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 is by [[substitution]]. We have | One way to solve this problem is by [[substitution]]. We have | ||
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===Solution 3=== | ===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>. | + | 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>. | 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>. | 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)(p^2+p-20) = 0 \implies p = -5, 1, 4 \implies \boxed{4}</math> | ||
== 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 10:17, 17 March 2023
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
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 |