AoPS Academy
be defined by![\[
a_1 = 1, \quad a_{n+1} = a_n + \frac{1}{\sqrt[2024]{a_n}} \quad \text{for } n \geq 1, \, n \in \mathbb{N}
\]](http://latex.artofproblemsolving.com/8/8/f/88f423d87c08b20dc552703fcf30f2a8e9585902.png)
Prove that![\[
a_n^{2025} >n^{2024}
\]](http://latex.artofproblemsolving.com/a/6/8/a6806c817edb7ce29a399981137c1a3e93f63c63.png)
for all positive integers 
.
is almost squarefree if there exists a prime number 
such that 
and 
is squarefree. Prove that for any almost squarefree positive integer 
the ratio 
is an integer.
be a positive integer and 
be a family of at most distinct subsets of the set 
with the following property: we can consider distinct points in the plane, labelled 
and draw segments connecting these points such that points 
and 
are connected if and only if 
belongs to subsets in for any 
Determine the maximal value that the sum of the cardinalities of the subsets in 
can take.
![$[ABCD]$](http://latex.artofproblemsolving.com/c/d/3/cd3361fb9586094387d885d23e9efce948e8b291.png)
be a convex quadrilateral with area 
, and let 
be a point on ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
and 
a point on ![$[AD]$](http://latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
such that triangles ![$[ABQ]$](http://latex.artofproblemsolving.com/f/2/b/f2bc924e20427dd4b9a1d3894f516f242edf134b.png)
and ![$[ADP]$](http://latex.artofproblemsolving.com/f/e/5/fe5380894cefa3066ca77b5ba48d58e0bcf518c7.png)
have area 
. Let 
be the intersection of ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
and ![$[PQ]$](http://latex.artofproblemsolving.com/2/1/c/21ca08816cf8b23ddf756ce9ae098ad327f2443d.png)
. Determine 
.
be the point of intersection of 
and 
and set 
The condition that ![$[DAP]=[BAQ]$](http://latex.artofproblemsolving.com/d/a/8/da80b9b5f58ae983920875a6c47578d143c9b67d.png)
leads to 
and thus 
Therefore 
and thus 
Now, ![$\frac{RC}{RA}=\frac{[RDC]}{[RAD]}=\frac{[ADC]}{[RAD]}-1=\frac{[ADL]+[LDC]}{[RAD]}-1=\frac{1}{1+\lambda}\cdot\frac{[ADB]+[DBC]}{[RAD]}-1=\frac{1}{1+\lambda}\cdot \frac{2014}{[RAD]}-1=\frac{2014}{1+\lambda}\cdot \frac{QA}{DA\cdot [AQR]}-1=\frac{2014}{1+\lambda}\cdot \frac{QA}{DA}\cdot \frac{1+\lambda}{[AQP]}-1=2014\cdot \frac{1}{[ADP]}-1=2014-1=2013.$](http://latex.artofproblemsolving.com/f/2/a/f2a72cabb0ac521b4e058c8fa67b5f8d242541d2.png)