Difference between revisions of "1983 AIME Problems/Problem 5"
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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? | 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? | ||
− | == | + | == Solutions == |
=== Solution 1 === | === Solution 1 === | ||
− | One way to solve this problem | + | One way to solve this problem is by [[substitution]]. We have |
<math>x^2+y^2=(x+y)^2-2xy=7</math> and | <math>x^2+y^2=(x+y)^2-2xy=7</math> and | ||
<math>x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10</math> | <math>x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10</math> | ||
− | + | Hence observe that we can write <math>w=x+y</math> and <math>z=xy</math>. | |
− | + | This reduces the equations to <math>w^2-2z=7</math> and | |
− | <math>w(7-z)=10</math> | + | <math>w(7-z)=10</math>. |
Because we want the largest possible <math>w</math>, let's find an expression for <math>z</math> in terms of <math>w</math>. | 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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<math>w^2-7=2z \implies z=\frac{w^2-7}{2}</math>. | <math>w^2-7=2z \implies z=\frac{w^2-7}{2}</math>. | ||
− | Substituting, <math>w^3-21w+20=0</math> | + | 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). |
The largest possible solution is therefore <math>x+y=w=\boxed{004}</math>. | The largest possible solution is therefore <math>x+y=w=\boxed{004}</math>. | ||
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<math>4a^3-21a+10=0</math> | <math>4a^3-21a+10=0</math> | ||
− | 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> | + | 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>. |
+ | |||
+ | ===Solution 3=== | ||
+ | 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>. | ||
+ | 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>. | ||
+ | |||
+ | ===Solution 4=== | ||
+ | <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)(x^2+5x-20) = 0 \implies p = -5, 1, 4 \implies \boxed{4}</math> | ||
+ | |||
+ | Minor edit: Jc426 | ||
== See Also == | == See Also == | ||
{{AIME box|year=1983|num-b=4|num-a=6}} | {{AIME box|year=1983|num-b=4|num-a=6}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 01:34, 29 January 2022
Contents
Problem
Suppose that the sum of the squares of two complex numbers and is and the sum of the cubes is . What is the largest real value that can have?
Solutions
Solution 1
One way to solve this problem is by substitution. We have
and
Hence observe that we can write and .
This reduces the equations to and .
Because we want the largest possible , let's find an expression for in terms of .
.
Substituting, , which factorizes as (the Rational Root Theorem may be used here, along with synthetic division).
The largest possible solution is therefore .
Solution 2
An alternate way to solve this is to let and .
Because we are looking for a value of that is real, we know that , and thus .
Expanding will give two equations, since the real and imaginary parts must match up.
Looking at the imaginary part of that equation, , so , and and are actually complex conjugates.
Looking at the real part of the equation and plugging in , , or .
Now, evaluating the real part of , which equals (ignoring the odd powers of , since they would not result in something in the form of ):
Since we know that , it can be plugged in for in the above equation to yield:
Since the problem is looking for to be a positive integer, only positive half-integers (and whole-integers) need to be tested. From the Rational Roots theorem, all fail, but does work. Thus, the real part of both numbers is , and their sum is .
Solution 3
Begin by assuming that and are roots of some polynomial of the form , such that by Vieta's Formulæ and some algebra (left as an exercise to the reader), and . Substituting , we deduce that , whose roots are , , and . Since is the sum of the roots and is maximized when , the answer is .
Solution 4
Also, Substituting our above into this, we get . Letting , we have that . Testing , we find that this is a root, to get
Minor edit: Jc426
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 |