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Difference between revisions of "2002 OIM Problems/Problem 5"

(Created page with "== Problem == In the square <math>ABCD</math>, let <math>P</math> and <math>Q</math> be points belonging to the sides <math>BC</math> and <math>CD</math> respectively, differe...")
 
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== Problem ==
 
== Problem ==
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In the square <math>ABCD</math>, let <math>P</math> and <math>Q</math> be points belonging to the sides <math>BC</math> and <math>CD</math> respectively, different from the ends, such that <math>BP = CQ</math>. Points <math>X</math> and <math>Y</math> are considered, belonging to the segments <math>AP</math> and <math>AQ</math> respectively. Show that, whatever <math>X</math> and <math>Y</math>, there exists a triangle whose sides have the lengths of the segments <math>BX</math>, <math>XY</math>, and <math>DY</math>.
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The sequence of real numbers <math>a1, a2, \cdots</math> is defined as:
  
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~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
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<cmath>a_1 = 56, a_{n+1} = a_n - \frac{1}{a_n}</cmath>
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for every integer <math>n \ge 1</math>.
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Prove that there exists an integer <math>k</math>, <math>1 \le k \le 2002</math>, such that <math>a_k < 0</math>.
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~translated into English by Tomas Diaz. orders@tomasdiaz.com
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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Notice that every time we apply the recursion, we are essentially subtracting <math>\frac{1}{a_n}</math> from the current term. Clearly, the sequence is decreasing, so for all <math>n</math>, <math>a_n\le56</math>, so <math>\frac{1}{a_n}\ge\frac{1}{56}</math>. Then, after every application of the recursion, the value of <math>a_i</math> will decrease by at least <math>\frac{1}{56}</math>; clearly, it will eventually reach negative numbers.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
  
 
== See also ==
 
== See also ==
 
https://www.oma.org.ar/enunciados/ibe18.htm
 
https://www.oma.org.ar/enunciados/ibe18.htm

Latest revision as of 18:23, 5 May 2025

Problem

The sequence of real numbers $a1, a2, \cdots$ is defined as:

\[a_1 = 56, a_{n+1} = a_n - \frac{1}{a_n}\]

for every integer $n \ge 1$.

Prove that there exists an integer $k$, $1 \le k \le 2002$, such that $a_k < 0$.

~translated into English by Tomas Diaz. orders@tomasdiaz.com

Solution

Notice that every time we apply the recursion, we are essentially subtracting $\frac{1}{a_n}$ from the current term. Clearly, the sequence is decreasing, so for all $n$, $a_n\le56$, so $\frac{1}{a_n}\ge\frac{1}{56}$. Then, after every application of the recursion, the value of $a_i$ will decrease by at least $\frac{1}{56}$; clearly, it will eventually reach negative numbers.

~ eevee9406

See also

https://www.oma.org.ar/enunciados/ibe18.htm

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