ka March Highlights and 2025 AoPS Online Class Information
jlacosta0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.
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Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.
Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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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!
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If that title occured already, it's definitely bad. And contest names aren't good either.
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[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.
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Source: MTRP 2019 Class 11-Multiple Choice Question: Problem 1 :-
Let is differentiable such that Then which of the following is correct?
[list=1]
[*] always exists but not necessarily zero.
[*] always exists and is equal to zero.
[*] may not exist.
[*] exists if is twice differentiable.
[/list]
Points with rational coordinates lie on a plane. It turned out that the distance between every pair of points is an integer. Prove that there exist points with integer coordinates such that ,,
A checkered board is cut into several squares with integer side length. The size of a square is it's side length.
What is the maximum amount of different sizes this squares can have?
For a sequence of integers, a pair with is called interesting if there exists a pair of integers with such that For each , find the largest possible number of interesting pairs in a sequence of length .
There are cards on which the numbers ,,, are written respectively. Alice and Bob play the following game: in each turn, Alice gives two cards to Bob, who must keep one card and discard the other. The game proceeds for four turns in total; in the first two turns, Bob cannot keep both of the cards with the larger numbers, and in the last two turns, Bob also cannot keep both of the cards with the larger numbers. Let be the sum of the numbers written on the cards that Bob keeps. Find the greatest positive integer for which Bob can guarantee that is at least .
Y byAmir Hossein, sa2001, AopsUser101, YC1math, ashrith9sagar_1, Commander_Anta78, Adventure10
The only functions which work are the identity and the zero-function. It is clear that both satisfy the equation. Now we show any valid function that is not zero is the identity. Suppose for some .
If then but is not a solution. So .
Put so . Put so for all ; hence is surjective over . Suppose then implies . So is injective at .
Now so . Put to get for all so for all integers . Now so for all integers . Consequently, is surjective over all of .
Plug to conclude . Put so hence for all . For , last claim is obvious. Now surjectivity shows for all so is multiplicative.
Thus, and hence preserves the sign of the argument. Now pick and so hence and hence so for all . Now suppose for some .
1. If .
Now for sufficiently large, hence which fails as .
2. If .
Again for sufficiently large, hence which fails as .
Finally, shifting down by large integers , we obtain for all as desired.
Y byAmir Hossein, ImbecileMathImbaTation, Adventure10
Easy for P3 (if problems order in difficult)and simple solution.
Let be the assertion of then or For all Indeed this solution work. such that them from
Then from both condition we can get
Lemma:
Proof: we know let show As desired.
From
From for all but contradiction. From lemma we get
Then (or )
Then our equation equivalent to
From
Also from then we get
From use we get
Then for all Indeed this solution work.
This post has been edited 2 times. Last edited by falantrng, Dec 28, 2018, 8:50 AM Reason: Typo
Nice problem. Here's my solution: Let denote the given assertion. Note that the only constant function which satisfies the given equation is the zero function. So from now on we assume that is non-constant.
CLAIM For , we have .
Proof: Note that we have Also, If , then using the first relation, we get that either or . But, when , we have , which contradicts the fact that is non-constant. That means implies . And, when , then (again from the first relation), we get . Summarizing the above, we can say that and are both true simultaneously.
Now, suppose that . Then .
Return to the problem at hand. Then we get
Now, . Using our Claim, we get that as . Then
Finally, Putting in , and using the above relation, we get that
Using and , we can easily find that . Thus, our final answers are Now one can easily verify that these solutions actually work. Hence, done.
This post has been edited 2 times. Last edited by math_pi_rate, Dec 20, 2018, 11:10 AM
I think this solution is more elementary than the others, although many parts are similar to the given above. Solution
Let denote the assertion.
gives us .
If , then we can put to be any real number, and the function would be constant , which clearly is false. Thus, .
now gives us , and we get or . From now on, we are supposing .
. But one can see easily that if , then implies . So the only possibility is here.
. This implies three important things: and using we get . (This is not really useful here, but funny ) It simplifies a lot our first functional equation, and we are now looking at . .
With and , setting integer , we get .
Now, it remains to prove that , for nonzero real . In fact, it would imply , and furthermore, , contradiction!
Thus, the solutions are . Verification is trivial.
Sorry for double posting, but I have a different ending after proving the claim in post #4. Here's my solution: As proved in post #4, we have and that is non-constant. Now, Again, as shown in post #4, one easily gets that . Thus, we have , i.e. is odd. We will show that is injective also. Suppose we have for some . Then However, we cannot have , as is odd. Hence, must be injective.
Now, as , so we can rewrite the problem condition as Then, gives that . This means that is surjective for . But, as , so we get that is in fact always surjective. Now, using injectivity, one can easily prove that , and so we have However, as is surjective, so we can take , giving that is multiplicative also. Let (As is surjective, so is also surjective). Then, using multiplicity, we have But, , and so we get where we use that . This means that is both additive and multiplicative, in which case it is well known that must be the identity function. Hence, done.
are the only functions. They can be trivially checked to be true.
Solution
Let be the assertion that . gives so or . If , then gives . Hence we have . Now gives . This gives or . If then gives or . This solution works. So assume otherwise that . Hence gives that . Now either or . Now if then gives contradiction. Hence, . If , then gives or . This also proves that only can be and no other value of gives . Hence . Now gives which gives for all integers. Now, where gives . Hence for rationals. Now, from applying we can see that . Put to get that . Now can be or . Suppose that there exists nonzero for which . Put to get that . Of course as is not equal to this equation implies that . This implies that or contradiction. Hence for nonzero . This with gives that the only solutions are and .
Edit: Just realised that my solution is similar to @2above.
This post has been edited 7 times. Last edited by Mathotsav, May 22, 2019, 11:45 AM
Answers: and .
Clearly both these functions satisfy, so we continue by proving them to be the only ones. Denote by the assertion . . Now, (which is a solution) or . Assuming to be non-constant,
take . Now suppose there exists such that . Then, , contradiction and thus is injective at is surjective over .
Now, . We now show that is injective. Indeed, if , then . So suppose . Then we get that . Using surjectivity, pick such that . Then we have , contradiction is injective. Now using similar arguments get that which implies is a bijection on . Then and using (obtainable from ) we establish which in turn implies and moreover, is multiplicative is Cauchy and multiplicative and thus is identity.
We show that the only functions satisfying this are and . These can be easily verified.
Clearly, if is a constant function, then we must have . So assume that is nonconstant for the rest of the solution.
Let denote the assertion Observe that and give and . Now gives which turns our equation into Denote the above assertion by . Note that is which implies that is injective. Now, by way of and , we find that Injectivity now implies that is odd. Also note that now is also surjective. Now from and we find that
Recall that we also had . Therefore, for any two reals we have so that is additive as well as multiplicative. Now it is not hard to see that , so we are done.
. It's easy to see that these are indeed solutions to the given FE. Let be the given assertion, we have if , we will have for some real number , plugging it into our FE, which means and for all real . If , then . If there exists a real number , such that , then which if for all real , then . So, is injective at and is not the zero function. Then, but since ,. Now, by comparing and , we have but again , so this also holds for all real . Plugging this back to our original FE, we have and Most importantly, is odd since Therefore, since is odd and , is surjective over . This means is multiplicative and in particular, . Finally, as this implies (it's a well-known FE).
This post has been edited 1 time. Last edited by Keith50, Mar 21, 2021, 5:25 AM
Easy, but nice.
Let denote the given assertion. is obviously a solution. Henceforth, assume that . Now since , we conclude that . So . Observe that and On comparing and we have that . Now note that for all Upon simplification, the above result is equivalent to . Also . Hence we have two solutions, i.e. Hence, we are done. Note
In the above solution, certain exceptions occur at certain places, where some statements are not valid at . But those things get fixed because we have separately calculated the values of at .
This post has been edited 4 times. Last edited by Wizard0001, Apr 25, 2021, 5:04 PM
Note that is a working function,so assume that there exist some other satisfying function.
So if substitute to get and if substitute to get ,so the two equations/statements are equivalent.
Now note that if there exists such that ,substitute to eliminate this case.
Now we substitute say assertion. ,substituiting this i the original relation gives, Now
Again, is an odd function.
Using the above fact (I think this turned redundant in my solution).
Finally to get rid of the annoying in the expression we try to use the equation from our armory.
Plug in to get We substitute in to get
So the working functions are and and its easy to check these work.The End
Hello, the equation in hand is f (x) f (y f(x) - 1) = x ^ 2 f (y) - f (x)
Assume there is some other function since f (x) =0
x = y = 0 => f (0) f (-1) = -f (0) => f(0) = 0 /f(-1) = -1
Taking f (0) = 0 and substitute y = 0
=> f (-1) = -1
Taking f (-1) = -1 and substitute y = 0
=> f (0) = 0
This proves that both statements provided before are equivalent
Considering that there exists an a which is not equal to 0 such that f (a) = 0,
Substituting x = a and y = -1. This will eliminate the case.
Substituting P (x, y)
P (-1, y) => -f (-y -1) = f(y) + 1 => f (y-1) = -f (-y) -1
Now substitute this in the original equation available to get
F (x) f (f (x)) = x^2
P(1, 1)
f(1) f (f(1) -1) = 0
f(1) = 1
Then, P (1,y) = f; This is an odd function
Use f (x +1) = f(x) +1 to get rid of the 1
Put x = x + 1, y = 1 to get
F (f(x)) = 2 x - f (x)
Now substitute f (x) f (f(x)) = x ^2 with f (f (x)) = 2x - f (x)
(f(x) - x) ^ 2 = 0
f(x) = x
We will prove that and are the only solutions.
Let be the assertion in .
It can be easily seen that the only constant function that works is , which is our first solution.
Let be non-constant.
Claim 1 : .
Proof :
Assume possible .
Then, .
As , as we vary over , also varies over .
Hence, .But doesn't satisfy given equation.
Contradiction!
Hence proved claim 1!
Claim 2 : injective.
Proof :
Let .
Let be such that . .
Hence, .
Assume that () is possible. .
If , we get . Contradiction.
If , we get . Contradiction!
Hence, if , is not possible.
Combining with , we get .
Hence proved claim 2!
Hence note that , .Claim 3 : .
Proof : .
(Here we used claim 1 and claim 2)
Hence Proved claim 3!
Claim 4 : .
Proof : . (Here we used claim 1 and claim 2)
Hence proved claim 4!
Claim 5 : The given equation reduces to .
Proof : . (Here we used claim 3) .
Substituting this in given equation, we get .
Hence proved claim 4!
Let be the assertion in .Claim 6 : and .
Proof : .
. .
Hence, (From last two equations).
Hence proved claim 6!
From claim 6, its well-known that is the solution. indeed works and is our second solution!
Hence, we are done
The answer is and only, which both work. Hence suppose . Let denote the assertion.
From we have . If , then for some nonzero constant . But this clearly doesn't work, so and thus . Then from we obtain .
If for some , by picking such that we find that . Then for , gives , hence the range of contains . Since , the range of also contains , so is surjective. Additionally, implies , hence .
Using , from we obtain . Thus we can rewrite the assertion as . From surjectivity, this becomes . Comparing with and using implies that is multiplicative. On the other hand, by plugging in we also get .
Since is multiplicative, we have for all , hence sends positive reals to positive reals. Thus let sending real numbers to real numbers, so is additive (by only considering the multiplicativity of over ). On the other hand, implies that for , hence for . Therefore is linear, so for some . Then we have , hence and as desired.