Difference between revisions of "2000 AIME I Problems/Problem 12"
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== Problem == | == Problem == | ||
| − | Given a function <math>f</math> for which | + | Given a [[function]] <math>f</math> for which |
<cmath> | <cmath> | ||
f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)</cmath> | f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)</cmath> | ||
| Line 22: | Line 22: | ||
To show that it is possible to have <math>f(23), f(24), \ldots, f(199)</math> distinct, we try to find a function which fulfills the given conditions. A bit of trial and error would lead to the [[cosine]] function: <math>f(x) = \cos \left(\frac{360}{352}(x-23)\right)</math> (in degrees). | To show that it is possible to have <math>f(23), f(24), \ldots, f(199)</math> distinct, we try to find a function which fulfills the given conditions. A bit of trial and error would lead to the [[cosine]] function: <math>f(x) = \cos \left(\frac{360}{352}(x-23)\right)</math> (in degrees). | ||
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== See also == | == See also == | ||
Revision as of 12:44, 16 June 2008
Problem
Given a function 
for which
![\[f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)\]](http://latex.artofproblemsolving.com/1/a/2/1a25d42e3138957558cb585af4e0f19dfc64192a.png)
holds for all real 
what is the largest number of different values that can appear in the list 
Solution
\begin{align*}f(2518 - x) = f(x) &= f(3214 - (2158 - x)) &= f(1056 + x)\\
f(398 - x) = f(x) &= f(2158 - (398 - x)) &= f(1760 + x)\end{eqnarray*} (Error compiling LaTeX. Unknown error_msg)
Since 
we can conclude that (by the Euclidean algorithm)
![\[f(x) = f(352 + x)\]](http://latex.artofproblemsolving.com/6/e/9/6e9149a031c90a84a16859cc9886eeec06f9a836.png)
So we need only to consider one period 
, which can have at most 
distinct values which determine the value of 
at all other integers.
But we also know that 
, so the values 
and 
are repeated. This gives a total of
![\[352 - (46 - 24 + 1) - (351 - 200 + 1) = \boxed{ 177 }\]](http://latex.artofproblemsolving.com/6/2/e/62e798c9e5afa71ead09fa48297384cd4df01524.png)
distinct values.
To show that it is possible to have 
distinct, we try to find a function which fulfills the given conditions. A bit of trial and error would lead to the cosine function: 
(in degrees).
See also
| 2000 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
