ka April Highlights and 2025 AoPS Online Class Information
jlacosta0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.
WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.
Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!
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Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:
To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.
More specifically:
For new threads:
a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.
Examples: Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿) Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"
b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.
Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".
c) Good problem statement:
Some recent really bad post was:
[quote][/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.
For answers to already existing threads:
d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve , 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.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.
e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.
To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!
Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).
The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
Given a sequence of real numbers , we can select a real number and transform the sequence into . This transformation can be performed multiple times, with each chosen real number potentially being different
(i) Prove that it is possible to transform the sequence into all zeros after a finite number of such transformations.
(ii) To ensure that the above result can be achieved for any given initial sequence, what is the minimum number of transformations required?
Kritesh manages traffic on a grid consisting of 2025 unit squares. Within each unit square is a car, facing either up, down, left, or right. If the square in front of a car in the direction it is facing is empty, it can choose to move forward. Each car wishes to exit the grid.
Kritesh realizes that it may not always be possible for all the cars to leave the grid. Therefore, before the process begins, he will remove cars from the grid in such a way that it becomes possible for all the remaining cars to eventually exit the grid.
What is the minimum value of that guarantees that Kritesh's job is possible?
Consider the isosceles triangle with and the circle of radius centered at Let be the midpoint of The line intersects a second time at Let be a point on such that Let be the intersection of and Prove that
Let a,b,c,d are positive real numbers, if the following inequality holds ab^2+ac^2>=5bcd. Find the minimum value of explanation: (a^2+b^2+c^2+d^2)(1/a^2+1/b^2+1/c^2+1/d^2)
I found the slip of paper a few days ago that I think I wrote when I was in kindergarten. It is just a sequence of numbers and you have to find the next number, the pattern is . I couldn't solve this and was wondering if any of you can find the pattern
s11 was dumb i added wrong
s18 was dumb how did i not see that
s23 was dumb how did i not see that
s24 was dumb how am i not able to read properly
s27 was dumb like the official solution was assume rectangle
t6 was dumb like i did not calcbash enough terms
t8 was dumb how did i not see that
This post has been edited 1 time. Last edited by c_double_sharp, Apr 4, 2025, 6:07 PM
s11 was dumb i added wrong
s18 was dumb how did i not see that
s23 was dumb how did i not see that
s24 was dumb how am i not able to read properly
s27 was dumb like the official solution was assume rectangle
t6 was dumb like i did not calcbash enough terms
t8 was dumb how did i not see that
s11 was dumb i added wrong
s18 was dumb how did i not see that
s23 was dumb how did i not see that
s24 was dumb how am i not able to read properly
s27 was dumb like the official solution was assume rectangle
t6 was dumb like i did not calcbash enough terms
t8 was dumb how did i not see that
I also assume rectangle because I hate rigorous geometry
Can someone explain why the expectation is the reciprocal?
Because it's kind of like saying it has a in chance of happening, so it happens twice every times. This means that it is expected to happen once every times.
Bro, sprint 20 took the most time for me in the first 20. I can't believe that I actually did coordinate geometry on the real test on this problem!
I also almost trolled this problem. I miscalculated by somehow getting the square root of 49/25 but I clutched and caught my mistake to change it to the square root of 81/25 which is 9/5.
This post has been edited 1 time. Last edited by ilikemath247365, Apr 5, 2025, 6:01 PM
s11 was dumb i added wrong
s18 was dumb how did i not see that
s23 was dumb how did i not see that
s24 was dumb how am i not able to read properly
s27 was dumb like the official solution was assume rectangle
t6 was dumb like i did not calcbash enough terms
t8 was dumb how did i not see that
totally agree
fr tho
This post has been edited 1 time. Last edited by PikaPika999, Apr 6, 2025, 12:44 AM
i had seen some extremely trivial solution for p4 target that didn't even require a calculator but i forgot it now
does anyone have a non-bash solution for that one
here are 5 stupid solutions to target #4
sol 1 (sort of bash, most logical)
let the legs be and . Then we have From the latter two equations, we see that Squaring both sides, we have so Substituting we get so the hypotenuse has length
sol 2 (cheese, no bash)
let the legs of the triangle be and . Then we have Additionally, by the triangle inequality, we know the hypotenuse is less than , and because it is the longest side, it is greater than Also, because the question simply asks for the length of the hypotenuse without "simplest radical form" or "common fraction" at the end of the problem, we know the hypotenuse is an integer. Thus, the only numbers that can be the length of the hypotenuse are ,,,,, and . Plugging values into the equation in line yields
sol 3 (graphing calc)
obtain from before. Plug in and into your graphing calculator and find the intersection. We find the graphs roughly intersect at We use our calculator to find the sums of the squares of these two numbers, and get that the hypotenuse has length
sol 4 (even more cheese)
From solution 2, we know that the hypotenuse must be equal to ,,,,, or . Notice that the longer the hypotenuse is, the larger the area of the triangle. Additionally, the area of this triangle is much less than the perimeter, so we can assume the hypotenuse is as large as possible. The length of the hypotenuse is maximized when it is equal to
sol 5 (sol 2+3 equals 5)
From the triangle inequality, we know the hypotenuse is the longest side of the triangle and is less than . We know that (from solution 3), the longer leg is roughly . This means that for the hypotenuse to be the longest side, it must equal
This post has been edited 3 times. Last edited by Bummer12345, Apr 7, 2025, 1:09 PM Reason: latex
The official solution works because the coins follow a Binomial Distribution, which means that we have that where is the expected value, is the probability, and is the number of trials. This also follows from linearity of expectation on each of the groups of . Then, setting and gives the desired answer.
Alternatively, another easy solution is to use recursion: suppose that we want the expected number of rolls, , until an event happens with probability . Then, we have that which rearranges to the desired .
edit: thinking about it now im not completely sure this is valid because expected amount of rolls between HTH and TTT is different but whatever
This post has been edited 3 times. Last edited by happypi31415, Apr 7, 2025, 8:32 PM
target seems interesting
personal difficulty ratings: 1 is trivial, 10 is nats t8(from a year not named 2024 or 2023)
t1: 1, trivial, just average 9876 and 5432.
t2: 1.5 to 2, trivial but annoying.
First two seem hard to silly, esp. with 1 being so easy.
t3: 1 to 1.5, also trivial.
t4: 2 to 2.5, too easy for the placement.
Another very sweepable round, as intended. Nowhere near the difficulty of last year's t4(which was the only one i got wrong last year).
t5: 1 to 1.5, just floor(202/24)
t6: 6 to 7, imo perfect placement. nontrivial. Albeit you don't even have to find the intended solution, simply approximate 2^2025-1 to 2^2025 and plugging the first ~13 terms into a calc should give a good approximation.
t7: 3 to 3.5, fair placement. simple crt.
t8: 4 to 9, really just depends on your perception. there are definitely people who just assumed you can always reciprocate, but there are definitely people who tried much, much harder methods.