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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

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0 replies
jwelsh
Jul 1, 2025
0 replies
Putnam 2002 A3: average of the elements of a set is integer
waver123   11
N 4 hours ago by dolphinday
Source: 0
Let $N$ be an integer greater than $1$ and let $T_n$ be the number of non empty subsets $S$ of $\{1,2,.....,n\}$ with the property that the average of the elements of $S$ is an integer.Prove
that $T_n - n$ is always even.
11 replies
waver123
Nov 2, 2011
dolphinday
4 hours ago
iterative functional equation
kyj050330   0
4 hours ago
does there exist a differentiable function such that for all real x
f(f(x))+f(x)=x^3


recursive formula for power series of f(x) is obtainable, but chances are that the radius of convergence is 0.
or is there another way to prove only the existence of such function without explicit form solution?
0 replies
kyj050330
4 hours ago
0 replies
Functional equation with conditions.
Synchrone   1
N 6 hours ago by paxtonw
Source: Math&Maroc Competition 2025 Day2 Problem 7
Let $f : \mathbb{N}_{\geq 1} \to \mathbb{R}_{\geq 0}$ be a function verifying :
- $\forall a,b \geq 1, f(ab) = f(a) + f(b)$
- $\exists n \geq 2,  f(n) = 0$
$1.$ Assume that $f(n+1) - f(n) \to_{n \to + \infty} 0$. Show that $\forall x \in \mathbb{N}_{\geq 1}$, $f(x) = 0$
$2.$ Assume that $\exists A > 0$, $\forall a,b \geq 1$, $|f(a) - f(b)| \leq A|a-b|$. Show that $\forall x \in \mathbb{N}_{\geq 1}$, $f(x) = 0$
1 reply
Synchrone
Today at 3:28 PM
paxtonw
6 hours ago
A nice identity
Synchrone   2
N Today at 3:38 PM by GreenKeeper
Source: Math&Maroc Competition 2025 Day1 Problem 3
For pairwise-distinct real numbers $a_1, \ldots, a_n$ prove that :
$$ \sum_{i =1}^n a_j^2 \prod_{k \neq j} \frac{a_j + a_k}{a_j - a_k} = (a_1 + \ldots + a_n)^2 $$
2 replies
Synchrone
Yesterday at 1:18 PM
GreenKeeper
Today at 3:38 PM
No more topics!
ISI UGB 2025 P1
SomeonecoolLovesMaths   11
N Jul 13, 2025 by Hello_Kitty
Source: ISI UGB 2025 P1
Suppose $f \colon \mathbb{R} \longrightarrow \mathbb{R}$ is differentiable and $| f'(x)| < \frac{1}{2}$ for all $x \in \mathbb{R}$. Show that for some $x_0 \in \mathbb{R}$, $f \left( x_0 \right) = x_0$.
11 replies
SomeonecoolLovesMaths
May 11, 2025
Hello_Kitty
Jul 13, 2025
ISI UGB 2025 P1
G H J
G H BBookmark kLocked kLocked NReply
Source: ISI UGB 2025 P1
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SomeonecoolLovesMaths
3455 posts
#1
Y by
Suppose $f \colon \mathbb{R} \longrightarrow \mathbb{R}$ is differentiable and $| f'(x)| < \frac{1}{2}$ for all $x \in \mathbb{R}$. Show that for some $x_0 \in \mathbb{R}$, $f \left( x_0 \right) = x_0$.
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alexheinis
10767 posts
#2 • 3 Y
Y by mqoi_KOLA, SatisfiedMagma, Alphaamss
Standard Banach contraction principle. Take $a_0\in R$ and define $a_{n+1}=f(a_n)$.
Then $|a_{n+2}-a_{n+1}|=|f(a_{n+1})-f(a_n)|\le 1/2|a_{n+1}-a_n|$ for all $n$ hence $|a_{k+1}-a_k|\le {{|a_1-a_0|}\over {2^k}}$ for all $k\ge 0$.
Hence $\sum_0^\infty (a_{k+1}-a_k)$ is absolutely convergent and $\sum_0^n (a_{k+1}-a_k)=a_{n+1}-a_0$ converges.
Hence $a_n\rightarrow a\in R$. Then $a_{n+1}=f(a_n)\rightarrow f(a)$ and $f(a)=a$.
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827681
163 posts
#3
Y by
SomeonecoolLovesMaths wrote:
Suppose $f \colon \mathbb{R} \longrightarrow \mathbb{R}$ is differentiable and $| f'(x)| < \frac{1}{2}$ for all $x \in \mathbb{R}$. Show that for some $x_0 \in \mathbb{R}$, $f \left( x_0 \right) = x_0$.

Take $h(x)=f(x)-x \implies \frac{-3}{2} < h'(x)<\frac{-1}{2}$
Now just apply lmvt to get $\frac{-3x}{2}+h(0) <h(x) <\frac{-x}{2}+h(0)$
Shift $x$ suitably and win.
This post has been edited 1 time. Last edited by 827681, May 11, 2025, 2:10 PM
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SomeonecoolLovesMaths
3455 posts
#4 • 1 Y
Y by Shiny_zubat
Define $g(x) = x - f(x)$. Note that $g$ is differentiable.
$g'(x) = 1 - f'(x)$ $\forall x \in \mathbb{R}$, so $\frac{3}{2} > g'(x) > \frac{1}{2}$ $\forall x \in \mathbb{R}$.
Appying LMVT on the functions $g$, $\forall x \in \mathbb{R}$, $\exists c$ such that $$g(x) - g(0) = g'(c) ( x - 0)$$Observe that, $\frac{x}{2} + g(0) < xg'(c) + g(0) = g(x) = xg'(c) + g(0) < \frac{3x}{2} + g(0)$.
For $x = \frac{-2g(0)}{3} - 1$, $g(x) < 0$.
For $x = -2g(0) +1$, $g(x) >0$.
Applying IVT on $h$, $\exists x_0 \in \mathbb{R}$ such that $g(x_0) = 0 = x_0 - f(x_0)$.
Giving us $\boxed{f(x_0) = x_0}$.
This post has been edited 1 time. Last edited by SomeonecoolLovesMaths, May 12, 2025, 5:10 PM
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ZeroAlephZeta
32 posts
#5 • 1 Y
Y by quasar_lord
Calculus was too easy this year
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Shiny_zubat
8 posts
#6
Y by
SomeonecoolLovesMaths wrote:
Define $g(x) = x - f(x)$. Note that $g$ is differentiable.
$g'(x) = 1 - f'(x)$ $\forall x \in \mathbb{R}$, so $\frac{3}{2} > g'(x) > \frac{1}{2}$ $\forall x \in \mathbb{R}$.
Appying LMVT on the functions $g$, $\forall x \in \mathbb{R}$, $\exists c$ such that $$g(x) - g(0) = g'(c) ( x - 0)$$Observe that, $\frac{x}{2} + g(0) < xg'(c) + g(0) = f(x) = xg'(c) + g(0) < \frac{3x}{2} + g(0)$.
For $x = \frac{-2g(0)}{3} - 1$, $g(x) < 0$.
For $x = -2g(0) +1$, $g(x) >0$.
Applying IVT on $h$, $\exists x_0 \in \mathbb{R}$ such that $g(x_0) = 0 = x_0 - f(x_0)$.
Giving us $\boxed{f(x_0) = x_0}$.

the Click to reveal hidden text
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SomeonecoolLovesMaths
3455 posts
#7 • 1 Y
Y by Shiny_zubat
Shiny_zubat wrote:
SomeonecoolLovesMaths wrote:
Define $g(x) = x - f(x)$. Note that $g$ is differentiable.
$g'(x) = 1 - f'(x)$ $\forall x \in \mathbb{R}$, so $\frac{3}{2} > g'(x) > \frac{1}{2}$ $\forall x \in \mathbb{R}$.
Appying LMVT on the functions $g$, $\forall x \in \mathbb{R}$, $\exists c$ such that $$g(x) - g(0) = g'(c) ( x - 0)$$Observe that, $\frac{x}{2} + g(0) < xg'(c) + g(0) = f(x) = xg'(c) + g(0) < \frac{3x}{2} + g(0)$.
For $x = \frac{-2g(0)}{3} - 1$, $g(x) < 0$.
For $x = -2g(0) +1$, $g(x) >0$.
Applying IVT on $h$, $\exists x_0 \in \mathbb{R}$ such that $g(x_0) = 0 = x_0 - f(x_0)$.
Giving us $\boxed{f(x_0) = x_0}$.

the Click to reveal hidden text

Yes that was an error from my end @above, thanks for pointing that out! Corrected!
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SatisfiedMagma
464 posts
#8 • 1 Y
Y by L13832
I honestly adore the problem quality which ISI have given this year, but the only problem I really have is that these problems are "kinda" well-known. This problem can be found in Art-n-Craft book by Paul Zeitz, but with a more general case where $\frac 1 2$ is replaced any constant $k < 1$ and $f'(x) {\color{red} \le} k$. In the exercises, it is even discussed why $k < 1$ is chosen and why would we not write $f'(x) < 1$ directly. It's a really good problem and everyone should look for it. I'll present the same solution as in the book and I acknowledge this solution isn't mine.

Although, honestly speaking, Banach FIxed Point Theorem was the first thing which striked my mind, but even this works wonderfully!
Generalized Problem Statement wrote:
Suppose $f$ is differentiable on $(-\infty, \infty)$ and there is a constant $k < 1$ such that $f'(x) \le k$ for all $x \in \mathbb{R}$. Show that $f$ must have a fixed point.

Solution: Consider the intuitive auxillary function $g(x) \coloneqq f(x) - x$. If there are two points $a,b$ such that $g(a)g(b) < 0$, then by the Intermediate Value Theorem one can conclude that $g$ must have a root, which would finish on the spot. Therefore we would assume that $g$ is either strictly positive, or strictly negative. For our solution, WLOG assume that $g > 0$.

Apply Lagrange's Mean Value Theorem on $[0,x]$ on the function $f$ (take $x > 0$). This yields that there exists some constant $c \in (0,x)$ such that
\[ \frac{f(x)-f(0)}{x} = f'(c)\]Since $g > 0$, we must have $f(x) > x$. This yields
\[k \ge f'(c) > 1 - \frac{f(0)}{x}. \]Taking $x$ large enough, we would get that $1-\frac{f(0)}{x} > k$ which would be a contradiction. Therefore $g$ can't be positive everytime, which show that $g$ must have a root, as desired.

The other case is omitted from the book. For the other case, basically do the same argument from left side. This time we will assume $f(x) < x$ for all $x$ apply MVT on the interval $[-x,0]$ where $x > 0$ again. This time you should get
\[ -k \le -f'(c) < -1 - \frac{f(0)}{x}\]multiplying everything by $-1$ and taking large $x$ suffices. $\blacksquare$

Exercise
Solution to the Exercise
This post has been edited 1 time. Last edited by SatisfiedMagma, May 14, 2025, 12:32 PM
Reason: banach
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mqoi_KOLA
112 posts
#10
Y by
Source of ISI 2025 paper-
UGA- JEE stuff
UGB-
P1 – (Banach Fixed Point Theorem)

P2 – (ISI BST 2011 P10)

P3 – (Math StackExchange link)

P4 – (China MO 1989 P3)

P5 – (Page No. 30 Problem 1.95)

P6 – (APMO 2007 P1)

P7 –(Orthic Triangle)

P8 – (Theorem 6)
This post has been edited 2 times. Last edited by mqoi_KOLA, Jul 8, 2025, 5:09 AM
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mqoi_KOLA
112 posts
#11 • 1 Y
Y by aidan0626
fvck isi
bye
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Cats_on_a_computer
81 posts
#13
Y by
By the Fundamental Theorem of Calculus, for every \(x\in\Bbb R\),
\[
f(x)
= f(0) + \int_0^x f'(t)\,dt
\quad\Longrightarrow\quad
\bigl|f(x)-f(0)\bigr|
\le \int_0^x \bigl|f'(t)\bigr|\,dt
< \tfrac12\,|x|.
\]Hence for all \(x\),
\[
f(0) - \tfrac12\,|x|\;\le\; f(x)\;\le\; f(0) + \tfrac12\,|x|.
\]Define
\[
h(x)=f(x)-x,
\]which is continuous. We now check the end‐behavior:

As \(x\to -\infty\), we have \(|x|=-x\), so
\[
    f(x)\;\ge\;f(0)-\tfrac12(-x)
    = f(0)+\tfrac12\,x,
  \]and hence
\[
    h(x)=f(x)-x
    \;\ge\;f(0)+\tfrac12x - x
    = f(0)-\tfrac12x
    \;\longrightarrow+\infty.
  \]As \(x\to +\infty\), we have \(|x|=x\), so
\[
    f(x)\;\le\;f(0)+\tfrac12x,
  \]and hence
\[
    h(x)=f(x)-x
    \;\le\;f(0)+\tfrac12x - x
    = f(0)-\tfrac12x
    \;\longrightarrow-\infty.
  \]

Since \(h\) is continuous and changes sign, the Intermediate Value Theorem gives an \(x_0\in\Bbb R\) with
\[
h(x_0)=0,
\]i.e, \(f(x_0)=x_0\).
This post has been edited 1 time. Last edited by Cats_on_a_computer, Jul 8, 2025, 8:03 AM
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Hello_Kitty
1909 posts
#14 • 1 Y
Y by mqoi_KOLA
$1/2\leq f'\leq 1/2\Rightarrow~\forall x, \; f_0-x/2\leq f_x\leq f_0+x/2$.

if $\forall x, \; f_x>x$ or $\forall x, \; f_x<x$ then above contradiction, IVT's thm concludes.
This post has been edited 1 time. Last edited by Hello_Kitty, Jul 13, 2025, 1:33 AM
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