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Difference between revisions of "2000 AIME I Problems/Problem 12"

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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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== Solution 2 ==
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    One can imagine that there must be multiple lines of symmetry for the function <math>f(x)</math>, as if a function can be expressed with <math>f(x)=f(a-x)</math> it must be symmetric against line <math>x=\frac{a}{2}</math>. Try this yourself by graphing a polynomial <math>f(x)=x^a</math>, then graphing <math>f(n-x)</math>. If <math>f(x)=f(n-x)</math>, their point of intersection at <math>x=\frac{n}{2}</math> must contain a line of symmetry.
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    For this particular function <math>f(x)</math>, it has 3 other values
  
 
== See also ==
 
== See also ==

Revision as of 14:59, 28 November 2024

Problem

Given a function $f$ for which \[f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)\] holds for all real $x,$ what is the largest number of different values that can appear in the list $f(0),f(1),f(2),\ldots,f(999)?$

Solution

\begin{align*}f(2158 - x) = f(x) &= f(3214 - (2158 - x)) &= f(1056 + x)\\ f(398 - x) = f(x) &= f(2158 - (398 - x)) &= f(1760 + x)\end{align*}

Since $\mathrm{gcd}(1056, 1760) = 352$ we can conclude that (by the Euclidean algorithm)

\[f(x) = f(352 + x)\]

So we need only to consider one period $f(0), f(1), ... f(351)$, which can have at most $352$ distinct values which determine the value of $f(x)$ at all other integers.

But we also know that $f(x) = f(46 - x) = f(398 - x)$, so the values $x = 24, 25, ... 46$ and $x = 200, 201, ... 351$ are repeated. This gives a total of

\[352 - (46 - 24 + 1) - (351 - 200 + 1) = \boxed{ 177 }\]

distinct values.

To show that it is possible to have $f(23), f(24), \ldots, f(199)$ distinct, we try to find a function which fulfills the given conditions. A bit of trial and error would lead to the cosine function: $f(x) = \cos \left(\frac{360}{352}(x-23)\right)$ (in degrees).

Solution 2

   One can imagine that there must be multiple lines of symmetry for the function $f(x)$, as if a function can be expressed with $f(x)=f(a-x)$ it must be symmetric against line $x=\frac{a}{2}$. Try this yourself by graphing a polynomial $f(x)=x^a$, then graphing $f(n-x)$. If $f(x)=f(n-x)$, their point of intersection at $x=\frac{n}{2}$ must contain a line of symmetry.

   For this particular function $f(x)$, it has 3 other values

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

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