Difference between revisions of "1983 AIME Problems/Problem 5"
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<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>. <math>w^2-7=2z \implies z=\frac{w^2-7}{2}</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>. |
| + | <math>w^2-7=2z \implies z=\frac{w^2-7}{2}</math>. | ||
Substituting, <math>w^3-21w+20=0</math>. Factored, <math>(w-1)(w+5)(w-4)=0</math> | Substituting, <math>w^3-21w+20=0</math>. Factored, <math>(w-1)(w+5)(w-4)=0</math> | ||
| − | The largest possible solution is therefore <math>x+y=w= | + | The largest possible solution is therefore <math>x+y=w=4</math>. |
== See also == | == See also == | ||
Revision as of 22:08, 26 February 2009
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
The best 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 
.
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 | ||
