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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
Power tower sum
Rijul saini   9
N 12 minutes ago by ihategeo_1969
Source: India IMOTC 2024 Day 3 Problem 2
Let $a$ and $n$ be positive integers such that:
1. $a^{2^n}-a$ is divisible by $n$,
2. $\sum\limits_{k=1}^{n} k^{2024}a^{2^k}$ is not divisible by $n$.

Prove that $n$ has a prime factor smaller than $2024$.

Proposed by Shantanu Nene
9 replies
Rijul saini
May 31, 2024
ihategeo_1969
12 minutes ago
Nice "if and only if" function problem
ICE_CNME_4   7
N 15 minutes ago by ICE_CNME_4
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )

Please do it at 9th grade level. Thank you!
7 replies
ICE_CNME_4
Yesterday at 7:23 PM
ICE_CNME_4
15 minutes ago
Cauchy-Schwarz 2
prtoi   8
N 16 minutes ago by mrtheory
Source: Handout by Samin Riasat
if $a^2+b^2+c^2+d^2=4$, prove that:
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\ge4$
8 replies
prtoi
Mar 26, 2025
mrtheory
16 minutes ago
D1037 : The super irreductibilty
Dattier   1
N 33 minutes ago by Dattier
Source: les dattes à Dattier
Let $P \in \mathbb Q[x]$, $P$ is super irreductible if $\forall Q \in \mathbb [x], \deg(Q)>0, P(Q(x))$ is irreductible.

Are there some $P \in \mathbb Q[x]$ super irreductible?
1 reply
Dattier
2 hours ago
Dattier
33 minutes ago
No more topics!
Problem 4 (second day)
darij grinberg   93
N Apr 25, 2025 by Ilikeminecraft
Source: IMO 2004 Athens
Let $n \geq 3$ be an integer. Let $t_1$, $t_2$, ..., $t_n$ be positive real numbers such that \[n^2 + 1 > \left( t_1 + t_2 + \cdots + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \cdots + \frac{1}{t_n} \right).\] Show that $t_i$, $t_j$, $t_k$ are side lengths of a triangle for all $i$, $j$, $k$ with $1 \leq i < j < k \leq n$.
93 replies
darij grinberg
Jul 13, 2004
Ilikeminecraft
Apr 25, 2025
Problem 4 (second day)
G H J
Source: IMO 2004 Athens
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darij grinberg
6555 posts
#1 • 11 Y
Y by Adventure10, chessgocube, megarnie, fluff_E, Mango247, bjump, buddyram, cubres, and 3 other users
Let $n \geq 3$ be an integer. Let $t_1$, $t_2$, ..., $t_n$ be positive real numbers such that \[n^2 + 1 > \left( t_1 + t_2 + \cdots + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \cdots + \frac{1}{t_n} \right).\] Show that $t_i$, $t_j$, $t_k$ are side lengths of a triangle for all $i$, $j$, $k$ with $1 \leq i < j < k \leq n$.
Attachments:
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Peter Scholze
644 posts
#2 • 5 Y
Y by chessgocube, Adventure10, Illuzion, Mango247, cubres
solution by induction: the hardest part is to show that it works for n=3. so assume $t_1\geq t_2\geq t_3$. it is easy to see that the right is strictly increasing in $t_3$. so if we show that the reverse inequality holds for $t_1=t_2+t_3$, we are done. a quick calculation is enough to show that(it is too boring to post). but if we know is true for n-1, then let's do the following computation:
$((t_1+...+t_{n-1})+t_n)(\frac{1}{t_1}+...+\frac{1}{t_{n-1}}+\frac{1}{t_n})\geq (t_1+...+t_{n-1})(\frac{1}{t_1}+...+\frac{1}{t_{n-1}})+2n-1$,
by AM-GM and AM-HM. so we get the inequality reduces to that one for n-1 variables.

Peter
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pbornsztein
3005 posts
#3 • 5 Y
Y by anantmudgal09, Adventure10, chessgocube, Mango247, sabkx
It reminds me the following problem from China 1987/88 :

Let $n \geq 3$ be an integer. Suppose the inequality
$(a_1^2 + a_2^2 + ... + a_n^2)^2 > (n-1)(a_1^4 + ... a_n^4)$
holds for some positive real numbers $a_1,...a_n$. Prove that $a_i,a_j,a_k$ are the three sides of some triangle for all $i<j<k$.

Even there is an elegant direct solution, the most common way is induction.

Pierre.
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jmerry
12096 posts
#4 • 16 Y
Y by GammaBetaAlpha, Polynom_Efendi, ETS1331, Adventure10, oneteen11, Illuzion, Mango247, aidan0626, EquationTracker, RobertRogo, and 6 other users
When I saw this, my immediate thought was- an elegant direct solution.

The inequality
$n^2 + 1 > \left( t_1 + t_2 + ... + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + ... + \frac{1}{t_n} \right)$
reduces to $\displaystyle{n^2+1>\sum_{j=1}^n \sum_{k=1}^n{\frac{t_j}{t_k}}}$
$\displaystyle{n^2+1>n+\sum_{1 \le j<k \le n}{\frac{t_j}{t_k}+\frac{t_k}{t_j}}}$
Each of the terms in the last sum is at least 2, so if the sum of some $m$ terms is at least $2m+1$, the whole sum will be at least $n^2-n+1$ and the inequality will be violated.

Suppose $t_i+t_j \le t_k$. Then $\frac{t_i}{t_k}+\frac{t_j}{t_k} \le 1$
If we can show that $a+b+\frac{1}{a}+\frac{1}{b} \ge 5$ when $a+b \le 1$, we will be done by the argument above.
First, we have $a+\frac{1}{a}=\frac{a^2+1}{a} \ge \frac{(a-\frac{1}{2})^2+a+\frac{3}{4}}{a} \ge 1+\frac{3}{4a}$
and $b+\frac{1}{b} \ge 1+\frac{3}{4b}$
so $a+b+\frac{1}{a}+\frac{1}{b} \ge 2+\frac{3}{4}(\frac{1}{a}+\frac{1}{b})$
By AM-HM, $\frac{2}{\frac{1}{a}+\frac{1}{b}} \le \frac{a+b}{2} \le \frac{1}{2}$, so $\frac{1}{a}+\frac{1}{b} \ge 4$ and $a+b+\frac{1}{a}+\frac{1}{b} \ge 5$

In fact the case n=3 is sufficient to prove the desired fact for all n - we can simply pair off the other terms and only worry about the ones in our non-triangle.
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Kantor
21 posts
#5 • 3 Y
Y by GammaBetaAlpha, Adventure10, Mango247
it is, indeed, a very elegant solution jmerry!

Kantor
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harazi
5526 posts
#6 • 3 Y
Y by Adventure10, Mango247, jolynefag
Well, I just adore this problem. So easy, beautiful and elegant. But try to think about the following problem, which I think it's much more difficult and interesting:
For iven n>2, find the greatest constant k (which may depend on n) such that if positive reals $ x_1,...,x_n$ verify $ k>(x_1+...+x_n)(\frac{1}{x_1}+...+\frac{1}{x_n}) $ then any three of them are sides of a triangle. I think I've managed to solve this problem and if my solution is correct, then it's a really hard and nice problem.
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Anh Cuong
431 posts
#7 • 2 Y
Y by Adventure10, Mango247
If my computation is not wrong then it should be
n^2+(2\sqrt10-6)n+19-6\sqrt{10}
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harazi
5526 posts
#8 • 3 Y
Y by Adventure10, Mango247, jolynefag
Yes, it is $ (n+\sqrt{10}-3)^2  $. I think this one is much more interesting.
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Valentin Vornicu
7301 posts
#9 • 4 Y
Y by adityaguharoy, Adventure10, Mango247, jolynefag
btw, the creator of this problem is Hojoo Lee :)
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billzhao
827 posts
#10 • 13 Y
Y by threehandsnal, Mateescu Constantin, Wizard0001, Adventure10, Periwinkle27, Jasurbek, Illuzion, Mango247, ehuseyinyigit, Bluecloud123, FairyBlade, and 2 other users
Another solution:

Note that

\[ (t_1+ \cdots + t_n) ( \frac{1}{t_1} + \cdots  + \frac{1}{t_n}) = n^2 + \sum_{i<j} (\sqrt{\frac{t_i}{t_j}} - \sqrt{\frac{t_j}{t_i}} )^2 \]

And so if three of the terms satisfy $a \geq b + c$ then

\[{{ (\sqrt{a/b} - \sqrt{b/a})^2 + (\sqrt{a/c} - \sqrt{c/a})^2 =
\frac{ (a-b)^2}{ab} } + \frac{ (a-c)^2}{ac} } 
\geq \frac{(a-b)c}{ab} + \frac{(a-c)b}{ac}
= c/b + b/c -(b+c)/a \geq 2 - 1 = 1 \]

And the result follows immediately.

(During the IMO, I accidentally forgot the square root, costing me a huge chunk of marks).
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vinoth_90_2004
301 posts
#11 • 2 Y
Y by Adventure10, Mango247
just wanted to say that this is *really* nice ...
btw, anh cuong/harazi could you please post the solution to the harder version (where n^2+1 is replaced by ( n + :sqrt: 10 - 3 ) ^ 2 )? thanks.
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harazi
5526 posts
#12 • 2 Y
Y by Adventure10, Mango247
Of course, Hojoo Lee is the official proposer. But I knew the case n=3 some four months ago, when I tried to create a problem for the junior TST in Romania.
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Sung-yoon Kim
324 posts
#13 • 2 Y
Y by Adventure10, Mango247
It's my solution in the 2004 IMO.

By Cauchy-Schwartz inequality,

(t_1+t_2+...+t_n)(1/t_1+1/t_2+...+1/t_n)=(3\times(t_1+t_2+t_3)/3 +t_4+...+t_n)(3\times(1/t_1+1/t_2+1/t_3)/3 +1/t_4+...+1/t_n)\geq(3\times \sqrt((t_1+t_2+t_3)/3\times(1/t_1+1/t_2+1/t_3)/3) +1+...+1)^2


So
n^2+1>(\sqrt((t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3)) +n-3)^2


=>
(t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3)<(\sqrt(n^2+1) -(n-3))^2




As
n\geq3
, it follows

(\sqrt(n^2+1) -(n-3))^2<10  <=>  2n^2 -6n<(2n-6) \sqrt(n^2+1)


So
(t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3)<10




Let
t_1 \geq t_2 \geq t_3
. Then
t_1+t_2 \geq t_3, t_1+t_3 \geq t_2


Assume that
t_1>t_2+t_3
. Then
(t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3)-((t_2+t_3)+t_2+t_3)(1/(t_2+t_3) +1/t_2+1/t_3) =(t_1-t_2-t_3)(1/t_2+1/t_3-1/t_1) \geq 0


so by Cauchy-Schwartz inequality,

10>(t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3) \geq 2(t_2+t_3)(1/(t_2+t_3)+1/t_2+1/t_3)   <=>   4>(t_2+t_3)(1/t_2+1/t_3) \geq 4


and contradiction.
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heartwork
308 posts
#14 • 2 Y
Y by Adventure10, Mango247
harazi wrote:
...find the greatest constant k (which may depend on n) such that if positive reals
x_1,x_2,...,x_n
verify
k(n)>(x_1+x_2+...+x_n)(1/x_1+1/x_2+...+1/x_n)
then any three of them are sides of a triangle.
harazi wrote:
Yes, it is
(n+\sqrt{10}-3)^2
. I think this one is much more interesting.
Consider
f(x_1,x_2,...,x_n)=(x_1+x_2+...+x_n)(1/x_1+1/x_2+...+1/x_n)
where
x_1<=x_2<=...<=x_n
.

Greatest constant
k(n)=(n+\sqrt{10}-3)^2
might be found elementary. From the solution of Sung-yoon Kim:
Sung-yoon Kim wrote:
...By Cauchy-Schwartz inequality,
..........
So
n^2+1>(\sqrt((x_1+x_2+x_3)(1/x_1+1/x_2+1/x_3))+(n-3)^2


=>
(x_1+x_2+x_3)(1/x_1+1/x_2+1/x_3)<(\sqrt(n^2+1) -(n-3))^2

instead of
(n^2+1)
, use that
k(n)
and instead of
x_3
, use
x_n
:
(x_1+x_2+x_n)(1/x_1+1/x_2+1/x_n)<(\sqrt(k(n)) -(n-3))^2

Assume
x_1+x_2<=x_n
then:
(x_1+x_2+x_n)(1/x_1+1/x_2+1/x_n)>=10
, so we must have
10>=(\sqrt(k(n)) -(n-3))^2
,
which leads to the wanted answer. Indeed if:
(x_1+x_2+x_n)(1/x_1+1/x_2+1/x_n)<(\sqrt(k(n)) -(n-3))^2
, then:
x_1+x_2>x_n
, so for any
1<= i, j, k <= n
:
x_i+x_j>x_k
.

Of course, there is a quite simple (but not elementary) "forged" solution, using optimization theory:
k(n)=min f(x_1,x_2,...,x_n)
,
when
x_1<=x_2<=...<=x_n
and
x_1+x_2<=x_n
, we may get:
k(n)=f(1,1,2\sqrt(2/5),...,2\sqrt(2/5),2)
, where
2\sqrt(2/5)
appears
(n-3)
times.

I think it's pretty interesting.
This post has been edited 2 times. Last edited by heartwork, Aug 3, 2004, 6:36 PM
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heartwork
308 posts
#15 • 2 Y
Y by Adventure10, Mango247
Problem may be generalised as it follows:
Instead of a triangle, use a polygon of "given"
M
edges ,
M>=3
so if the same inequality holds for any
n>=M
positive reals (of course with something like
k(n,M)
in the RHS) then any
M
from our
n
real numbers can be the side lengths of a polygon with
M
edges. Solution is basically the same.
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G
H
=
a