[/quote]
,
are non-negative reals such that 
.
.
a tangencial quadrilateral that is not a rhombus, let 
be lengths of tangents from 
to the incircle of the quadrilateral which center is 
.
be the midpoints of 
resp. Prove that![\[ \frac{MI}{IN}=\frac{a+c}{b+d} \]](http://latex.artofproblemsolving.com/e/7/0/e709acb8b17c0466a385ba8e7a89bb1bf16c72af.png)
is given. It turns out that for every positive integer number 
,
and 
are positive, they have the same number of divisors. Is it necessarily true that 
?
with 
be the intersection of 
and 
and define 
in a similar way. Furthermore, let 
.
,
are concyclic.
positive integers. Product of any one of them with sum of remaining numbers increased by 
is divisible with sum of all 
![$P \in \mathbb R[x,y]$](http://latex.artofproblemsolving.com/8/a/0/8a0576299e976c51607959fb68c7c742b2a49c6a.png)
with ![$P \not\in \mathbb R[x] \cup \mathbb R[y]$](http://latex.artofproblemsolving.com/a/f/7/af71ba8c643d5bd8a2928fa6714ccd91713bd641.png)
and 
homeomorphismes of 
,
is an homoemorphisme too.
,
.
be the second intersection point of the circumcircles of the triangles 
and 
.
,
and 
are the circumcenters of the triangles 
are concyclic.
where 
is a prime.(Don't get mad at me,I've used the search function and did not see a correct and complete solution anywhere.)
![\[
2^a + 3^b + 1 = 6^c
\]](http://latex.artofproblemsolving.com/1/9/e/19eb5efa2ccdf54cd277339f8d3b99f75168779d.png)
Y by deduck, clarkculus, Alex-131, ld414think, CyclicISLscelesTrapezoid, Leo.Euler, P_Groudon, dolphinday, anantmudgal09, bjump, Shreyasharma, megahertz13, happypi31415, crazyeyemoody907, amaops1123, exp-ipi-1, FIREDRAGONMATH16, eg4334, mahaler, Marcus_Zhang, Jack_w, youlost_thegame_1434, Danielzh, Andyluo, vincentwant, zhenghua, megarnie, ryanbear, Mathdreams, evt917, mathlove_13520, gladIasked, Sedro, gracemoon124, HamstPan38825, Awesomeness_in_a_bun, NaturalSelection, aidan0626, J55406, Rice_Farmer, aliz, EpicBird08, bluecornbot, studymoremath, Grolarbear, DS768, ESAOPS, Maximilian113, abeot, giratina3, OronSH, KnowingAnt, alan9535, peppapig_, puffypundo, David_He, NoSignOfTheta, joshualiu315, trk08, chloekyan, vrondoS, pingpongmerrily, fuzimiao2013, MathLoverYeah, discretemathcombinatorics, IMUKAT, channing421, bachkieu, chiaroscuro, zhoujef000, dragon888, NicoN9
Satisfactory. Keep cooking.
Problems are posted at https://web.evanchen.cc/mockaime.html#current now!
Like last year, we're running the OTIS Mock AIME 2025 again, except this time there will actually be both a I and a II because we had enough problems to pull it off. However, the two versions will feel quite different from each other:
Details are written out at https://web.evanchen.cc/mockaime.html, but to highlight important info:
Problems are posted at https://web.evanchen.cc/mockaime.html#current now!
Like last year, we're running the OTIS Mock AIME 2025 again, except this time there will actually be both a I and a II because we had enough problems to pull it off. However, the two versions will feel quite different from each other:
- The OTIS Mock AIME I is going to be tough. It will definitely be harder than the actual AIME, by perhaps 2 to 4 problems. But more tangibly, it will also have significant artistic license. Problems will freely assume IMO-style background throughout the test, and intentionally stretch the boundary of what constitutes an “AIME problem”.
- The OTIS Mock AIME II is meant to be more practically useful. It will adhere more closely to the difficulty and style of the real AIME. There will inevitably still be some more IMO-flavored problems, but they’ll appear later in the ordering.
Details are written out at https://web.evanchen.cc/mockaime.html, but to highlight important info:
- Free, obviously. Anyone can participate.
- Both tests will release sometime Dec 19th. You can do either/both.
- If you'd like to submit for scoring, you should do so by January 20th at 23:59 Pacific time (same deadline for both). Please hold off on public spoilers before then.
- Solutions, statistics, and maybe some high scores will be published shortly after that.
This post has been edited 1 time. Last edited by v_Enhance, Dec 19, 2024, 6:43 PM
, yeah AIME I is very olympiad-focused. The raw "difficulty" of AIME II is actually very similar to a harder AIME like 2023, but demands more knowledge towards the end.
[14 points] Yonghao Jin
[13 points] Michael Zhao, person1324.990, tenth
[12 points] Advaith Tumiki, Eric Dai, lethan3, sami
