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.
is called lucky if it has at least two distinct prime divisors and can be written in the form:![\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]](http://latex.artofproblemsolving.com/7/4/4/744a5ccaeb9476ebd7d999c395762cb6e99a7a71.png)
where 
are distinct prime numbers that divide . (Note: it is possible that has other prime divisors not among .) Prove that for every prime number 
, there exists a lucky number such that 
.
be a sequence of positive real numbers such that for every 
, we have:![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
, the following holds:![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
, 
, 
, 
, where and 
are odd primes and![\[2^ap^b=(p+2)^c-1.\]](http://latex.artofproblemsolving.com/5/8/5/58518a8ebed14836b72af743ff100eae2efba5e2.png)
, such that for any positive integer 
and non-negative reals 
, the following inequality always holds:
Here 
and 
denote the median and arithmetic mean of 
, respectively.
be an isosceles triangle with 
, and 
an arbitrary point on side 
. The internal angle bisector of 
meets the circumcircle of 
again at 
, and the internal angle bisector of 
meets it again at 
. Show that lines 
, 
and 
are concurrent.
be non-negative real numbers. Prove that
be a triangle such that 
. Points 
lie on segments 
, respectively, such that lines 
and 
are the angle bisectors of triangle 
. Find the measure of angle 
.
satisfy 
prove in equality.
has circumcircle 
and circumcenter 
. A circle 
with center 
intersects the segment at points 
and 
, such that 
, , , and 
are all different and lie on line in this order. Let 
and 
be the points of intersection of and , such that , , , , and lie on in this order. Let 
be the second point of intersection of the circumcircle of triangle 
and the segment 
. Let 
be the second point of intersection of the circumcircle of triangle 
and the segment 
.
and 
are different and intersect at the point 
. Prove that lies on the line 
.
be a point inside a triangle 
such that![\[ \angle APB - \angle ACB = \angle APC - \angle ABC.
\]](http://latex.artofproblemsolving.com/a/6/e/a6e3dc27d0457682fa22af4f918f58d9cd8403bc.png)
, 
be the incenters of triangles 
, 
, respectively. Show that the lines 
, 
, 
meet at a point.
with 
. Let 
points on sides 
, respectively, such that 
are angle bisectors on . is isosceles if and only if 
is right-angled isosceles.
with 
is isosceles. Let 
with sides 
respectively represented by the lines 
. We next obtain 
along with: \Rightarrow y = (2-\sqrt{3})x +(2\sqrt{3}-3)k$](http://latex.artofproblemsolving.com/b/e/7/be732d400d9bdf97feb4c8f5dba288b4b423dc7a.png)
(i); \Rightarrow y = (\sqrt{3}-2)x +(2\sqrt{3}-3)k$](http://latex.artofproblemsolving.com/9/4/8/9487847eaee3e420f696030070e9ca97b1618c3a.png)
(ii);
and 
. Finally, we observe that the product of the slopes:
;
;
is isosceles 
is right-isosceles.
Prove that
,such that 
P
satisfying
Prove
