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

Latest revision as of 03:45, 14 December 2023

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

This problem needs a solution. If you have a solution for it, please help us out by adding it.

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 == 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>.
+The sequence of real numbers <math>a1, a2, \cdots</math> is defined as:
-~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
+<cmath>a_1 = 56, a_{n+1} = a_n - \frac{1}{a_n}</cmath>
+for every integer <math>n \ge 1</math>.
+Prove that there exists an integer <math>k</math>, <math>1 \le k \le 2002</math>, such that <math>a_k < 0</math>.
+~translated into English by Tomas Diaz. orders@tomasdiaz.com
 == Solution ==