[/quote]
,
be a sequence of positive integers with 
,
and![\[a_n = a_{n-1}^{a_{n-1}a_{n-2}}-1\]](http://latex.artofproblemsolving.com/1/b/f/1bf1a88979e7d2b33530de979d66a4e81032953d.png)
for all 
.
is a prime less than 
for some positive integer 
,
such that 
.
be a triangle with circumcircle 
,
.
,
,
with 
,
,
,
and lines 
and 
intersect at 
.
and 
intersect again at 
.
and 
intersect the circumcircle of triangle 
again at 
and 
,
lies on 
.
for which
(use in a problem for junior) to practice. Can anyone share to me?
be an acute angled scalene triangle with circumcircle 
.
be the circumcenter and orthocenter of 
respectively. Let 
and 
be points on segments 
and 
Prove that 
and 
intersect on the circumcircle of 
.
grid with 
rows and 
columns, and each square of the grid is filled with a non-negative integer. Let 
be the average of all the numbers in the grid. Prove that if 
the there exist two identical rows (meaning same numbers in the same order).
is called feirense if it is possible to write on a sheet of paper some integers such that every positive divisor of 
side length and with center 
.
is drawn where point 
.
,
where 
is the number of times the number 
appears in the list. So, for example, for the list 
,
.
numbers. Then he asked him to apply the program to the resulting list, and so on, until a number greater than or equal to 
,
table is colored black or white. Every cell on table border is black. It is known, that in every 
square there are cells of two colors. Prove, that exist 
and 
are acute and 
.
,
.
Y by Adventure10, Mango247
Let 
be a convex quadrilateral, let 
and 
be the midpoints of the sides 
and 
, respectively. The segment 
meets 
in 
. Show that if the lines 
and 
divide the side 
in 3 equal parts, then is a parallelogram.
be a convex quadrilateral, let 
and 
be the midpoints of the sides 
and 
, respectively. The segment 
meets 
in 
. Show that if the lines 
and 
divide the side 
in 3 equal parts, then is a parallelogram.This post has been edited 1 time. Last edited by carlosbr, May 12, 2006, 3:43 AM
![$2[TDB]=[TBC]=2[TFB]$](http://latex.artofproblemsolving.com/b/d/2/bd232b181a02cc9f944df1c99611c97ac5a43c46.png)
and so ![$[TDB]=[TFB]$](http://latex.artofproblemsolving.com/9/e/b/9ebdd89caad59b5ba9577ed3de4ea7e90e41831a.png)
.
.
and so 
is a parallelogramm because the diagonals bisects each other. But also 
.
the intersection beetween 
and 
.![$ 2[TDB] = [TBC]$](http://latex.artofproblemsolving.com/c/c/1/cc19b318362381976f19f0e3d7650b1527ed6d52.png)
.
FS
DC and S is AO 
and 
so DK=KL=LC= x
