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 of <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 of <math>x + y</math> can have? | ||
− | == Solution == | + | == Solution 1== |
− | + | One way to solve this problem seems to be by [[brute force]]. | |
<math>x^2+y^2=(x+y)^2-2xy=7</math> and | <math>x^2+y^2=(x+y)^2-2xy=7</math> and | ||
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The largest possible solution is therefore <math>x+y=w=4</math>. | The largest possible solution is therefore <math>x+y=w=4</math>. | ||
+ | |||
+ | == Solution 2== | ||
+ | An alternate way to solve this is to let <math>x=a+bi</math> and <math>y=c+di</math>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | <math>(a^2+c^2-2b^2)+(2ab-2cb)i=7+0i</math> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>): | ||
+ | |||
+ | <math>a^3+3a(bi)^2+a^3+3a(-bi)^2=10</math> | ||
+ | |||
+ | <math>2a^3-6ab^2=10</math> | ||
+ | |||
+ | Since we know that <math>2b^2=2a^2-7</math>, it can be plugged in for <math>b^2</math> in the above equataion to yield: | ||
+ | |||
+ | <math>2a^3-3a(2a^2-7)=10</math> | ||
+ | |||
+ | <math>-4a^3+21a=10</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> | ||
== See also == | == See also == |
Revision as of 04:27, 29 December 2009
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 of
can have?
Solution 1
One way to solve this problem seems to be by brute force.
and
Because we are only left with and
, substitution won't be too bad. Let
and
.
We get and
Because we want the largest possible , let's find an expression for
in terms of
.
.
Substituting, . Factored,
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 we know that , it can be plugged in for
in the above equataion 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
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 |