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.
numbers 
, calculates the list 
where 
is the number of times the number 
appears in the list. So, for example, for the list 
, the program returns 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 
appears in the list. Find the largest value of for which, whatever the initial list of positive integers 
, it is possible for Alexander to do what the teacher asked him to do.
of convex polygons (in the plane). It turns out that every point of the plane lies in the same number of elements of 
as elements of 
. Prove that 
. or .
. Let the perpendicular line from 
to 
meets 
at points 
, respectively, and the foot from 
to 
is 
. 
intersects line 
at 
, 
intersects line at 
, and lines 
intersect at 
.
is tangent to 
.
+1 w
be pairwise coprime positive integers with 
being prime and 
. On the segment ![$I = [0, a_1 a_2 \cdots a_n ]$](http://latex.artofproblemsolving.com/a/2/b/a2bce96b048c9b8fc94926db80aba37fd5037b4a.png)
of the real line, mark all integers that are divisible by at least one of the numbers 
. These points split 
into a number of smaller segments. Prove that the sum of the squares of the lengths of these segments is divisible by .
such that for all 
In the 
plane with the origin 
, the line 
intersects the lines 
, and 
at the points 
, respectively. Find the maximum value of the area of 
be positive real numbers.![$$\sqrt[3]{\left(\frac{a^{2022}-a}{b}+\frac{2021}{a^{\frac{a}{b}}}+1\right)\left(\frac{b^{2022}-b}{c}+\frac{2021}{b^{\frac{b}{c}}}+1\right)\left(\frac{c^{2022}-c}{a}+\frac{2021}{c^{\frac{c}{a}}}+1\right)}$$](http://latex.artofproblemsolving.com/c/4/9/c4923b16be8a8faf86422477a6ed6b4b60eb2552.png)
satisfying ![$$(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge \frac{9}{4}+k\cdot\frac{a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)}{(a+b+c)^{3}}$$](http://latex.artofproblemsolving.com/f/6/e/f6ed10f7fff1cc94edd8f451e75718a0916a8bfa.png)
holds for all 
be a triangle, and let 
and 
be the orthogonal projections of onto the internal bisectors from and 
. Prove that 
is parallel to .
be the incenter. Then 
from 
concyclic. Then 
so 
, so 
.
hits at 
, then is the reflection of on the -bisector and -bisector, resp., and consequently is the image of 
after a homothety from with ratio 
, implying 
.
be feet of on angle bisectors of and respectively. We have ![\[\angle IXY = \angle IAY = \angle CAY - \angle CAI = 90^{\circ} - \frac{C}{2} - \frac{A}{2} = \frac{B}{2}.\]](http://latex.artofproblemsolving.com/0/3/4/034c95103a410189f239bf1a9969a66fdfff6cc6.png)
= 
.
+ 
= 
.
= 
= 
- 
. = 
- 
and = - 
- 
. (from the sum of the angles of the triangle FAC) + 
+
= (*) = 2
= 2
,...(**)
=
, so we're done.
such that 
Prove that 
and 
Prove that
