Difference between revisions of "1983 AIME Problems/Problem 5"
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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> | 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== | ||
+ | I started by assuming x and y were roots of some polynomial of the form <math>w^2+bw+c</math> | ||
+ | So then <math>b^2-2c=7</math> and <math>3bc-b^3=10</math> | ||
+ | Substituting <math>c=\frac{b^2-7}{2}</math> we arrive at the polynomial <math>b^3-21b-20=0</math> | ||
+ | From rational root theorem we find the roots to be <math>-4,-1,5</math> | ||
+ | Since <math>-b</math> is the sum of the roots and is maximized when b is -4, we the answer is -(-4)=<math>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]] |
Revision as of 15:11, 8 August 2014
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?
Solution
Solution 1
One way to solve this problem seems to be by substitution.
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 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
I started by assuming x and y were roots of some polynomial of the form So then and Substituting we arrive at the polynomial From rational root theorem we find the roots to be Since is the sum of the roots and is maximized when b is -4, we the answer is -(-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 |