[/quote]
,
(use in a problem for junior) to practice. Can anyone share to me?
, such that 
Prove that :
equality cases : ?
distinct positive integers such that adding the product of any two of them to 
yields a perfect square?
,
.
,
,
with 
,
,
,
and lines 
and 
intersect at 
.
and 
intersect again at 
.
and 
intersect the circumcircle of triangle 
again at 
and 
,
lies on 
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 
.
for which
be an acute angled scalene triangle with circumcircle 
.
and 
be the circumcenter and orthocenter of 
respectively. Let 
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 
be a square with 
side length and with center 
.
is drawn where point 
lies on line 
.
,
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 
,
Y by
Let 
be an acute triangle with 
. 
are foots of the altitudes drawn from 
respectively. Let 
be the midpoint of segment 
. The tangent at 
to the circumcircle of cuts in 
and 
cuts the parallel to from at 
. Prove that 
is perpendicular to 
.
be an acute triangle with 
. 
are foots of the altitudes drawn from 
respectively. Let 
be the midpoint of segment 
. The tangent at 
to the circumcircle of cuts in 
and 
cuts the parallel to from at 
. Prove that 
is perpendicular to 
.This post has been edited 1 time. Last edited by matinyousefi, May 12, 2020, 8:56 PM
