

$\text{Let }\,x + y \:=\:a$ .[1]

Square [1]:
. . $(x+y)^2 \:=\:a^2 \quad\Rightarrow\quad \underbrace{x^2 + y^2}_{\text{This is 7}} + 2xy \:=\:a^2 \quad\Rightarrow\quad 7 + 2xy \:=\:a^2$
. . $\text{Hence: }\,xy \:=\:\frac{a^2-7}{2}$ .[2]

Cube [1]:
. . $(x+y)^3 \:=\:a^3 \quad\Rightarrow\quad x^3 + 3x^2y + 3xy^2 + y^3 \:=\:a^3$

. . $\underbrace{x^3 + y^3}_{\text{This is 10}} + 3\underbrace{(xy)}_{[2]}\underbrace{(x+y)}_{[1]} \:=\:a^2 \quad\Rightarrow\quad 10 + 3\left(\frac{a^2-7}{2}\right)(a) \:=\:a^3$

. . $a^3 - 21a + 20 \:=\:0 \quad\Rightarrow\quad (a-1)(a-4)(a+5) \:=\:0$

. . $\text{Hence: }\,a \;=\;-5,\:1,\:4$

$\text{Therefore, the maximum real value of }x+y\text{ is: }\:a \:=\:4$

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AIME 1983 problem

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