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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
no of ways of selecting 8 integers a_i such that 1<a_1<=...<=a_8<=8
parmenides51   12
N 5 minutes ago by FrancoGiosefAG
Source: Mexican Mathematical Olympiad 1988 OMM P4
In how many ways can one select eight integers $a_1,a_2, ... ,a_8$, not necesarily distinct, such that $1 \le  a_1 \le ... \le a_8 \le 8$?
12 replies
parmenides51
Jul 27, 2018
FrancoGiosefAG
5 minutes ago
Easy geometry problem
dwrty   0
9 minutes ago

Let ABCD be a square with center O. Triangles BJC and CKD are constructed outward from the square such that BJ = CJ = CK = DK. Let M be the midpoint of CJ. Prove that the lines OM and BK are perpendicular.

0 replies
dwrty
9 minutes ago
0 replies
Prove n is square-free given divisibility condition
CatalanThinker   3
N 27 minutes ago by CHESSR1DER
Source: 1995 Indian Mathematical Olympiad
Let \( n \) be a positive integer such that \( n \) divides the sum
\[
1 + \sum_{i=1}^{n-1} i^{n-1}.
\]Prove that \( n \) is square-free.
3 replies
CatalanThinker
Today at 3:05 AM
CHESSR1DER
27 minutes ago
China Northern MO 2009 p4 CNMO
parkjungmin   4
N an hour ago by exoticc
Source: China Northern MO 2009 p4 CNMO
China Northern MO 2009 p4 CNMO

The problem is too difficult.
Is there anyone who can help me?
4 replies
parkjungmin
Apr 30, 2025
exoticc
an hour ago
No more topics!
IMO 2010 Problem 1
canada   121
N May 15, 2025 by maromex
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
121 replies
canada
Jul 7, 2010
maromex
May 15, 2025
IMO 2010 Problem 1
G H J
G H BBookmark kLocked kLocked NReply
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dolphinday
1328 posts
#134 • 1 Y
Y by cubres
Our two solutions are $f(x) = 0$ and $f(x) = c \forall c \in [1, 2]$.
We can plug in $(0, 0)$ to get $f(0) = f(0)\left\lfloor f(0)\right\rfloor$ which implies $f(0) = 0$ or $\left\lfloor f(0)\right\rfloor = 1$. If the latter is true then plug in $(t, 0)$ then plug in $f(0) = f(t) = c \in [1, 2]$. If $f(0) = 0$ we can plug in $(1, 1)$ to get $f(1) = f(1)\left\lfloor f(1)\right\rfloor \implies f(1) = 0 \veebar \left\lfloor f(1)\right\rfloor = 1$. If $f(1) = 0$ then $(1, 0)$ gives $f(y) = 0$, done. If $\left\lfloor f(1)\right\rfloor = 1$ then $(x, 1) \implies f(\left\lfloor x\right\rfloor) = f(x)$. Now plug in $(1434, \frac{1}{1434})$ to get $f(1) = f(1434)f\left(\left\lfloor \frac{1}{1434} \right\rfloor\right) = 0$, contradiction. So our solutions are $f(x) = 0$ and $\left\lfloor f(x)\right\rfloor = 1$.
This post has been edited 1 time. Last edited by dolphinday, May 16, 2024, 3:53 AM
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Sedro
5850 posts
#135 • 1 Y
Y by cubres
We claim the only solutions are $f\equiv c$, where $c\in \{0\}\cup [1,2)$. It is easy to see all such functions work; we know prove they are the only ones.

Let $P(x,y)$ denote the assertion. From $P(0,0)$, we have that $f(0) = f(0)\lfloor f(0) \rfloor$. Thus, $f(0) = 0$ or $1\le f(0) < 2$. We proceed by casework.

Case 1: $f(0)=0$. Fix $x\in (0,1)$, and we have $f(0) = 0 = f(x)\lfloor f(y) \rfloor$. Suppose there exists some real $y$ such that $f(y) \notin [0,1)$. Then, $f(x)=0$ for all $x\in (0,1)$. Then, let $x$ vary over the reals and $y$ over $(0,1)$ to find that $f(\lfloor x \rfloor y) = 0$. But since $\lfloor x \rfloor y$ can be any real number, $f\equiv 0$, contradiction. Thus, we always have $f(x)\in [0,1)$. Thus, we conclude $f(\lfloor x \rfloor y) = 0$, which implies $f\equiv 0$, as desired.

Case 2: $f(0)\in [1,2)$. If this is the case, then by $P(x,0)$, we have $f(0) = f(x)$, as desired, and we are done. $\blacksquare$
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Bryan0224
60 posts
#137 • 1 Y
Y by cubres
Put $\displaystyle x=y=0$.
We have $\displaystyle f(0)=0$ or $\left\lfloor f(0)\right\rfloor=1$.
We split into two cases.
$(i)\left\lfloor f(0)\right\rfloor=1$
Putting $\displaystyle y=0$ gives $f\equiv f(0)$,so $f\equiv c$ where $1\le c \le2$
$\displaystyle (ii)f(0)=0$.
Put $\displaystyle x=y=1$.
We have $\displaystyle f(1)=0$ or $\left\lfloor f(1)\right\rfloor=1$.
Claim. $\displaystyle f(1)=0$
$\emph{Proof.}$ We use proof by contradiction.
Assume that $\left\lfloor f(1)\right\rfloor=1$.
Puttting $\displaystyle y=1$ gives $f(\left\lfloor x\right\rfloor)=f(x)$
Now $f(1)=f(2)\left\lfloor f(\frac{1}{2})\right\rfloor=f(2)\left\lfloor f(0)\right\rfloor=0$ since $\left\lfloor \frac{1}{2}\right\rfloor=0$.$\blacksquare$
Put $\displaystyle x=1,y=t$.
$\displaystyle f(t)=0$
Combining these cases together, we have $f\equiv c$, where $c\in \{0\}\cup [1,2)$
This post has been edited 2 times. Last edited by Bryan0224, Oct 15, 2024, 7:49 AM
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Jndd
1417 posts
#138 • 1 Y
Y by cubres
Let $P(x,y)$ denote the assertion. First, we have that $P(0,y)$ implies $f(0)=f(0)\lfloor f(y)\rfloor$, so either $f(0)=0$ or $\lfloor f(1)\rfloor = 1$ for all $y$.

In the latter case, we have $f(\lfloor x\rfloor y)=f(x)\cdot 1$, so setting $x=1$, we get $f(y)=f(1)$. Plugging in $f(x)=c$ for all $x$ into the equation, with $c=f(1)$, we see that either $c=0$ or $\lfloor c\rfloor = 1$.

Now, we deal with the case where $f(0)=0$. $P(1,1)$ gives us $f(1) = f(1)\lfloor f(1)\rfloor$, so either $f(1)=0$ or $\lfloor f(1)\rfloor = 1$. If $f(1)=0$, then $P(1,y)$ gives $f(y)=f(1)\lfloor f(y)\rfloor = 0$ for all $y$. Otherwise if $\lfloor f(1)\rfloor = 1$, then $P(x,1)$ gives \[f(\lfloor x\rfloor) = f(x)\lfloor f(1)\rfloor = f(x).\]Then, for any $0\leq y<1$, we have \[f(\lfloor x\rfloor y) = f(x)\lfloor f(y)\rfloor = f(x) \lfloor f(\lfloor y\rfloor)\rfloor = f(x)\cdot 0 = 0,\]so since for any $a$, we can select some $x>a+1$ and let $0\leq y=\frac{a}{\lfloor a+1\rfloor } < 1$, we have $f(a)=0$ for all $a$.

Since we have addressed all our cases, the solutions are $f(x)=c$ for all $x$ where $c\in \{0\}\cup [1, 2)$, and it's easy to check that all such functions satisfy our equation.
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Warideeb
59 posts
#142 • 1 Y
Y by cubres
Warideeb wrote:
Let $P(x,y)$ be the assertion of
$f(\lfloor x \rfloor y)=f(x) \lfloor f(y) \rfloor$
Our \(Claim\) is that $f \equiv 0$ or $f \equiv c$ where $1\leq c < 2$
Now $f \equiv 0$ is clearly a solution. Now we assume $f \not\equiv 0$
Now $P(0,x)$ gives us $f(0)=f(0) \lfloor f(x) \rfloor$ Now if $f(0) \ne 0$ then this gives us $f \equiv c$ for some $ 1\leq c < 2$
Now lets assume $f(0)=0$
$P(1,1)$ gives us $f(1)=c$ where $1 \leq c < 2$ now $P(x,1)$ gives us $f(x)=f(\lfloor x \rfloor)$ define a number $0 \leq A <1$ Now
$P(x,A)$ and subsituting $f(A)$ with $f(\lfloor A \rfloor)$ gives us $f(\lfloor x \rfloor A)=0$ and give is $f \equiv 0$ and here is our solution
This post has been edited 1 time. Last edited by Warideeb, Aug 15, 2024, 2:30 PM
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ezpotd
1279 posts
#144 • 1 Y
Y by cubres
bad formatting because i copied my solution from mathdash

vary x along [n, n+1), then we see for all x in this interval we must either have f(y) in [0,1) (to make the floor zero), or if the floor is not zero we are forced that all x in the interval have the same output when f is applied.



we can now divide into two cases: f(y) lies in [0,1) for all y. then we get f(floor(x)y) = 0, so letting x = 1 gives f(y) = 0, which trivially satisfies the fe.



otherwise, take some value of y for which this is not true. then we get for all integers n, if a,b in [n, n+1), then f(a) = f(b). now e = n, 0 < f - e < 1. then consider varying y over [e,f], then the right side is constant, the left side must be as well, so f(floor(x)y) is constant for all e <= y <= f. with e = 0, f = (m - 1)/m for some large integer m, set x = m, we then get f(my) is constant for all 0 <= y <= m - 1, take m to infinity and we see f is constant over positives. then set x = -1, vary y over all positives and see f is constant over negatives. we show the constant value for postiives and negatives must be the same. firstly, if the positive value is zero set x = 1 and vary y to get negative constant is also zero, so we can dsicrard this, same for other case except do x = -1. now neither constant value is 0, so set the value for x to 0 and we then get 1 = floor(f(y)), so floor(f(0)) = 1. then the positive constant value is in [1,2), then substite negative x, y = 0, then we instantly get the negative constant is also the same as f(0), so the other solution is f = k for k in [1,2), which obviously satisfies the fe.
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Maximilian113
575 posts
#146 • 1 Y
Y by cubres
Let $P(x, y)$ denote the assertion. First, suppose that $f(0)=0.$ Then for all $0 \leq x < 1,$ we have that $$f(x) \lfloor f(y) \rfloor = 0.$$If there is some $x$ in that range such that $f(x) \neq 0,$ we have that $\lfloor f(y) \rfloor = 0$ for all $y$ so $f(x) = 0$ for all $x.$ Otherwise, for all $0 \leq x < 1$ we have $f(x)=0.$ Then $$P(2, 0.5) \implies f(1)=0 \implies P(1, x) \implies f(x)=0 \, \forall x \in \mathbb R.$$
Now assume that $f(0) \neq 0.$ Thus $P(0, 0) \implies \lfloor f(0) \rfloor = 1,$ so $P(x, 0) \implies f(x) = f(0)=c.$ Clearly $1 \leq c < 2$ and we see easily that this solution works.

Therefore, the only solutions are $f(x) = 0, c \in [1, 2).$
This post has been edited 1 time. Last edited by Maximilian113, Mar 16, 2025, 4:51 PM
Reason: oops
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Bardia7003
22 posts
#147 • 1 Y
Y by cubres
Let $P(x,y)$ be the assertion for $f([x]y) = f(x)[f(y)]$.
By $P(0,0)$ we get: $f(0) = f(0)[f(0)]$.
Here there are two cases:
Case 1: $f(0) \neq 0$
So we can cross out $f(0)$ and have $[f(0)] = 1$
Now by $P(x,0)$ we have: $f(0) = f(x)$ so $f$ is constant. Also by $[f(0)] = 1$ we have that if $f(x)=c$ then $[c]$ should equal to 1. We can easily check that for any such $c$ the equality holds.

Case 2: $f(0)=0$
Now by $P(1,1)$ we have $f(1) = f(1)[f(1)]$. If $f(1) = 0$ then $P(1,x)$ gives $f(x) = 0$ which we can check that it is a solution. So we continue by supposing $f(1) \neq 0$ to check the other case. As a result, we can cross out $f(1)$ and have that $[f(1)]=1$. Then by $P(x,1)$ we have $f([x])=f(x)$.
Now $\forall x>2: P(x,\frac{1}{[x]})$ gives us $f(1) = f(x).[f(\frac{1}{[x]}]$ and as we know $[f(\frac{1}{[x]})] = [f([\frac{1}{[x]}])] = [f(0)] = 0$. so $f(1) = 0$, contradiction.

To sum up, we proved that the only functions in which the given equality holds are $f(x) = c, \forall x \in \mathbb{R}, \forall c \in [1,2)$ and $f(x) = 0, \forall x \in \mathbb{R}$. And we can easily check that these two functions work. :)
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TestX01
341 posts
#148 • 2 Y
Y by MihaiT, cubres
ily >_<

Lets take $x=0$. Then, $f(0)=f(0)\lfloor f(y)\rfloor$. If $f(0)\neq 0$, then $\lfloor f(y) \rfloor=1$. If so, $f(\lfloor x \rfloor y)=f(x)$. Fix $x>1434$. Then $y$ times a scalar not zero gives all reals, hence $f$ is constant.

Now, if $f(0)=0$, then take $0\leq x<1$, so $0=f(x)\lfloor f(y)\rfloor$. We vary $y$ over all reals now. If at any point $\lfloor f(y)\rfloor\neq 0$, then pick that $y$, and thus $f(x)=0$ for all $0\leq x<1$. Now, in the initial equation let $0\leq y<1$, so $f(y)=0$, and $f(\lfloor x\rfloor y)=0$. Now say we want to achieve $z$ with $\lfloor x\rfloor y$. If $z>0$, then make $\lfloor x\rfloor>z$, and scale down by $y$. If $z<0$ then make $\lfloor x \rfloor < z$ and scale down by $y$. Of course $f(0)=0$ already so nothing is to be done at $0$. Hence $f$ is simply $0$.

Else, $\lfloor f(y)\rfloor=0$ for all $y$. Then, $f(\lfloor x\rfloor y)=0$. Now, fix $x>1+4+3+4$, and vary $y$ to get all reals, so $f$ is zero again.

Now, if $f$ was just constant, then we have $c=c\lfloor c\rfloor$. Thus, $c=0$ or $\lfloor c\rfloor =1$. This means that
$f(x)=0$ or $f(x)=c$, where $1\leq c<2$.
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blueprimes
356 posts
#149 • 1 Y
Y by cubres
We claim the only solutions are $f(x) \equiv c$ where $c = 0$ or $\lfloor c \rfloor = 1$, which obviously work. Now we prove they are the only solutions, let $x = y = 0$, we obtain $f(0) = f(0) \lfloor f(0) \rfloor$. We now consider cases:

$\textbf{Case 1: }$ $\lfloor f(0) \rfloor = 1$
Consider $y = 0$, we obtain $f(0) = f(x)$ so $f$ is constant. Clearly the only function that works is $f(x) \equiv c$ where $\lfloor c \rfloor = 1$.

$\textbf{Case 2: }$ $f(0) = 0$.
Asserting $x = y = 1$ yields $f(1) = f(1) \lfloor f(1) \rfloor$. If $f(1) = 0$, then $x = 1$ gives $f(y) = 0$ for all $y$ so $f(x) \equiv 0$. On the other hand, if $\lfloor f(1) \rfloor = 1$ then $y = 1$ implies $f( \lfloor x \rfloor) = f(x)$. So $(x, y) = \left(2, \dfrac{1}{2} \right)$ gives
\[ f(1) = f(2) f \left(\dfrac{1}{2} \right) = f(2) f(0) = 0 \]which fails.

Altogether our only solutions are $f(x) \equiv c$ where $c = 0$ or $\lfloor c \rfloor = 1$ as wanted.
This post has been edited 1 time. Last edited by blueprimes, Feb 28, 2025, 1:25 PM
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Marcus_Zhang
980 posts
#150 • 1 Y
Y by cubres
Finished an IMO problem for IMO 2010 - the year I was born yay

Storage
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Ilikeminecraft
658 posts
#151 • 1 Y
Y by cubres
I claim the answer is $f(x) = c,$ where $c\in\{0\}\cup[0,1)$ is a constant.
Let $P(x, y)$ denote our assertion. From $P(0, x),$ we see that $f(0) = f(0)\lfloor f(y)\rfloor.$ There exist 2 cases from here, if $f(0) = 0$ or if $1\leq f(y) < 2\forall y\in\mathbb R.$

We handle $f(0) = 0$ first. From $P(x, y)$ where $\lfloor x \rfloor = 0,$ we see that $0 = f(x)\lfloor f(y) \rfloor.$ From here, notice that in both cases, $\lfloor f(x) \rfloor = 0.$ Now, take $P(x, y)$ where $0 < y < 1.$ We see that $f(\lfloor x \rfloor y) = 0,$ and hence, for all $x\in\mathbb R_{\geq0},$ $f(x) = 0.$ Finally, take $P(-1, y)$ for $y > 0$ to see that $f(-y) = 0.$

Now, look at when $1\leq f(y) < 2$ for all $y.$ From $P(1, y),$ we see that $f(y) = f(1)\lfloor f(y)\rfloor = f(1).$ Thus, we have that $f(x)$ is constant.
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lpieleanu
3001 posts
#152 • 2 Y
Y by cubres, lrjr23
Solution
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chenghaohu
72 posts
#153
Y by
$P(0, y) => f(0) = f(0)\cdot \lfloor f(y) \rfloor$. This produces two cases. Either $f(0) = 0$ or $\lfloor f(y) \rfloor = 1$ for all reals $y$. Now we continue with the two cases.

Case 1: If $f(0) = 0$

Let $0< x< 1$, $P(x, y) => 0 = f(x) \cdot \lfloor f(y)\rfloor$. This gives that $f(x) = 0$ for all $0 < x < 1$, or for some $0 < a < 1$, $f(a) \neq 0$ and $\lfloor f(y) \rfloor = 0$ for all reals $y$. We split this into two more subcases.

Subcase 1: $f(x) = 0$ for all $0< x < 1$

Let $0 < a < 1$, $P(x, a) => f(\lfloor x \rfloor a) = 0$ for all reals $x$. Varying the value of $x$ and $a$, we see that all reals can be represented as $\lfloor x \rfloor a$ for some values of $x$ and $a$. Therefore, this case gives that $f(x) \equiv 0$ as its only solution.

Subcase 2: For some $0 < a < 1$, $f(a) \neq 0$ and $\lfloor f(y) \rfloor = 0$ for all reals $y$.

$P(x, y) => f(\lfloor x \rfloor y) = 0$ for all reals $x$ and $y$ again, forcing $f(x) \equiv 0$. However, this violates the condition that for at least one value of $x$, $f(x) \neq 0$. Therefore this subcase produces no solutions.

Case 2: $\lfloor f(y) \rfloor = 1$ for all reals $y$

$P(1, y) => f(y) = f(1) \cdot \lfloor f(y) \rfloor = f(1)$. Therefore, $f(x) \equiv c$ for some $1\le c < 2$ are the solutions from this case.

It is easy to see that all solutions we obtained works. Therefore, our solution set of $f(x)$ is $f(x) \equiv 0$ and $f(x) \equiv c$, for $c$ a real constant such that $1\le c < 2$, and we are done with the problem.
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maromex
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Let $0 \le x < 1$, we get $f(0) = f(x)\lfloor f(y) \rfloor$ for all such $x, y$. We can vary $y$ freely. This implies that, either $\lfloor f(y) \rfloor$ is constant, or $f(x) = 0$ for all $x \in [0, 1)$.

Case 1: Have $\lfloor f(y) \rfloor = c_0$. Then we get $f(\lfloor x \rfloor y) = c_0f(x).$ Varying $y$ and fixing $x$ where $\lfloor x \rfloor \neq 0$ will tell us that $f$ is constant. Let's check constant solutions. We see and check that $c = c\lfloor c \rfloor$ has solutions: $c = 0$ or $\lfloor c \rfloor = 1$.

Case 2: Have $f(x) = 0$ for all $x \in [0, 1)$. Then, let $y \in [0, 1)$, so we get $f(\lfloor x \rfloor y) = 0$. Let $x$ be arbitrarily large and vary $y$, we get $f(a) = 0$ for all $a \in [0, x)$. Let $x$ be arbitrarily small and vary $y$, we get $f(a) = 0$ for all $a \in (x, 0]$. Therefore, $f(x) = 0$ for all $x$.

So the only solutions turn out to be the constant solutions $f(x) = c$ where $c = 0$ or $1 \le c < 2$.
This post has been edited 2 times. Last edited by maromex, May 15, 2025, 4:07 PM
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