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Difference between revisions of "1983 AIME Problems/Problem 5"
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== Problem ==
 
== Problem ==
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Suppose that the sum of the squares of two complex numbers <math>x</math> and <math>y</math> is <math>7</math> and the sum of the cubes is <math>10</math>. What is the largest real value of <math>x + y</math> can have?
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Suppose that the sum of the squares of two complex numbers <math>x</math> and <math>y</math> is <math>7</math> and the sum of the cubes is <math>10</math>. What is the largest real value that <math>x + y</math> can have?
  
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== Solution ==
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== Solutions ==
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The best way to solve this problem seems to be by [[brute force]].
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=== Solution 1 ===
 +
One way to solve this problem is by [[substitution]]. We have
  
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<math>w^2+z^2=(w+z)^2-2wz=7</math> and
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<math>x^2+y^2=(x+y)^2-2xy=7</math> and
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<math>w^3+z^3=(w+z)(w^2+z^2)-wz(w+z)=(w+z)(7)-wz(w+z)=(7-wz)(w+z)=10</math>
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<math>x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10</math>
  
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Because we are only left with <math>w+z</math> and <math>wz</math>, [[substitution]] won't be too bad. Let <math>x=w+z</math> and <math>y=wz</math>.
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Hence observe that we can write <math>w=x+y</math> and <math>z=xy</math>.
  
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We get <math>x^2-2y=7</math> and
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This reduces the equations to <math>w^2-2z=7</math> and
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<math>x(7-y)=10</math>
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<math>w(7-z)=10</math>.
  
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Because we want the largest possible <math>x</math>, let's find an expression for <math>y</math> in terms of <math>x</math>. <math>x^2-7=2y \implies y=\frac{x^2-7}{2}</math>.
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Because we want the largest possible <math>w</math>, let's find an expression for <math>z</math> in terms of <math>w</math>.
  
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Substituting, <math>x^3-21x+20=0</math>. Factored, <math>(x-1)(x+5)(x-4)=0</math>
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<math>w^2-7=2z \implies z=\frac{w^2-7}{2}</math>.
  
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The largest possible solution is therefore <math>4</math>.
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Substituting, <math>w^3-21w+20=0</math>, which factorizes as <math>(w-1)(w+5)(w-4)=0</math> (the [[Rational Root Theorem]] may be used here, along with synthetic division).
  
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----
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The largest possible solution is therefore <math>x+y=w=\boxed{004}</math>.
  
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* [[1983 AIME Problems/Problem 4|Previous Problem]]
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===Solution 2 ===
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* [[1983 AIME Problems/Problem 6|Next Problem]]
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An alternate way to solve this is to let <math>x=a+bi</math> and <math>y=c+di</math>.
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* [[1983 AIME Problems|Back to Exam]]
 
  
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== See also ==
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Because we are looking for a value of <math>x+y</math> that is real, we know that <math>d=-b</math>, and thus <math>y=c-bi</math>.
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* [[AIME Problems and Solutions]]
 
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* [[American Invitational Mathematics Examination]]
 
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* [[Mathematics competition resources]]
 
  
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[[Category:Intermediate Complex Numbers Problems]]
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Expanding <math>x^2+y^2=7+0i</math> will give two equations, since the real and imaginary parts must match up.
 +
 
 +
<math>(a+bi)^2+(c-bi)^2=7+0i</math>
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<math>(a^2+c^2-2b^2)+(2ab-2cb)i=7+0i</math>
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 +
Looking at the imaginary part of that equation, <math>2ab-2cb=0</math>, so <math>a=c</math>, and <math>x</math> and <math>y</math> are actually complex conjugates.
 +
 
 +
Looking at the real part of the equation and plugging in <math>a=c</math>, <math>2a^2-2b^2=7</math>, or <math>2b^2=2a^2-7</math>.
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Now, evaluating the real part of <math>(a+bi)^3+(a-bi)^3</math>, which equals <math>10</math> (ignoring the odd powers of <math>i</math>, since they would not result in something in the form of <math>10+0i</math>):
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<math>a^3+3a(bi)^2+a^3+3a(-bi)^2=10</math>
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<math>2a^3-6ab^2=10</math>
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Since we know that <math>2b^2=2a^2-7</math>, it can be plugged in for <math>b^2</math> in the above equation to yield:
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<math>2a^3-3a(2a^2-7)=10</math>
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<math>-4a^3+21a=10</math>
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<math>4a^3-21a+10=0</math>
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 +
Since the problem is looking for <math>x+y=2a</math> to be a positive integer, only positive half-integers (and whole-integers) need to be tested.  From the Rational Roots theorem, <math>a=10, a=5, a=\frac{5}{2}</math> all fail, but <math>a=2</math> does work.  Thus, the real part of both numbers is <math>2</math>, and their sum is <math>\boxed{004}</math>.
 +
 
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===Solution 3===
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Begin by assuming that <math>x</math> and <math>y</math> are roots of some polynomial of the form <math>w^2+bw+c</math>, such that by [[Vieta's Formulæ]] and some algebra (left as an exercise to the reader), <math>b^2-2c=7</math> and <math>3bc-b^3=10</math>.
 +
Substituting <math>c=\frac{b^2-7}{2}</math>, we deduce that <math>b^3-21b-20=0</math>, whose roots are <math>-4</math>, <math>-1</math>, and <math>5</math>.
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Since <math>-b</math> is the sum of the roots and is maximized when <math>b=-4</math>, the answer is <math>-(-4)=\boxed{004}</math>.
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===Solution 4===
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<math>x^3 + y^3 = 10 = (x+y)(x^2-xy+y^2) = (x+y)(7-xy) \implies xy = 7 - \frac{10}{x+y}.</math>  Also, <math>(x+y)^3 = x^3 + 3x^2y+3xy^2+y^3 = 10 + 3xy(x+y).</math>  Substituting our above into this, we get <math>10 + 3(7-\frac{10}{x+y})(x+y) = 21x+21y-20 = (x+y)^3</math>.  Letting <math>p = x+y</math>, we have that <math>p^3 - 21p + 20 = 0</math>.  Testing <math>p = 1</math>, we find that this is a root, to get <math>(p-1)(p^2+p-20) = 0 \implies p = -5, 1, 4 \implies \boxed{4}</math>
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== See Also ==
 +
{{AIME box|year=1983|num-b=4|num-a=6}}
 +
[[Category:Intermediate Algebra Problems]]

Latest revision as of 09:17, 17 March 2023

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 that $x + y$ can have?

Solutions

Solution 1

One way to solve this problem is by substitution. We have

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

Hence observe that we can write $w=x+y$ and $z=xy$.

This reduces the equations to $w^2-2z=7$ 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, $w^3-21w+20=0$, which factorizes as $(w-1)(w+5)(w-4)=0$ (the Rational Root Theorem may be used here, along with synthetic division).

The largest possible solution is therefore $x+y=w=\boxed{004}$.

Solution 2

An alternate way to solve this is to let $x=a+bi$ and $y=c+di$.

Because we are looking for a value of $x+y$ that is real, we know that $d=-b$, and thus $y=c-bi$.

Expanding $x^2+y^2=7+0i$ will give two equations, since the real and imaginary parts must match up.

$(a+bi)^2+(c-bi)^2=7+0i$

$(a^2+c^2-2b^2)+(2ab-2cb)i=7+0i$

Looking at the imaginary part of that equation, $2ab-2cb=0$, so $a=c$, and $x$ and $y$ are actually complex conjugates.

Looking at the real part of the equation and plugging in $a=c$, $2a^2-2b^2=7$, or $2b^2=2a^2-7$.

Now, evaluating the real part of $(a+bi)^3+(a-bi)^3$, which equals $10$ (ignoring the odd powers of $i$, since they would not result in something in the form of $10+0i$):

$a^3+3a(bi)^2+a^3+3a(-bi)^2=10$

$2a^3-6ab^2=10$

Since we know that $2b^2=2a^2-7$, it can be plugged in for $b^2$ in the above equation to yield:

$2a^3-3a(2a^2-7)=10$

$-4a^3+21a=10$

$4a^3-21a+10=0$

Since the problem is looking for $x+y=2a$ to be a positive integer, only positive half-integers (and whole-integers) need to be tested. From the Rational Roots theorem, $a=10, a=5, a=\frac{5}{2}$ all fail, but $a=2$ does work. Thus, the real part of both numbers is $2$, and their sum is $\boxed{004}$.

Solution 3

Begin by assuming that $x$ and $y$ are roots of some polynomial of the form $w^2+bw+c$, such that by Vieta's Formulæ and some algebra (left as an exercise to the reader), $b^2-2c=7$ and $3bc-b^3=10$. Substituting $c=\frac{b^2-7}{2}$, we deduce that $b^3-21b-20=0$, whose roots are $-4$, $-1$, and $5$. Since $-b$ is the sum of the roots and is maximized when $b=-4$, the answer is $-(-4)=\boxed{004}$.

Solution 4

$x^3 + y^3 = 10 = (x+y)(x^2-xy+y^2) = (x+y)(7-xy) \implies xy = 7 - \frac{10}{x+y}.$ Also, $(x+y)^3 = x^3 + 3x^2y+3xy^2+y^3 = 10 + 3xy(x+y).$ Substituting our above into this, we get $10 + 3(7-\frac{10}{x+y})(x+y) = 21x+21y-20 = (x+y)^3$. Letting $p = x+y$, we have that $p^3 - 21p + 20 = 0$. Testing $p = 1$, we find that this is a root, to get $(p-1)(p^2+p-20) = 0 \implies p = -5, 1, 4 \implies \boxed{4}$

See Also

1983 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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