AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
on the board. For each procces, the stutent deletes the number x and replaces it with either 
or 
or 
. Is this possible to make the number 
on the board?
has circumcircle 
and incenter 
. The incircle of triangle touches side 
at 
respectively. Circumcircle of triangle 
intersects again at 
distinct from 
, circumcircle of triangle 
intersects again at 
distinct from 
. Prove that the 
points 
are concyclic.
+1 w
is defined by 
and for all positive integers 
, 
. Show that for every positive integer 
, there exists a suitable positive integer 
such that 
is a multiple of . lies inside triangle so that 
. The point is such that 
is a parallelogram. Prove that 
.
, point 
on 
satisfies 
and 
.
be the point where the perpendicular from 
to line 
meets the angle bisector of 
.
holds. is an isosceles triangle.
be a positive integer, and let its positive divisors be![\[
d_1 < d_2 < \cdots < d_k.
\]](http://latex.artofproblemsolving.com/6/3/e/63ec82aba0d6c4eece89af6a1247a89498616281.png)
Define 
to be the number of ordered pairs 
with 
such that 
.
. when![\[
n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k},
\]](http://latex.artofproblemsolving.com/2/7/2/272dc2cc33ebbf695311cb61567fa0297e1299b6.png)
where the 
are distinct primes and the 
are positive integers.
such that 
.
that verify the following equation for all 
:
be a positive integer. Determine all sequences 
of positive integers such that
for all positive integers 
.
and 
be two circles with respective centres 
and 
intersecting in two distinct points 
and such that 
is an obtuse angle. Let the circumcircle of 
intersect and respectively in points 
and 
. Let the line 
intersect in 
; let the line 
intersect in 
. Prove that, the points 
are concyclic.
be an acute scalene triangle with circumcenter 
. Let 
be the feet of the altitudes from 
to 
respectively. Let 
be the midpoint of 
. Let 
be the circle with diameter 
. Let 
be the intersection of 
and 
. Let 
be the orthocenter of . Let 
be the intersection of 
and . Let 
be the lines through tangent to 
respectively. Let be the intersection of 
and 
. Let be the intersection of 
and 
. Let 
be the line through parallel to 
and let 
be the reflection of across . Prove that 
is tangent to 
.
be a positive integer. The ordered pairs 
where 
are integers in 
are each labeled with a positive integer less than or equal to 
such that every label is used exactly times and there exist integers 
and 
such that the following property holds: For any two lattice points 
and 
that are both labeled 
, there exists an integer such that 
and 
are both divisible by . How many such labelings exist?
be an odd prime. Define 
over the positive integers as follows:
be chosen such that there exists an ordered pair of positive integers 
where 
such that 
. Prove that there exists at least three integers 
such that 
and 
is a perfect square.
denote the set of all positive integers less than 
that are relatively prime to . Let 
. Is the polynomial 
reducible over the rational numbers, given that it has integer coefficients?
be a non-obtuse triangle. For each point in segment let in segment 
and 
in segment 
, such that triangle 
has minimum perimeter. Prove that if all the lines 
are concurrent (when moves along 
) then the angle 
is right.
be a non-obtuse triangle. For each point in segment let in segment and in segment , such that triangle has minimum perimeter. Prove that if all the lines are concurrent (when moves along ) then the angle is right.
and :
be a line and 
and 
are points which are on the same side of . 
and 
are projections of and on . 
is a point on such that lies between and and 
. 
be another point on . Then 
be any point on . Because of wee just choose , 
is fixed. So we need to find locus of such that 
is 
.
and 
be projections of and on ,respectively. From Lemma is a point such that 
. Similarly do same \ thing for . So,after some easy angle-chasing we get that is 
of 
. Moreover, 
and similarly 
. and in a easy form (i.e. not with length ,just with angles).
