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Bonza functions
KevinYang2.71   65
N 43 minutes ago by numbertheory97
Source: 2025 IMO P3
Let $\mathbb{N}$ denote the set of positive integers. A function $f\colon\mathbb{N}\to\mathbb{N}$ is said to be bonza if
\[ f(a)~~\text{divides}~~b^a-f(b)^{f(a)} \]for all positive integers $a$ and $b$.

Determine the smallest real constant $c$ such that $f(n)\leqslant cn$ for all bonza functions $f$ and all positive integers $n$.

Proposed by Lorenzo Sarria, Colombia
65 replies
KevinYang2.71
Jul 15, 2025
numbertheory97
43 minutes ago
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IMO ShortList 1998, algebra problem 1
orl   41
N 43 minutes ago by lpieleanu
Source: IMO ShortList 1998, algebra problem 1
Let $a_{1},a_{2},\ldots ,a_{n}$ be positive real numbers such that $a_{1}+a_{2}+\cdots +a_{n}<1$. Prove that

\[ \frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}. \]
41 replies
orl
Oct 22, 2004
lpieleanu
43 minutes ago
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Isogonal lines
Giahuytls2326   5
N an hour ago by Giahuytls2326
Source: my cute teacher
Given a triangle \( ABC \) inscribed in a circle \( (O) \), let the altitudes \( AD, BE, CF \) concur at the orthocenter \( H \). Let \( X \) be the circumcenter of triangle \( DEF \) , and let \( Y \) be the circumcenter of triangle \( BOC \). Let \( U \) and \( V \) be the two intersection points of the circles centered at \( X \) and \( Y \). Prove that \( AU \) and \( AV \) are isogonal lines with respect to \( \angle BAC \).
5 replies
Giahuytls2326
6 hours ago
Giahuytls2326
an hour ago
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IMO Shortlist 2012, Combinatorics 1
lyukhson   77
N an hour ago by happypi31415
Source: IMO Shortlist 2012, Combinatorics 1
Several positive integers are written in a row. Iteratively, Alice chooses two adjacent numbers $x$ and $y$ such that $x>y$ and $x$ is to the left of $y$, and replaces the pair $(x,y)$ by either $(y+1,x)$ or $(x-1,x)$. Prove that she can perform only finitely many such iterations.

Proposed by Warut Suksompong, Thailand
77 replies
lyukhson
Jul 29, 2013
happypi31415
an hour ago
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Cyclic sum of 1/((3-c)(4-c))
v_Enhance   25
N an hour ago by lpieleanu
Source: ELMO Shortlist 2013: Problem A6, by David Stoner
Let $a, b, c$ be positive reals such that $a+b+c=3$. Prove that \[18\sum_{\text{cyc}}\frac{1}{(3-c)(4-c)}+2(ab+bc+ca)\ge 15. \]Proposed by David Stoner
25 replies
v_Enhance
Jul 23, 2013
lpieleanu
an hour ago
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Junior Balkan Mathematical Olympiad 2024- P2
Lukaluce   19
N an hour ago by lendsarctix280
Source: JBMO 2024
Let $ABC$ be a triangle such that $AB < AC$. Let the excircle opposite to A be tangent to the lines $AB, AC$, and $BC$ at points $D, E$, and $F$, respectively, and let $J$ be its centre. Let $P$ be a point on the side $BC$. The circumcircles of the triangles $BDP$ and $CEP$ intersect for the second time at $Q$. Let $R$ be the foot of the perpendicular from $A$ to the line $FJ$. Prove that the points $P, Q$, and $R$ are collinear.

(The excircle of a triangle $ABC$ opposite to $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)

Proposed by Bozhidar Dimitrov, Bulgaria
19 replies
Lukaluce
Jun 27, 2024
lendsarctix280
an hour ago
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Inequality
SunnyEvan   5
N an hour ago by JARP091
Let $ a,b,c \in R $ such that: $ abc>0 $ and $a^2+b^2+c^2=4(ab+bc+ca)$
Prove that: $$\frac{abc(a^3+b^3+c^3+ab(a+b)+bc(b+c)+ca(c+a))}{730a^2b^2c^2+k((a-b)(b-c)(c-a))^2} \leq \frac{2}{\sqrt{\frac{730}{3}k-9k^2}-3k} $$Where $ k\in(0,\frac{365(2-\sqrt2)}{54}]. $


$$ \frac{abc(a^3+b^3+c^3+ab(a+b)+bc(b+c)+ca(c+a))}{730a^2b^2c^2+k((a-b)(b-c)(c-a))^2} \leq \frac{18(\sqrt2+1)}{365} $$Where $ k\in[\frac{365(2-\sqrt2)}{54}, +\infty). $


$$ \frac{abc(a^3+b^3+c^3+ab(a+b)+bc(b+c)+ca(c+a))}{730a^2b^2c^2+k((a-b)(b-c)(c-a))^2} \geq \frac{18(\sqrt2+1)}{365} $$Where $ k\in(-\infty ,0). $
5 replies
SunnyEvan
Jul 1, 2025
JARP091
an hour ago
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Number theory
truongngochieu   1
N an hour ago by truongngochieu
Prove that \[\tau\big(\varphi(n)\big) \geq \varphi\big(\tau(n)\big)\]with all integer n.
1 reply
truongngochieu
Yesterday at 5:06 AM
truongngochieu
an hour ago
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sumcif..
teomihai   1
N an hour ago by Royal_mhyasd
Let $a=123456789^{123456789}$ ,$a_{1}=sumcif\{a\}$ ,$a_{2}=sumcifa\{1\}$...
Find number $a_{k}$ with one digit.
1 reply
teomihai
2 hours ago
Royal_mhyasd
an hour ago
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2024 SL C5
Twoisaprime   7
N an hour ago by NTguy
Source: 2024 IMO Shortlist C5
Let $N$ be a positive integer. Geoff and Ceri play a game in which they start by writing the numbers $1, 2, \dots, N$ on a board. They then take turns to make a move, starting with Geoff. Each move consists of choosing a pair of integers $(k, n)$, where $k \geq 0$ and $n$ is one of the integers on the board, and then erasing every integer $s$ on the board such that $2^k \mid n - s$. The game continues until the board is empty. The player who erases the last integer on the board loses.

Determine all values of $N$ for which Geoff can ensure that he wins, no matter how Ceri plays.
7 replies
Twoisaprime
Jul 16, 2025
NTguy
an hour ago
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Function equation
LeDuonggg   6
N May 2, 2025 by MathLuis
Find all functions $f: \mathbb{R^+} \rightarrow \mathbb{R^+}$ , such that for all $x,y>0$:
\[ f(x+f(y))=\dfrac{f(x)}{1+f(xy)}\]
6 replies
LeDuonggg
May 1, 2025
MathLuis
May 2, 2025
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Function equation
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LeDuonggg
3 posts
LeDuonggg
#1 h  May 1, 2025, 2:59 PM • 3 Y
Y by User141208, MathLuis, cubres
Find all functions $f: \mathbb{R^+} \rightarrow \mathbb{R^+}$ , such that for all $x,y>0$:
\[ f(x+f(y))=\dfrac{f(x)}{1+f(xy)}\]
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luutrongphuc
68 posts
luutrongphuc
#3 h  May 2, 2025, 2:41 AM • 1 Y
Y by cubres
Can anyone check my solution, pls
Let $f(1)=t$
Suppose there exist $a>b$ such that $f(a)=f(b)$. Let $T=\dfrac{a}{b} >1$
$P(x,a)+P(x,b): f(xa)=f(xb) \rightarrow f(xT)=f(x), \forall x \in \mathbb{R}^+$
By induction, we have: $f\left(xT^k\right)=f(x)$. Fix $x$, let $k \to \infty$, we have $f(x)=0$(contrdiction)
So that, $f$ is injective $(1)$
Suppose there exist $u>v$ such that $f(u) >f(v)$
$P\left(\dfrac{x}{y},y\right):f\left(\dfrac{x}{y}+f(y)\right)=\dfrac{f\left(\dfrac{x}{y}\right)}{1+f(x)}$
Let $x_0=uv. \dfrac{f(u)-f(v)}{u-v}$
Put $x=x_0,y=u$ in $(2)$: $f\left(v. \dfrac{f(u)-f(v)}{u-v}+f(u)\right)=\dfrac{f\left(v. \dfrac{f(u)-f(v)}{u-v}\right)}{1+f(x_0)}$
Put $x=x_0,y=v$ in $(2)$: $f\left(u. \dfrac{f(u)-f(v)}{u-v}+f(v)\right)=\dfrac{f\left(u. \dfrac{f(u)-f(v)}{u-v}\right)}{1+f(x_0)}$
So that: $f\left(v. \dfrac{f(u)-f(v)}{u-v}\right)=f\left(u. \dfrac{f(u)-f(v)}{u-v}\right)$
Since $f$ is injective, we have contradiction.
Therefore, $\forall x>y$, we have $f(x) \le f(y)$. From $(1)$, we have: $\forall x>y$, we have $f(x) < f(y) (2)$
$(1)$ and $(2)$ give us $f$ is surjective.
$P(1,y): f(1+f(y))=\dfrac{t}{1+f(y)} \rightarrow f(x)= \dfrac{t}{x}, \forall x>1$
Fix $y\le1$. Then there exist $x>1$ such that $xy>1$
$P(x,y): \dfrac{t}{x+f(y)}=\dfrac{t}{x+\dfrac{t}{y}} \rightarrow f(x) =\dfrac{t}{x}, \forall x \le 1$
Therefore, all solution to this problem are:
$$\boxed{f(x)=\dfrac{t}{f(x)}, \forall x \in \mathbb{R}^+}$$
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jasperE3
11446 posts
jasperE3
#4 h  May 2, 2025, 4:02 AM • 1 Y
Y by cubres
luutrongphuc wrote:
Can anyone check my solution, pls
Let $f(1)=t$
Suppose there exist $a>b$ such that $f(a)=f(b)$. Let $T=\dfrac{a}{b} >1$
$P(x,a)+P(x,b): f(xa)=f(xb) \rightarrow f(xT)=f(x), \forall x \in \mathbb{R}^+$
By induction, we have: $f\left(xT^k\right)=f(x)$. Fix $x$, let $k \to \infty$, we have $f(x)=0$(contrdiction)
So that, $f$ is injective $(1)$
Suppose there exist $u>v$ such that $f(u) >f(v)$
$P\left(\dfrac{x}{y},y\right):f\left(\dfrac{x}{y}+f(y)\right)=\dfrac{f\left(\dfrac{x}{y}\right)}{1+f(x)}$
Let $x_0=uv. \dfrac{f(u)-f(v)}{u-v}$
Put $x=x_0,y=u$ in $(2)$: $f\left(v. \dfrac{f(u)-f(v)}{u-v}+f(u)\right)=\dfrac{f\left(v. \dfrac{f(u)-f(v)}{u-v}\right)}{1+f(x_0)}$
Put $x=x_0,y=v$ in $(2)$: $f\left(u. \dfrac{f(u)-f(v)}{u-v}+f(v)\right)=\dfrac{f\left(u. \dfrac{f(u)-f(v)}{u-v}\right)}{1+f(x_0)}$
So that: $f\left(v. \dfrac{f(u)-f(v)}{u-v}\right)=f\left(u. \dfrac{f(u)-f(v)}{u-v}\right)$
Since $f$ is injective, we have contradiction.
Therefore, $\forall x>y$, we have $f(x) \le f(y)$. From $(1)$, we have: $\forall x>y$, we have $f(x) < f(y) (2)$
$(1)$ and $(2)$ give us $f$ is surjective.
$P(1,y): f(1+f(y))=\dfrac{t}{1+f(y)} \rightarrow f(x)= \dfrac{t}{x}, \forall x>1$
Fix $y\le1$. Then there exist $x>1$ such that $xy>1$
$P(x,y): \dfrac{t}{x+f(y)}=\dfrac{t}{x+\dfrac{t}{y}} \rightarrow f(x) =\dfrac{t}{x}, \forall x \le 1$
Therefore, all solution to this problem are:
$$\boxed{f(x)=\dfrac{t}{f(x)}, \forall x \in \mathbb{R}^+}$$

why is $f$ surjective? there are injective and strictly increasing functions that aren't surjective (note that strictly increasing implies injective), like x/(x+1)?
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luutrongphuc
68 posts
luutrongphuc
#5 h  May 2, 2025, 6:05 AM • 1 Y
Y by cubres
jasperE3 wrote:
luutrongphuc wrote:
Can anyone check my solution, pls
Let $f(1)=t$
Suppose there exist $a>b$ such that $f(a)=f(b)$. Let $T=\dfrac{a}{b} >1$
$P(x,a)+P(x,b): f(xa)=f(xb) \rightarrow f(xT)=f(x), \forall x \in \mathbb{R}^+$
By induction, we have: $f\left(xT^k\right)=f(x)$. Fix $x$, let $k \to \infty$, we have $f(x)=0$(contrdiction)
So that, $f$ is injective $(1)$
Suppose there exist $u>v$ such that $f(u) >f(v)$
$P\left(\dfrac{x}{y},y\right):f\left(\dfrac{x}{y}+f(y)\right)=\dfrac{f\left(\dfrac{x}{y}\right)}{1+f(x)}$
Let $x_0=uv. \dfrac{f(u)-f(v)}{u-v}$
Put $x=x_0,y=u$ in $(2)$: $f\left(v. \dfrac{f(u)-f(v)}{u-v}+f(u)\right)=\dfrac{f\left(v. \dfrac{f(u)-f(v)}{u-v}\right)}{1+f(x_0)}$
Put $x=x_0,y=v$ in $(2)$: $f\left(u. \dfrac{f(u)-f(v)}{u-v}+f(v)\right)=\dfrac{f\left(u. \dfrac{f(u)-f(v)}{u-v}\right)}{1+f(x_0)}$
So that: $f\left(v. \dfrac{f(u)-f(v)}{u-v}\right)=f\left(u. \dfrac{f(u)-f(v)}{u-v}\right)$
Since $f$ is injective, we have contradiction.
Therefore, $\forall x>y$, we have $f(x) \le f(y)$. From $(1)$, we have: $\forall x>y$, we have $f(x) < f(y) (2)$
$(1)$ and $(2)$ give us $f$ is surjective.
$P(1,y): f(1+f(y))=\dfrac{t}{1+f(y)} \rightarrow f(x)= \dfrac{t}{x}, \forall x>1$
Fix $y\le1$. Then there exist $x>1$ such that $xy>1$
$P(x,y): \dfrac{t}{x+f(y)}=\dfrac{t}{x+\dfrac{t}{y}} \rightarrow f(x) =\dfrac{t}{x}, \forall x \le 1$
Therefore, all solution to this problem are:
$$\boxed{f(x)=\dfrac{t}{f(x)}, \forall x \in \mathbb{R}^+}$$

why is $f$ surjective? there are injective and strictly increasing functions that aren't surjective (note that strictly increasing implies injective), like x/(x+1)?
Oh, tks i will fix it
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mashumaro
46 posts
mashumaro
#6 h  May 2, 2025, 11:48 AM • 2 Y
Y by cubres, navi_09220114
luutrongphuc wrote:
Fix $x$, let $k \to \infty$, we have $f(x)=0$(contrdiction)
Why?

Anyway,
1. You can prove injective by using this sub instead.
2. You can prove surjectivity using the following:
Claim 1: $\lim_{x\to+\infty} f(x) = 0$
Proof

Claim 2: $f$ is continuous
Proof

Claim 3: $\lim_{x\to 0^+} f(x) = +\infty$
Proof
Alternative proof by ja.

These 3 claims imply $f$ is surjective.
This post has been edited 1 time. Last edited by mashumaro, May 2, 2025, 2:05 PM
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luutrongphuc
68 posts
luutrongphuc
#7 h  May 2, 2025, 12:09 PM • 1 Y
Y by cubres
mashumaro wrote:
luutrongphuc wrote:
Fix $x$, let $k \to \infty$, we have $f(x)=0$(contrdiction)
Why?

Anyway,
1. You can prove injective by using this sub instead.
2. You can prove surjectivity using the following:
Claim 1: $\lim_{x\to+\infty} f(x) = 0$
Proof

Claim 2: $f$ is continuous
Proof

Claim 3: $\lim_{x\to 0^+} f(x) = +\infty$
Proof

These 3 claims imply $f$ is surjective.
Thank you for spotting my mistake.
I tried to fix it but failed
This post has been edited 2 times. Last edited by luutrongphuc, May 2, 2025, 12:20 PM
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MathLuis
1600 posts
MathLuis
#8 h  May 2, 2025, 8:13 PM • 3 Y
Y by cubres, mashumaro, LeDuonggg
Denote $P(x,y)$ the assertion of the given F.E. (New color pog)
Claim 1: $f$ is injective.
Proof: Suppose FTSOC there existed $a>b$ with $f(a)=f(b)$ then from $P(x,a)-P(x,b)$ we get that $P(ax)=f(bx)$ and thus $f(x)=f \left( \left(\frac{a}{b} \right)^k x \right)$ for all $k \in \mathbb Z$ and so $P \left(\frac{bx}{a}, y \right)$ gives $f \left(\frac{bx}{a}+f(y) \right)=\frac{f(x)}{1+f(xy)}$ but here now we can let $x=\frac{af(y)}{a-b}$ to get that $f \left(\frac{ayf(y)}{a-b} \right)=0$ which is of course a contradiction, therefore $f$ is injective as desired.
Claim 2: $f$ is strictly decreasing.
Proof: Suppose FTSOC there existed $m>n$ with $f(m)>f(n)$ then from $P \left(\frac{m(f(m)-f(n)}{m-n}, n \right)-P \left(\frac{n(f(m)-f(n)}{m-n}, m \right)$ we get that $m=n$ are cancelling everything and using Claim 1 and thus a contradiction, meaning $f$ is indeed strictly decreasing as $f(n)=f(m)$ can't happen.
Claim 3: $\lim_{n \to \infty} f(a_n)=0$ for all sequences $a_n>nf(1)$ for all $n>N$.
Proof: Notice $P(x,1)$ gives $f(x+f(1))=\frac{f(x)}{1+f(x)}$ and inductive here using the jump $x \to x+f(1)$ gives that $f(x+nf(1))=\frac{f(x)}{nf(x)+1}$ for all $n \in \mathbb Z_{>0}$ and obviously taking $n \to \infty$ is sufficient to prove the claim (the more relevant part is to see how it decreases at a slow but sure pace, which we will use next).
Claim 4: $f$ is continuos.
Proof: Fix $x$ and consider $c_n>\max \{nf(1), xnf(1) \}$ for all $n>N$ then by $P(x,c_n)$ and setting $n \to \infty$ makes it clear why this claim holds true, as the inside of LHS becomes closer and closer to $x$ while on the RHS we can see how the whole thing is closer to $f(x)$ so by epsilon-delta definition the claim is true to the right. Now to check its continuos to the left again fix a constant $x$ the same sequence $c_n$ but in addition we instead consider all $n>M$ for which $x-f(c_n)>C$ for some constant $x>C$ and thus $P(x-f(c_n), c_n)$ and the fact that $f(Cc_n)$ will still become arbitrarily small eventually will give from $n \to \infty$ that $f$ is also continuos to the left, finish the claim from epsilon-delta definition.
Claim 5: $f$ is unbounded.
Proof: Suppose $L=\lim_{x \to 0^+} f(x)$ was bounded then on $P(x,y)$ for taking $x \to 0^+$ and a fixed $y$ gives that $f(f(y))=\frac{L}{L+1}$ however $f$ is injective so this can't happen, thus proving the claim.
Claim 6: $f$ is surjective and $f(x)=\frac{f(1)}{x}$ for all $x>1$.
Proof: $f$ is surjective follows from Claims 2,3,4,5 (mainly continuity carrying lol) and now from $P(1,x)$ and our surjectivity we have that $f(1+t)=\frac{f(1)}{1+t}$ for all $t>0$ as desired.
The finish: Now fix some $x$ then taking the identity on Claim 3 for some large enough $nf(1)>1$ we have that $\frac{f(x)}{nf(x)+1}=\frac{1}{x+nf(1)}$ now if $x>1$ then we have that $\frac{1}{n+x}=\frac{1}{x+nf(1)}$ so $f(1)=1$ and now setting $0<x<1$ would give that $nf(x)+1=xf(x)+nf(1)$ so $xf(x)=1$ and thus combining all this gives $f(x)=\frac{1}{x}$ for all $x \in \mathbb R_{>0}$ thus we are done :cool:.
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