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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
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!
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0 replies
jlacosta
Mar 2, 2025
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AMC- IMO preparation
asyaela.   9
N 21 minutes ago by Schintalpati
I'm a ninth grader, and I recently attempted the AMC 12, getting 18 questions correct and leaving 7 empty. I started working on Olympiad math in November and currently dedicate about two hours per day to preparation. I'm feeling a bit demotivated, but if it's possible for me to reach IMO level, I'd be willing to put in more time. How realistic is it for me to get there, and how much study would it typically take?
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asyaela.
3 hours ago
Schintalpati
21 minutes ago
Tennessee Math Tournament (TMT) Online 2025
TennesseeMathTournament   29
N 27 minutes ago by NashvilleSC
Hello everyone! We are excited to announce a new competition, the Tennessee Math Tournament, created by the Tennessee Math Coalition! Anyone can participate in the virtual competition for free.

The testing window is from March 22nd to April 5th, 2025. Virtual competitors may participate in the competition at any time during that window.

The virtual competition consists of three rounds: Individual, Bullet, and Team. The Individual Round is 60 minutes long and consists of 30 questions (AMC 10 level). The Bullet Round is 20 minutes long and consists of 80 questions (Mathcounts Chapter level). The Team Round is 30 minutes long and consists of 16 questions (AMC 12 level). Virtual competitors may compete in teams of four, or choose to not participate in the team round.

To register and see more information, click here!

If you have any questions, please email connect@tnmathcoalition.org or reply to this thread!
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TennesseeMathTournament
Mar 9, 2025
NashvilleSC
27 minutes ago
AIME score for college apps
Happyllamaalways   75
N 37 minutes ago by hashbrown2009
What good colleges do I have a chance of getting into with an 11 on AIME? (Any chances for Princeton)

Also idk if this has weight but I had the highest AIME score in my school.
75 replies
Happyllamaalways
Mar 13, 2025
hashbrown2009
37 minutes ago
AMC 8 discussion
Jaxman8   42
N an hour ago by mpcnotnpc
Discuss the AMC 8 below!
42 replies
Jaxman8
Jan 29, 2025
mpcnotnpc
an hour ago
No more topics!
average FE
KevinYang2.71   75
N 5 hours ago by Marcus_Zhang
Source: USAJMO 2024/5
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy
\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]for all $x,y\in\mathbb{R}$.

Proposed by Carl Schildkraut
75 replies
KevinYang2.71
Mar 21, 2024
Marcus_Zhang
5 hours ago
average FE
G H J
Source: USAJMO 2024/5
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KevinYang2.71
392 posts
#1 • 7 Y
Y by peelybonehead, megarnie, Rounak_iitr, Blue_banana4, buddyram, deduck, ItsBesi
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy
\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]for all $x,y\in\mathbb{R}$.

Proposed by Carl Schildkraut
This post has been edited 1 time. Last edited by KevinYang2.71, Mar 21, 2024, 3:13 PM
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eibc
595 posts
#2 • 3 Y
Y by Aryan_Agarwal, aidan0626, Jack_w
Solution avoiding pointwise trap:

The solutions are $f \equiv 0, f \equiv x^2, f \equiv -x^2$. We can verify these all work, so now we prove they are the only ones.

Let $P(x, y)$ denote the original assertion. From $P(0, 0)$ we have $f(f(0)) = 0$. Then, from $P(x, \tfrac{x^2}{2})$ we get $x^2f(x) = f(f(x))$.

Claim 1: $f$ is even

Proof: Substituting $x^2f(x) = f(f(x))$ into the original equation, we have
$$f(x^2 - y) + 2yf(x) = x^2f(x) + f(y).$$Then, substituting $x \mapsto -x$ yields
$$f(x^2 - y) + 2yf(-x) = x^2f(-x) + f(y).$$Subtracting the two equations, we have
$$x^2f(x) - x^2f(-x) - 2yf(x) + 2yf(-x) = 0 \implies (f(x) - f(-x))(x^2 - 2y) = 0,$$so taking $y$ such that $x^2 \neq 2y$ gives $f(x) = f(-x)$. $\Box$

Now, taking $P(0, y)$ for any nonzero $y$ yields $2y f(0) = 0$, so $f(0) = 0$. Then, from $P(x, 0)$, we have $f(x^2) = f(f(x))$.

Claim 2: There exists a constant $k$ such that $f(x) = kx^2$ for all $x$

Proof: Since $f$ is even and $f(0) = k \cdot 0^2$ for any constant $k$, it suffices to prove the claim for positive $x$. Notice that $xf(\sqrt{x}) = f(f(\sqrt{x})) = f(x)$, so from $P(\sqrt{x}, y)$ for positive $y$ we have
$$f(x - y) + \frac{2yf(x)}{x} = f(x) + f(y).$$Similarly, $P(\sqrt{y}, x)$ yields
$$f(y - x) + \frac{2xf(y)}{y} = f(y) + f(x).$$Since $f$ is even, we have $f(x - y) = f(y - x)$, so $\tfrac{2yf(x)}{x} = \tfrac{2xf(y)}{y} \implies \tfrac{f(x)}{x^2} = \tfrac{f(y)}{y^2}$, and from varying $y$ it follows that $f(x) = kx^2$ for some constant $k$. $\Box$

Now, notice that
$$kx^4 = f(x^2) = f(f(x)) = k^3x^4,$$so by taking nonzero $x$ we have $k^3 = k \implies k \in \{-1, 0, 1\}$, which implies the solution set. $\blacksquare$
This post has been edited 3 times. Last edited by eibc, Mar 21, 2024, 3:21 PM
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OronSH
1718 posts
#3 • 1 Y
Y by Aryan_Agarwal
$f(x^2-y)+2yf(x)=f(f(x))+f(y)$

We claim the solutions are $f(x)=0,f(x)=x^2,f(x)=-x^2.$

Denote the assertion as $P(x,y).$
Take $P(x,0)$ to get $f(x^2)=f(f(x))+f(0).$
Take $P(x,x^2)$ to get $f(0)+2x^2f(x)=f(f(x))+f(x)^2=2f(f(x))+f(0).$ Thus $x^2f(x)=f(f(x)).$ In particular, $f(f(0))=0.$
From $P(x,0)$ and $P(-x,0)$ we see $f(f(x))=f(f(-x)).$ Thus $x^2f(x)=x^2f(-x),$ so for $x\ne 0$ we have $f(x)=f(-x)$ and thus $f$ is even.
Take $P(0,y)$ to get $f(-y)+2yf(0)=f(f(0))+f(y).$ This simplifies to $2yf(0)=0,$ so $f(0)=0.$

Now suppose $f(k)=0.$ From $P(k,0)$ we get $f(k^2)=0.$ Taking $P(k,x^2)$ gives $f(x^2)=f(k^2-x^2)=f(x^2-k^2).$ Then $P(x,k^2)$ gives $f(x^2-k^2)+2kf(x)=f(f(x))=f(x^2),$ so $2kf(x)=0,$ so either $f$ is identically zero or $k=0.$

Now suppose $f(a)=f(b)\ne 0.$ From $x^2f(x)=f(f(x))$ we have $a^2f(a)=f(f(a))=f(f(b))=b^2f(b)$ so $a^2=b^2$ and $a=\pm b.$ From $P(x,0)$ we get $f(x)=\pm x^2$ for all $x.$

Taking $P(x,f(y))$ gives $f(x^2-f(y))=f(f(x))+f(f(y))-2f(x)f(y).$ This is symmetric, so $f(x^2-f(y))=f(y^2-f(x))$ so $x^2-f(y)=\pm(y^2-f(x)).$

Now suppose $f(a)=a^2,f(b)=-b^2.$ Taking $x=a,y=b$ in the above, we get $a^2+b^2=\pm(b^2-a^2),$ but clearly this only holds when either $a=0$ or $b=0.$

Thus the only possible solutions are $f(x)=0,f(x)=x^2,f(x)=-x^2$ and we may easily check that these work.
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GrantStar
812 posts
#4 • 1 Y
Y by Jack_w
how many points if i got $x^6f(x)=x^2f(x)^3$ and said this means $f(x)=0,x^2$ and did pointwise trap on these ( i forgot about $-x^2$)
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MathLuis
1448 posts
#5 • 2 Y
Y by ACBY, KevinYang2.71
Let $P(x,y)$ the assertion of the following F.E., i claim that $f(x)=0,x^2,-x^2$ work.
Indeed it is trivial by replacing, now we prove those are the only ones that work.
$P(0,0)$ gives $f(f(0))=0$, $P(x,f(y))$ with symetry gives $f(x^2-f(y))=f(y^2-f(x))$ call this $Q(x,y)$.
Now $P(0,x)$ gives $f(-x)-f(x)=-2xf(0)$, $P(x,0)$ gives $f(x^2)=f(f(x))+f(0)$ which gives $f (f(x))=f(f(-x))$.
Also $P(-x,y)$ for $y \ne 0$ gives $2yf(x)=2yf(-x)$ therefore $f$ is even and thus $f (0)=0$.
Now we have $f(x^2)=f(f(x))$ as well, if $f(a)=f(b)$ y $P(a,x)-P(b,x)$ we get $f(a^2-x)=f (b^2-x)$.
Therefore if $f$ is periodic (say its period is $T$), by $P(x,y)-P(x,y+T)$ we get $2yf(x)=2 (y+T)f(x)$.
From here we can conclude that $f(x)=0$ or $a^2=b^2$, suppose the latter then $f(x)=x^2$ or $f(x)=-x^2$.
Suppose $f(a)=a^2$ and $f(b)=-b^2$ for $a,b \ne 0$, then by $Q(a,b)$ we get $f(a^2+b^ 2)=f(a^2-b^2)$
If $f(x)=x^2$ for both then $b=0$, if $f(x)=-x^2$ then $b=0$ so suppose $(a^2+b^2)^2=-(a ^2-b^2)^2$, then $a=b$ which cannot happen, therefore $f(x)=x^2$ or $f(x)=-x^2$ as desired thus we are done :cool:.
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EpicBird08
1730 posts
#6 • 2 Y
Y by WizardOfChess2011, Tem8
Here is my cursed solution which somehow got a 7/7.

The only solutions are $\boxed{f(x) = 0}, \boxed{f(x) = x^2},$ and $\boxed{f(x) = -x^2}.$ These can all be sought out to satisfy our functional equation.

Now we show that these are the only solutions. Let $P(x,y)$ denote the given equation.

$P\left(x,\frac{x^2}{2}\right)$ gives
\[
f(f(x)) = x^2 f(x).
\]$P(x,0)$ gives
\[
f(x^2) = f(f(x)) + f(0).
\]On the other hand, $P(-x,0)$ gives
\[
f(x^2) = f(f(-x)) + f(0).
\]Combining these two gives
\[
f(f(-x)) = f(f(x)).
\]Using $f(f(x)) = x^2 f(x)$ gives
\[
x^2 f(x) = x^2 f(-x) \rightarrow \boxed{f(x) = f(-x) \text{ for all } x}.
\]Then $P(0,x)$ gives
\[
f(-x) + 2x f(0) = f(f(0)) + f(x).
\]Since $f(f(0)) = f(0) \cdot 0^2 = 0$ and $f(-x) = f(x),$ we get $2x f(0) = 0,$ so $f(0) = 0.$ Plugging into $f(x^2) = f(f(x)) + f(0)$ thus yields
\[
\boxed{f(f(x)) = f(x^2) = x^2 f(x)}.
\]We now find the value of $f(f(f(x)))$ in two different ways. The first is to use the above identity on $f(x)$ to get
\[
f(f(f(x)) = f(x)^2 f(f(x)) = f(x)^2 \cdot x^2 \cdot f(x) = f(x)^3 \cdot x^2.
\]The second way is to find $f(f(x))$ and stick it into $f.$ We get
\[
f(f(f(x)) = f(f(x^2)) = (x^2)^2 f(x^2) = x^4 \cdot x^2 \cdot f(x) = x^6 \cdot f(x).
\]Therefore,
\[
x^6 \cdot f(x) = x^2 \cdot f(x)^3.
\]This implies that $f(x) = 0, x^2, -x^2$ for all individual $x.$

We now deal with the pointwise trap. There are $3$ cases that we consider:

Case 1: $f(a) = a^2$ and $f(b) = 0$ for nonzero $a$ and $b.$ Then $P(b,a)$ yields
\[
f(b^2 - a) + 2a f(b) = f(f(b)) + f(a) \rightarrow f(b^2 - a) = f(0) + a^2 = a^2,
\]while $P(b,-a)$ yields
\[
f(b^2+a) - 2a f(b) = f(f(b)) + f(-a) \rightarrow f(b^2 + a) = 0 + f(a) = a^2.
\]Since $a^2 > 0,$ we require that $f(b^2 - a) = (b^2 - a)^2 = a^2$ and $f(b^2+a) = (b^2 + a)^2 = a^2.$ Thus $(b^2 - a)^2 = (b^2 + a)^2,$ which simplifies down to $4a^2 b = 0.$ This is impossible since $a,b \ne 0.$

Case 2: $f(a) = -a^2$ and $f(b) = 0$ for nonzero $a$ and $b.$ Then $P(b,a)$ yields
\[
f(b^2 - a) + 2a f(b) = f(f(b)) + f(a) \rightarrow f(b^2 - a) = f(0) - a^2 = -a^2,
\]while $P(b,-a)$ yields
\[
f(b^2+a) - 2a f(b) = f(f(b)) + f(-a) \rightarrow f(b^2 + a) = 0 + f(a) = -a^2.
\]Since $-a^2 < 0,$ we require that $f(b^2 - a) = -(b^2 - a)^2 = -a^2$ and $f(b^2+a) = -(b^2 + a)^2 =- a^2.$ Thus $(b^2 - a)^2 = (b^2 + a)^2,$ which simplifies down to $4a^2 b = 0.$ This is impossible since $a,b \ne 0.$

Case 3: $f(a) = a^2$ and $f(b) = -b^2$ for nonzero $a$ and $b.$ Then $P(a,b)$ yields
\[
f(a^2 - b) + 2b f(a) = f(f(a)) + f(b) \rightarrow f(a^2-b) = f(a^2) - 2ba^2 - b^2 = a^2 f(a) - 2ba^2 - b^2 = a^4 - 2ba^2 - b^2
\]while $P(a,-b)$ yields
\[
f(a^2 + b) - 2b f(a) = f(f(a)) + f(-b) \rightarrow f(a^2+b) = f(a^2) + 2ba^2 - b^2 = a^4 + 2ba^2 - b^2.
\]If both of these are equal to $0,$ this clearly implies that $4ba^2 = 0,$ which is impossible. Thus there are $4$ subcases from here:

Subcase 3a: $f(a^2 - b) = (a^2-b)^2 = a^4 - 2ba^2 - b^2.$ This simplifies to $2b^2 = 0,$ impossible.

Subcase 3b: $f(a^2 + b) = (a^2 + b)^2 = a^4 + 2ba^2 - b^2.$ This also yields $2b^2 = 0,$ impossible.

Subcase 3c: $f(a^2 - b) = -(a^2 - b)^2 = a^4 - 2ba^2 - b^2.$ This simplifies to $2a^4 - 4ba^2 = 0,$ so $a^2 = 2b.$ Plugging into $P(a,-b)$ gives
\[
f(3b) = (2b)^2 + 2b(2b) - b^2 = 7b^2 \ne 0, 9b^2, -9b^2,
\]which is a contradiction.

Subcase 3d: $f(a^2 + b) = -(a^2 + b)^2 = a^4 + 2ba^2 - b^2.$ This yields $2a^4 + 4ba^2 = 0,$ so $a^2 = -2b.$ Plugging into $P(a,b)$ gives
\[
f(-3b) = (-2b)^2 - 2b(-2b) - b^2 = 7b^2 \ne 0, 9b^2, -9b^2,
\]which is impossible.

We have exhausted through all cases, so Case $3$ is impossible.

Therefore, the pointwise trap is impossible, and so the only solutions to the given functional equation are those given at the beginning of the solution, as desired.
This post has been edited 2 times. Last edited by EpicBird08, Oct 10, 2024, 4:46 PM
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pi_is_3.14
1437 posts
#7
Y by
Here is my solution: no pointwise :D
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Technodoggo
1928 posts
#8
Y by
How many points for getting the right answer and most of the proof??
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krithikrokcs
142 posts
#9
Y by
I was nearly trapped by pointwise and fixed it in last 15 min; answer is still $$f(x)=x^2,-x^2,0$$
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Orthogonal.
585 posts
#10
Y by
how many points for doing everything like post #5 but only providing like 70 percent detail in the proof of no pointwise trap
This post has been edited 1 time. Last edited by Orthogonal., Apr 29, 2024, 1:20 AM
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Kevin_Liu
33 posts
#11 • 2 Y
Y by Professor33, elasticwealth
oopsy daisies
This post has been edited 1 time. Last edited by Kevin_Liu, Sep 30, 2024, 8:50 PM
Reason: dumb q
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Mathandski
706 posts
#12
Y by
Solution Sketch:

P(0, 0) -> f(f(0)) = 0
Also get,
f(x^2) = x^2 f(x) = f(f(x))
So f(0) = 0

$f$ is even from taking $x = 0$.

If f(n) = 0 for some $n \neq 0$, we plug in $x = n$ to get,
$f(n^2 - y) = f(y)$
Then, plug in $y = x^2 - n^2$ to eventually get $-2n^2 f(x) = 0 \iff f(x) = 0$ for all $x$.

Prove injectivity over positives by FTSOC f(m) = f(n) but $m \neq n > 0$...

Now look at $f(f(x)) = f(x^2)$ to get $f(x) = \pm x^2$ depending on its sign (use even).

$f(x) \in {x^2, -x^2}$ pretty straight-forward pointwise trap.
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Danielzh
476 posts
#13
Y by
how many points for showing $x^6*f(x)=x^2f(x)^3$ :what?:
This post has been edited 1 time. Last edited by Danielzh, Mar 21, 2024, 4:29 AM
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Technodoggo
1928 posts
#14
Y by
I showed that $f(f(x))=f(x^2)$, and I wrote that $f(x)=x^2$, $0$, or $-x^2$. I also made a reference to the pointwise trap and wrote a little bit of garbage about avoiding it; I think it's obvious enough that if you have $f(a)=a^2$ and $f(b)=0$ or $f(a)=a^2$ and $f(b)=-b^2$ or $f(a)=-a^2$ and $f(b)=0$, then it doesn't satisfy the original equation, but I don't think I made that clear enough since I had literally 1 minute left while writing that. :oops:

I also showed that $f(0)=0$. I talked about how if $f(x)$ was injective, then $f(x)=x^2$, but clearly $f$ is not injective since we have $f(x)=f(-x)$. I couldn't figure out how to rigorously prove that $f$ could only be $x^2$, $0$, and $-x^2$ though :noo: :noo:
hoping for a 5 or somehow 6 on this one :noo: :noo:
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gracemoon124
872 posts
#15 • 1 Y
Y by Jack_w
i missed -x^2 help

but i proved f(0)=0, f(1)=0 or 1, f(x^2-x)=(x-1)^2f(x), f(x)=f(-x), and f(x^2)=x^2f(x)=f(x^2). how many points would i get?
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