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Saudi Arabian Mathematical Competitions

Algebra Difficulty 4.9 AIME Prove it Saudi Arabia

Find the greatest positive real number $M$ such that for all positive real sequence $\left(a_{n}\right)$ and for all real number $m < M$ , it is possible to find some index $n \geq 1$ that satisfies the inequality
$a_{1} + a_{2} + a_{3} + \cdots + a_{n} + a_{n+1} > m a_{n}.$

Solution

Denote $S, T$ as the midpoints of $A M, A Q$ respectively. Hence, $K S$ is the perpendicular bisector of $A M$ and $T L$ is the perpendicular bisector of $A Q$ .

These imply that $A I, S K, T L$ are concurrent at the circumcenter $G$ of triangle $A M Q$ .
Note that $B I$ is the perpendicular bisector of $M N$ then $K \in B I, D I$ is the perpendicular bisector of $P Q$ then $L \in D I$ .
Consider two triangles IBD and GST with BS, DT, IG are concurrent at A then by applying the Desargues theorem, we have $S, T, R$ are collinear.
In other word, $R$ belongs to $S T$ which is the perpendicular bisector of $A J$ then $R A=R J$ .

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