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

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== Problem ==
 
== 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, 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:
  
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
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<math></math>a_1 = 56, a_{n+1} = a_n - \frac{1}{a_n}
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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 ==

Revision as of 04:45, 14 December 2023

Problem

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

$$ (Error compiling LaTeX. Unknown error_msg)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

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See also

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