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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Need help with barycentric
Sadigly   0
13 minutes ago
Hi,is there a good handout/book that explains barycentric,other than EGMO?
0 replies
Sadigly
13 minutes ago
0 replies
Combinatorics
P162008   3
N 36 minutes ago by P162008
Let $m,n \in \mathbb{N}.$ Let $[n]$ denote the set of natural numbers less than or equal to $n.$

Let $f(m,n) = \sum_{(x_1,x_2,x_3, \cdots, x_m) \in [n]^{m}} \frac{x_1}{x_1 + x_2 + x_3 + \cdots + x_m} \binom{n}{x_1} \binom{n}{x_2} \binom{n}{x_3} \cdots \binom{n}{x_m} 2^{\left(\sum_{i=1}^{m} x_i\right)}$

Compute the sum of the digits of $f(4,4).$
3 replies
P162008
4 hours ago
P162008
36 minutes ago
Find min and max
lgx57   0
an hour ago
Source: Own
$x_1,x_2, \cdots ,x_n\ge 0$,$\displaystyle\sum_{i=1}^n x_i=m$. $k_1,k_2,\cdots,k_n >0$. Find min and max of
$$\sum_{i=1}^n(k_i\prod_{j=1}^i x_j)$$
0 replies
lgx57
an hour ago
0 replies
Find min
lgx57   0
an hour ago
Source: Own
$a,b>0$, $a^4+ab+b^4=60$. Find min of
$$4a^2-ab+4b^2$$
$a,b>0$, $a^4-ab+b^4=60$. Find min of
$$4a^2-ab+4b^2$$
0 replies
lgx57
an hour ago
0 replies
No more topics!
IMO 2010 Problem 1
canada   119
N Today at 1:20 AM by lpieleanu
Find all function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all $x,y\in\mathbb{R}$ the following equality holds \[
f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor \] where $\left\lfloor a\right\rfloor $ is greatest integer not greater than $a.$

Proposed by Pierre Bornsztein, France
119 replies
canada
Jul 7, 2010
lpieleanu
Today at 1:20 AM
IMO 2010 Problem 1
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canada
134 posts
#1 • 26 Y
Y by MSTang, Davi-8191, anantmudgal09, bel.jad5, Ramanujan_math, Gerninza, itslumi, TheMathCruncher_007, samrocksnature, centslordm, HWenslawski, megarnie, Amoonguss, ImSh95, David-Vieta, Mogmog8, Adventure10, Mango247, seansean01347, aidan0626, cubres, PikaPika999, and 4 other users
Find all function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all $x,y\in\mathbb{R}$ the following equality holds \[
f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor \] where $\left\lfloor a\right\rfloor $ is greatest integer not greater than $a.$

Proposed by Pierre Bornsztein, France
This post has been edited 1 time. Last edited by djmathman, Apr 27, 2015, 2:58 PM
Reason: formatting
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socrates
2105 posts
#2 • 54 Y
Y by mad, plwseven, Pythagorasauras, AdithyaBhaskar, droid347, Sx763_, JasperL, Wizard_32, Illuzion, Severus, MathbugAOPS, Promi, Pluto1708, A_Math_Lover, Minusonetwelfth, ayan_mathematics_king, green_leaf, Bassiskicking, OlympusHero, itslumi, Smathematician, Aurn0b, kirillnaval, Muaaz.SY, samrocksnature, centslordm, mijail, Sush0, megarnie, Amoonguss, myh2910, skyguy88, David-Vieta, Zaro23, Stuffybear, Adventure10, Mango247, Sedro, aidan0626, dxd29070501, AlexCenteno2007, cosdealfa, poirasss, EpicBird08, cubres, benren30, PikaPika999, and 7 other users
Put $x=y=0$. Then $f(0)=0$ or $\lfloor f(0) \rfloor=1$.

$\bullet$ If $\lfloor f(0) \rfloor=1$, putting $y=0$ we get $f(x)=f(0)$, that is f is constant.
Substituing in the original equation we find $f(x)=0, \ \forall x \in \mathbb{R}$ or $f(x)=a, \ \forall x \in \mathbb{R}$, where $a \in [1,2)$.

$\bullet$ If $f(0)=0$, putting $x=y=1$ we get $f(1)=0$ or $\lfloor f(1) \rfloor=1$.

For $f(1)=0$, we set $x=1$ to find $f(y)=0 \ \forall y$, which is a solution.


For $\lfloor f(1) \rfloor=1$, setting $y=1$ yields $f(\lfloor x \rfloor)=f(x), \ (*)$.

Putting $x=2, y=\frac{1}{2}$ to the original we get $f(1)=f(2)\lfloor f(\frac{1}{2}) \rfloor$.
However, from $(*)$ we have $f(\frac{1}{2})=f(0)=0$, so $f(1)=0$ which contradicts the fact $\lfloor f(1) \rfloor=1$.

So, $f(x)=0, \ \forall x$ or $f(x)=a, \ \forall x, \ a \in [1,2)$.

Hope I'm right.

:)
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wangsacl
182 posts
#3 • 9 Y
Y by Pythagorasauras, Illuzion, samrocksnature, centslordm, Amoonguss, Adventure10, cubres, and 2 other users
Let $P(x,y)$ be the assertation of $f([x]y)=f(x)[f(y)]$
$P(0,0): f(0)=f(0)[f(0)]$
There are 2 cases:
1. $[f(0)]=1$. Then,
$P(x,0): f(0)=f(x)$
Making $f(x)=c$ for some constant $c$. Substituting back to $P(x,y)$, we get:
$c=c[c]$
or $[c]=1$.
Then, $1\leq c<2$.
2. $f(0)=0$.
$P(1,1): f(1)=f(1)[f(1)]$
Again, there are 2 subcases:
a. $f(1)=0$.
$P(1,y): f(y)=0$.
After checking back to the f.q., we get $f(x)=0$ is also a solution.
b. $[f(1)]=1$.
$P(x,1): f(x)=f([x])$.
Consider:
$P(2010,\frac1{2010}): f(1)=f(2010)[f(\frac{1}{2010})]=f(2010)[f(0)]=0$.
Contradiction, because $1=[f(1)]=[0]=0$.

Then the function which satisfies $P(x,y)$ is:
$f(x)=0\forall x\in\mathbb{R}$
$f(x)=c\forall x\in\mathbb{R}, 1\leq c<2$.

@above: we almost have the same solution.. hope we're right..
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m.candales
186 posts
#4 • 6 Y
Y by samrocksnature, Adventure10, Mango247, cubres, and 2 other users
Solution to Problem 1:
$f([x]y)=f(x)[f(y)]$ (1)

Substituting $y=0$ we have $f(0) = f(x) [f(0)]$.
If $[f0)] \ne 0$ then $f(x) = \frac{f(0)}{[f(0)]}$. Then $f(x)$ is constant. Let $f(x)=c$. Then substituting that in (1) we have $c=c[c] \Rightarrow c(1-[c])=0 \Rightarrow c=0$, or $[c]=1$. Therefore $f(x)=c$ where $c=0$ or $c \in [1,2)$

If $[f(0)] = 0$ then $f(0)=0$.
Now substituting $x=1$ we have $f(y)=f(1)[f(y)]$.
If $f(1) \ne 0$ then $[f(y)] = \frac{f(y)}{f(1)}$ and substituting this in (1) we have $f([x]y)=\frac{f(x)f(y)}{f(1)}$.
Then $f([x]y)=f(x[y])$.
Substituting $x=1/2, y=2$ we get $f(0)=f(1)$. Then $f(1)=0$, which is a contradiction
Therefore $f(1)=0$. and then $f(y)=0$ for all $R$

Then the only solutions are $f(x)=0$ or $f(x)=c$ where $c \in [1,2)$
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Jorge Miranda
157 posts
#5 • 6 Y
Y by samrocksnature, Adventure10, Mango247, cubres, and 2 other users
Easy, but nice
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MellowMelon
5850 posts
#6 • 8 Y
Y by samrocksnature, ZETA_in_olympiad, Adventure10, Mango247, cubres, and 3 other users
The answers are $f(x) = c$ for all $x$ and constant $c$, where $c \in \{0\} \cup [1,2)$.

Put in $x = 0$ to get $f(0) = f(0) \lfloor f(y) \rfloor$. This implies either (a) $\lfloor f(y) \rfloor = 1$ for all $y$, or (b) $f(0) = 0$.

If (a) is true, then the original equation reduces to $f(\lfloor x \rfloor y) = f(x)$. Set $y = 0$ and we get $f(0) = f(x)$, or rather that $f$ is constant. We know $\lfloor f(0) \rfloor = 1$, and we can verify for any $a \in [1,2)$, $f(x) = a$ for all $x$ is a solution.

Otherwise we have (b), $f(0) = 0$. Let $0 < x < 1$. Then $\lfloor x \rfloor = 0$ and the equation becomes $f(x) \lfloor f(y) \rfloor = 0$. Either (c) $\lfloor f(y) \rfloor = 0$ for all $y$, or (d) $f(x) = 0$ for all $0 < x < 1$.

If (c) is true, then the original equation becomes $f(\lfloor x \rfloor y) = 0$, and $x = 1$ gives $f(y) = 0$ for all $y$.

If (d) is true, then let $0 < y < 1$ in the original equation. We get $f( \lfloor x \rfloor y) = 0$. For any real $a$ with $\lfloor a \rfloor \neq 0$, take $x = 2a$ and $y = \frac{a}{\lfloor 2a \rfloor}$, and note that $0 < y < 1$. Then $f( \lfloor x \rfloor y) = f(a) = 0$, and so $f$ is constant at $0$.

EDIT: Alternate finish since the end is a bit sloppy. Use $x = 2, y = 1/2$ to get $f(1) = 0$, then plug in $x = 1$ to the original equation.
This post has been edited 1 time. Last edited by MellowMelon, Jul 7, 2010, 10:45 PM
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Justanotherone
10 posts
#7 • 6 Y
Y by samrocksnature, Dhruv777, Adventure10, cubres, and 2 other users
good problem..hope all the indian members could solve it
This post has been edited 3 times. Last edited by Justanotherone, Jul 8, 2010, 7:22 AM
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ralexamigos
5 posts
#8 • 4 Y
Y by samrocksnature, Adventure10, Mango247, cubres
Ohhhh i'm sure that the most of the bolivian team didn't do this problem, i think just Arran did it ^_^

the bolivian team was not tried tried :(
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mavropnevma
15142 posts
#9 • 13 Y
Y by FlakeLCR, samrocksnature, DofL, ZETA_in_olympiad, Adventure10, AlexCenteno2007, cubres, and 6 other users
Clearly $f(\left\lfloor x\right\rfloor y) = f(\left\lfloor \lfloor x \rfloor \right\rfloor y) = f(\lfloor x \rfloor)\left\lfloor f(y)\right\rfloor$, so $(f(x) - f(\lfloor x \rfloor))\left\lfloor f(y)\right\rfloor = 0$ for all $x,y\in\mathbb{R}$. If $\left\lfloor f(y)\right\rfloor = 0$ for all $y \in \mathbb{R}$, then by taking $x=1$ we get $f(y)=f(1)\left\lfloor f(y)\right\rfloor = 0$, so $f$ is identically null (which checks). If, contrariwise, $\left\lfloor f(y_0)\right\rfloor \neq 0$ for some $y_0 \in \mathbb{R}$, it follows $f(x) = f(\lfloor x \rfloor)$ for all $x \in \mathbb{R}$.
Now it immediately follows $f(x) = f(\lfloor x \rfloor  \cdot 1) = f(x)\lfloor f(1) \rfloor$, hence $f(x)(1 - \lfloor f(1) \rfloor) = 0$. For $x=y_0$ this implies $\lfloor f(1) \rfloor = 1$.
Assume $\lfloor f(0) \rfloor=0$; then $1 \leq f(1) = f\left ( 2\cdot \dfrac {1} {2} \right ) = f(2)\left \lfloor f \left ( \dfrac {1} {2} \right ) \right \rfloor = f(2)\left \lfloor f \left ( \left \lfloor \dfrac {1} {2} \right \rfloor \right ) \right \rfloor = f(2)\left \lfloor f(0) \right \rfloor = 0$, absurd. Therefore $\lfloor f(0) \rfloor \neq 0$, and now $y=0$ in the given functional equation yields $f(0) = f(x)\lfloor f(0) \rfloor$ for all $x \in \mathbb{R}$, therefore $f(x) = c \neq 0$ constant, with $\lfloor c \rfloor = \lfloor f(1) \rfloor = 1$, i.e. $c \in [1,2)$ (which obviously checks).
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rustam
348 posts
#10 • 4 Y
Y by samrocksnature, Adventure10, cubres, and 1 other user
That is a nice problem for number 1 of day 1.
If i could solve it in 40 minutes, I hope everyone from Uzbekistan team managed to solve it =p
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abacadaea
2176 posts
#11 • 6 Y
Y by samrocksnature, Adventure10, Mango247, cubres, and 2 other users
hopefully correct
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SnowEverywhere
801 posts
#12 • 6 Y
Y by samrocksnature, Adventure10, Mango247, AlexCenteno2007, cubres, and 1 other user
Yay, I solved an IMO problem. Is it just me or does it seem as though this year's #1 and #4 are much easier than normal?

Solution

We claim that either $f(x)=k$ such that $k \in [1,2)$ or $f(x)=0$.

We can assume that $f(1) \neq 0$. Because otherwise, by the equality, we have that $f(y)=f(1) \lfloor f(y) \rfloor =0$ and the claim is satsfied. Let $f(1)=k \neq 0$ where $k \in [1,2)$.

Now we will establish several results about $f(x)$.

$(1)$ Let $x=y=1$. Then since $f(1) \neq 0$, it follows that $f(1) \lfloor f(1) \rfloor \quad \Rightarrow \quad \lfloor f(1) \rfloor = 1$.

$(2)$ Let $x=1$. Then $f(y)=f(1) \lfloor f(y) \rfloor = k \lfloor f(y) \rfloor$.

$(3)$ Let $y=1$. Then by (1), $f(\lfloor x \rfloor)=f(x) \lfloor f(1) \rfloor =f(x)$.

Now we will divide into two cases.

Case 1: $f(0) \neq 0$.

Let $x=0$. Then it follows that $f(0)=f(0) \lfloor f(y) \rfloor \quad \Rightarrow \quad \lfloor f(y) \rfloor =1$. Combining this with (2) yields that $f(x)=k \lfloor f(x) \rfloor =k$ where $k \in [1,2)$ and the claim is satisfied.

Case 2: $f(0)=0$.

Let $y$ be such that $y \in [0,1)$. Then it follows that $f(\lfloor x \rfloor y)=f(x) \lfloor f(y) \rfloor$. Now note that by (3), we have that $f(y)=f(\lfloor y \rfloor)=f(0)=0$. Hence $f(\lfloor x \rfloor y)=f(x) \lfloor 0 \rfloor=0$.

Now we will prove that for each real number $a$, we have $x$ and $y$ such that $y \in [0,1)$ and $a= \lfloor x \rfloor y$. If $a>0$, then let $x=a+1$ and $0 \le y=a/\lfloor a+1 \rfloor < 1$. It is clear that from here, we have that $a= \lfloor x \rfloor y$. If $a<0$, then let $x=a-1$ and $0 \le y = a/\lfloor a-1 \rfloor <1$. Hence $a= \lfloor x \rfloor y$ This yields that for all real $a$, it follows that $f(a)=0$ and the claim is satisfied.

Now we will show that both of the claimed functions satisfy the equation for all real $x$ and $y$.

$(1)$ $f(x)=k$ where $k \in [1,2) \quad \Rightarrow \quad f(\lfloor x \rfloor y)=k=f(x) \lfloor f(y) \rfloor$

$(2)$ $f(x)=0 \quad \Rightarrow \quad  f(\lfloor x \rfloor y)=0=f(x) \lfloor f(y) \rfloor$

Therefore the claim has been proven.
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Sylphaen
151 posts
#13 • 4 Y
Y by samrocksnature, Adventure10, Mango247, cubres
My Solution :
For : $x=0 $ we get :

$f(0)=f(0)\left \lfloor f(y) \right \rfloor $
So either $ f(0)=0 $ or $\left \lfloor f(y) \right \rfloor=1$ for all y

$\left \lfloor f(y) \right \rfloor=1$ for all y is indeed a solution .

Now if $f(0)=0$ : Letting $y=1$ and $x=n \in \mathbb N^* $ we get :
$f(n)=f(n)\left \lfloor f(1) \right \rfloor$
if $f(n)=0$ for all n then for $x=n$ we must have $f(yn)=0$ for all y so $f(y)=0 $which is indeed a solution .

If $f(0)=0$ and $ \left \lfloor f(1) \right \rfloor=1 $ and for $x=n $ :
$f(ny)=f(n)\left \lfloor f(y) \right \rfloor $
for $x=n+p$ where $n\in \mathbb Z $ and $p \in \left( 0,1 \right )$:
$f(ny)=f(n+p)\left \lfloor f(y) \right \rfloor $

so either : $f(n)=f(n+p)$ for $p  \in \left( 0,1 \right )$ or $\left \lfloor f(y) \right \rfloor=0$ for all y but $\left \lfloor f(1) \right \rfloor=1$ so :
$f(n)=f(n+p) for p  \in \left( 0,1 \right )$

Let n be an integer $n<0 $
we have :
$f(n+1)=f(\left \lfloor n \right \rfloor\frac{n+1}{n})=f(n)\left \lfloor f(1+\frac{1}{n}) \right \rfloor=f(n).\left \lfloor f(0) \right \rfloor=0$
So $ f(x) = 0 $ for all $x\leq-2$
Now let n be an integer $n>1$ :
We have :
$f(n+1)=f(\left \lfloor n \right \rfloor\frac{n+1}{n})=f(n)\left \lfloor f(1+\frac{1}{n}) \right \rfloor=f(n).\left \lfloor f(1) \right \rfloor=f(n)=c$
so $f(x)=c $for all $x\geq2$
for$ x=-2$ and $y=-1 $ we get $c=0$
Using the fact that $ f(n)=f(n+p) $ for$ p\in (0,1)$ we will have :
$f(x)=0 $for $x\in(-\infty,-1)\cup [0,1)\cup (2,+\infty)$
$f(x)=f(-1) $for $x\in [-1,0) $
$f(x)=f(1) $for$ x \in [1,2) $

but we have :
$f(1)=f( 2 .\frac{1}{2})=f(2).\left \lfloor f(\frac{1}{2} )\right \rfloor =0 $
whish gives a contradiction .
So the solutions are :
$f(x)=0$
$f(x)=c , c \in [1,2[ $
Hope it's correct :) !
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stsamster
18 posts
#14 • 4 Y
Y by samrocksnature, Adventure10, Mango247, cubres
$ f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor $-
Q


case 1 : there is $a$ s.t. $\left\lfloor a\right\rfloor$ is not zero and $f(a)=0$.

$ f(\left\lfloor a\right\rfloor y)=f(a)\left\lfloor f(y)\right\rfloor $.
so $ f(\left\lfloor a\right\rfloor y)=0 $. put $y=\frac{z}{\left\lfloor a\right\rfloor}$. so we get $f(z)=0$ for all $z$.


case 2 : take the negation of case 1. so if $f(x)=0$ then $x$ belongs to $[0,1)$.

case 2.1 : $f(0)$ is not equal to $0$.
put $x=0$ to equation Q. so we get $\left\lfloor f(y)\right\rfloor=1$. thus equation Q gives $ f(\left\lfloor x\right\rfloor y)=f(x) $. put $x=1$. then $f(y)=f(1)$ for each $y$.
thus $f(x)=b$ for each $x$ where $b$ is a number in $[1,2)$.

case 2.2 : $f(0)=0$.

case 2.2.1 : there is $b$ in $[1,0)$ s.t. $f(b)$ is not zero.
so $f(\left\lfloor b\right\rfloor y)=f(b)\left\lfloor f(y)\right\rfloor$. thus $\left\lfloor f(y)\right\rfloor=0$. by Q we get $f(\left\lfloor x\right\rfloor y)=0$. putting $x=1$ gives $f(y)=0$ for each $y$.

case 2.2.2 : negation of case 2.2.1.
put $y=0.5$ to Q. then since $f(0.5)=0$ we get $f(\frac{\left\lfloor x\right\rfloor}{2})=0$. putting $x=2$ will give a contradiction since $f(1)$ is not zero.


thus the solutions are $f(x)=0$ for each $x$, $f(x)=c$ for each $x$ where $c$ is a constsnt in $[1,2)$.
we can easily verify them.
$Q.E.D$
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Dr N0
131 posts
#15 • 4 Y
Y by samrocksnature, Adventure10, Mango247, cubres
I will try to do it differently, maybe it's wrong I dont know.
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