that satisfy![\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]](http://latex.artofproblemsolving.com/f/b/b/fbb693770cd90d8a144026608074122e54da5802.png)
for all 
.
board is filled with some nonnegative integer. Two numbers in the filling are said to be adjacent if their cells share a common side. (Note that two numbers in cells that share only a corner are not adjacent). The filling is called a garden if it satisfies the following two conditions:
or 
.
and 
.
,
,
and point 
lies inside triangle 
.
and 
meet 
and 
,
intersects 
and 
at points 
and 
,
is cyclic if and only if 
bisects 
.
be a scalene triangle with circumcircle 
and incenter 
.
meets 
at 
and meets 
;
cuts 
.
and 
meet at 
,
is the midpoint of 
.
and 
intersect at points 
and 
.
or 
.
and 
are written on a blackboard, where 
and 
are relatively prime positive integers. At any point, Evan may pick two of the numbers 
and 
written on the board and write either their arithmetic mean 
or their harmonic mean 
on the board as well. Find all pairs 
such that Evan can write 1 on the board in finitely many steps.
be the set of positive integers. A function 
satisfies the equation ![\[\underbrace{f(f(\ldots f}_{f(n)\text{ times}}(n)\ldots))=\frac{n^2}{f(f(n))}\]](http://latex.artofproblemsolving.com/2/e/7/2e73ff2f75b103e9d6a4004a42e4ed7f4ee64a67.png)
for all positive integers 
.
Y by mathmonkey14
1: Little bit harder than most Nr 1's, but you can just bash this out either way.
2. Kinda annoying, but once u break it down it isnt that bad. Average nr2.
3. Cool problem, just break it down into the 3 cases and it isnt too bad. Tiny bit easy for a n3.
4. Dividing by xy, x^2, or y^2 makes this problem a lot easier. You can also factor. avg no 4.
5. Easily sillyable again, kinda annoying for a nr 5. Honestly I would switch no 5 and 6.
6. Pretty simple if you use pithot. If not it can be difficult. Avg no 6.
7. Very hard for a nr7. The whole problem itself is not bad, just it is extremely sillyable
8. Very bashy, and not super easy to solve, but putting this one on a cartesian plane makes it easier. Harder than last years nr8 by a mile.
9. Easy if you see the trick, impossible if not. Without the trick, this problem becomes super bashy, but probably avg nr9. I spent like 1hr on this to absolutely no avail. I got into an equation with degree 6 bc I didnt see the trick.
10. Not very easy, as you have to break it down well for the solution to flow nicely. Quite hard for a nr10 imo
11. Difficult to understand, but once you have the hang of it down it is not horrible. Despite this, many struggled to understand it in the first place. Fairly hard for a nr 11.
12. Lotta people struggle to graph inequalities in 3d planes (So do I). little hard for a nr 12.
13. Very confusing for a bunch of people (including me). Avg for a nr 13 tho.
14. Super hard problem, but extremely elegant. Fermats point is a cool concept here. Hard for a nr 14 tho
15. LTE helps a lot. Honestly I would switch 14 and 15.
Overall, I think problems 1-6 were avg for an aime, but after that the problem got significantly harder. Much harder than last yrs imo. Tough to say what cutoffs will be given all that has gone last year. Tbf I predict sub 220 for both 10a and 10b, but I could be wrong. We kinda just have to wait and see.
2. Kinda annoying, but once u break it down it isnt that bad. Average nr2.
3. Cool problem, just break it down into the 3 cases and it isnt too bad. Tiny bit easy for a n3.
4. Dividing by xy, x^2, or y^2 makes this problem a lot easier. You can also factor. avg no 4.
5. Easily sillyable again, kinda annoying for a nr 5. Honestly I would switch no 5 and 6.
6. Pretty simple if you use pithot. If not it can be difficult. Avg no 6.
7. Very hard for a nr7. The whole problem itself is not bad, just it is extremely sillyable
8. Very bashy, and not super easy to solve, but putting this one on a cartesian plane makes it easier. Harder than last years nr8 by a mile.
9. Easy if you see the trick, impossible if not. Without the trick, this problem becomes super bashy, but probably avg nr9. I spent like 1hr on this to absolutely no avail. I got into an equation with degree 6 bc I didnt see the trick.
10. Not very easy, as you have to break it down well for the solution to flow nicely. Quite hard for a nr10 imo
11. Difficult to understand, but once you have the hang of it down it is not horrible. Despite this, many struggled to understand it in the first place. Fairly hard for a nr 11.
12. Lotta people struggle to graph inequalities in 3d planes (So do I). little hard for a nr 12.
13. Very confusing for a bunch of people (including me). Avg for a nr 13 tho.
14. Super hard problem, but extremely elegant. Fermats point is a cool concept here. Hard for a nr 14 tho
15. LTE helps a lot. Honestly I would switch 14 and 15.
Overall, I think problems 1-6 were avg for an aime, but after that the problem got significantly harder. Much harder than last yrs imo. Tough to say what cutoffs will be given all that has gone last year. Tbf I predict sub 220 for both 10a and 10b, but I could be wrong. We kinda just have to wait and see.
This post has been edited 1 time. Last edited by isache, Feb 11, 2025, 3:36 AM
Reason: because
Reason: because
) 3 cases and boom
,
