Difference between revisions of "1983 AIME Problems/Problem 5"
m (→Solution 3) |
m (→Solution 4) |
||
| (One intermediate revision by one other user not shown) | |||
| Line 59: | Line 59: | ||
===Solution 4=== | ===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)( | + | <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 09:17, 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 | ||
