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

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(wrong problem? - fix problem, add tweaked solution by t0rajir0u)
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== Problem ==
 
== Problem ==
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A sphere is inscribed in the tetrahedron whose vertices are <math>\mathrm {A}=(6,0,0)</math>, <math>\mathrm {B}=(0,4,0)</math>, <math>\mathrm {C}=(0,0,2)</math>,  and <math>\mathrm {D}=(0,0,0)</math>. The radius of the sphere is <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
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Given a function <math>f</math> for which
 +
<cmath>
 +
f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)</cmath>
 +
holds for all real <math>x,</math> what is the largest number of different values that can appear in the list <math>f(0),f(1),f(2),\ldots,f(999)?</math>
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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<cmath>\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*}</cmath>
 +
 
 +
Since <math>\mathrm{gcd}(1056, 1760) = 352</math> we can conclude that (by the [[Euclidean algorithm]])
 +
 
 +
<cmath>f(x) = f(352 + x)</cmath>
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 +
So we need only to consider one period <math>f(0), f(1), ... f(351)</math>, which can have at most <math>352</math> distinct values which determine the value of <math>f(x)</math> at all other integers. 
 +
 
 +
But we also know that <math>f(x) = f(46 - x) = f(398 - x)</math>, so the values <math>x = 24, 25, ... 46</math> and <math>x = 200, 201, ... 351</math> are repeated.  This gives a total of
 +
 
 +
<cmath>352 - (46 - 24 + 1) - (351 - 200 + 1) = \boxed{ 177 }</cmath>
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distinct values.
 +
 
 +
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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{{incomplete|solution}}
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2000|n=I|num-b=11|num-a=13}}
 
{{AIME box|year=2000|n=I|num-b=11|num-a=13}}
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 +
[[Category:Intermediate Algebra Problems]]

Revision as of 11:05, 1 January 2008

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(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 $\mathrm{gcd}(1056, 1760) = 352$ 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).

Template:Incomplete

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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