[/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 
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 
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 
be a square with 
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 
,
Y by
Acute triangle 
with 
is inscribed in circle 
. Points 
and 
, that lie on 
and 
are the feet of altitudes from 
and 
in triangle , and 
is the midpoint of the segment 
. Half-line 
intersects the circle for the second time in 
. Show that the circumcenter of triangle 
lies on the line 
.
with 
is inscribed in circle 
. Points 
and 
, that lie on 
and 
are the feet of altitudes from 
and 
in triangle , and 
is the midpoint of the segment 
. Half-line 
intersects the circle for the second time in 
. Show that the circumcenter of triangle 
lies on the line 
.
![[asy]
size(7cm);
defaultpen(fontsize(10pt));
pair A = dir(100), B = dir(220), C = dir(320), H = orthocenter(A,B,C), D = foot(A,B,C), E = foot(C,A,B), M = (E+D)/2, N = intersectionpoint(M--M+dir(A--M)*100,unitcircle), Ha = 2D-H, Hb = 2*foot(B,C,A)-H, K = extension(A,A+dir(D--E),Hb,foot(Hb,B,C));
draw(A--B--C--A);
draw(unitcircle);
draw(A--Ha^^B--Hb^^C--E, gray);
draw(D--E^^A--N);
draw(N--Hb, dashed);
draw(A--K--Hb);
dot("$A$", A, dir(100));
dot("$B$", B, dir(220));
dot("$C$", C, dir(320));
dot("$H_A$", Ha, dir(240));
dot("$N$", N, dir(230));
dot("$D$", D, dir(315));
dot("$E$", E, dir(150));
dot("$M$", M, dir(180));
dot("$H_B$", Hb, dir(30));
dot("$K$", K, dir(330));
dot("$H$", H, dir(60));
[/asy]](http://latex.artofproblemsolving.com/3/a/2/3a2a27cfdb386222d37c1091684bdc5686358fc4.png)
,
.
be such that 
.
.
which means 
is an isosceles trapezoid, implying 
we deduce 
are collinear. It follows that 
which implies the conclusion.
,
collinear where 
is the second intersection of 
and 
.
.
is symmedian in 
wrt 
is 
point. So, 
cyclic, where 
is the foot of altitude from 
to 
.
which means that 
,
shows that 
collinear.
