Difference between revisions of "2007 AIME II Problems/Problem 14"
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== Solution == | == Solution == | ||
| − | + | : ''Note:The following solution(s) are non-rigorous.'' | |
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| + | Substitute the values <math>x = \pm i</math>. We find that <math>\displaystyle f(i)f(-2) = f(-i)</math>, and that <math>\displaystyle f(-i)f(-2) = f(i)</math>. This means that <math>f(i)f(-i)f(-2)^2 = f(i)f(-i) \Longrightarrow f(i)f(-i)(f(-2) - 1) = 0</math>. This suggests that <math>\pm i</math> are roots of the polynomial, and so <math>\displaystyle (x - i)(x + i) = x^2 + 1</math> will be a root of the polynomial. | ||
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| + | The polynomial is likely in the form of <math>f(x) = (x^2 + 1)g(x)</math>; <math>g(x)</math> appears to satisfy the same relation as <math>f(x)</math>, so it also probably has the same roots. Thus, <math>f(x) = (x^2 + 1)^nh(x)</math> is the solution. Guessing values for <math>h(x)</math>, try <math>h(x) = 1</math>. Checking a couple of values shows that <math>f(x) = (x^2 + 1)^2</math> works, and so the solution is <math>f(5) = 676</math>. | ||
== See also == | == See also == | ||
Revision as of 09:32, 5 April 2007
Problem
Let 
be a polynomial with real coefficients such that 
and for all 
, 
Find 
Solution
- Note:The following solution(s) are non-rigorous.
Substitute the values 
. We find that 
, and that 
. This means that 
. This suggests that 
are roots of the polynomial, and so 
will be a root of the polynomial.
The polynomial is likely in the form of 
; 
appears to satisfy the same relation as , so it also probably has the same roots. Thus, 
is the solution. Guessing values for 
, try 
. Checking a couple of values shows that 
works, and so the solution is 
.
See also
| 2007 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
