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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Cyclic Quads and Parallel Lines
gracemoon124   16
N 7 minutes ago by ohiorizzler1434
Source: 2015 British Mathematical Olympiad?
Let $ABCD$ be a cyclic quadrilateral. Let $F$ be the midpoint of the arc $AB$ of its circumcircle which does not contain $C$ or $D$. Let the lines $DF$ and $AC$ meet at $P$ and the lines $CF$ and $BD$ meet at $Q$. Prove that the lines $PQ$ and $AB$ are parallel.
16 replies
1 viewing
gracemoon124
Aug 16, 2023
ohiorizzler1434
7 minutes ago
Radical Center on the Euler Line (USEMO 2020/3)
franzliszt   37
N 27 minutes ago by Ilikeminecraft
Source: USEMO 2020/3
Let $ABC$ be an acute triangle with circumcenter $O$ and orthocenter $H$. Let $\Gamma$ denote the circumcircle of triangle $ABC$, and $N$ the midpoint of $OH$. The tangents to $\Gamma$ at $B$ and $C$, and the line through $H$ perpendicular to line $AN$, determine a triangle whose circumcircle we denote by $\omega_A$. Define $\omega_B$ and $\omega_C$ similarly.
Prove that the common chords of $\omega_A$,$\omega_B$ and $\omega_C$ are concurrent on line $OH$.

Proposed by Anant Mudgal
37 replies
franzliszt
Oct 24, 2020
Ilikeminecraft
27 minutes ago
Functional equation with powers
tapir1729   13
N 28 minutes ago by ihategeo_1969
Source: TSTST 2024, problem 6
Determine whether there exists a function $f: \mathbb{Z}_{> 0} \rightarrow \mathbb{Z}_{> 0}$ such that for all positive integers $m$ and $n$,
\[f(m+nf(m))=f(n)^m+2024! \cdot m.\]Jaedon Whyte
13 replies
tapir1729
Jun 24, 2024
ihategeo_1969
28 minutes ago
Powers of a Prime
numbertheorist17   34
N an hour ago by KevinYang2.71
Source: USA TSTST 2014, Problem 6
Suppose we have distinct positive integers $a, b, c, d$, and an odd prime $p$ not dividing any of them, and an integer $M$ such that if one considers the infinite sequence \begin{align*}
		ca &- db \\
		ca^2 &- db^2 \\
		ca^3 &- db^3 \\
		ca^4 &- db^4 \\
&\vdots
	\end{align*} and looks at the highest power of $p$ that divides each of them, these powers are not all zero, and are all at most $M$. Prove that there exists some $T$ (which may depend on $a,b,c,d,p,M$) such that whenever $p$ divides an element of this sequence, the maximum power of $p$ that divides that element is exactly $p^T$.
34 replies
+1 w
numbertheorist17
Jul 16, 2014
KevinYang2.71
an hour ago
q(x) to be the product of all primes less than p(x)
orl   18
N an hour ago by happypi31415
Source: IMO Shortlist 1995, S3
For an integer $x \geq 1$, let $p(x)$ be the least prime that does not divide $x$, and define $q(x)$ to be the product of all primes less than $p(x)$. In particular, $p(1) = 2.$ For $x$ having $p(x) = 2$, define $q(x) = 1$. Consider the sequence $x_0, x_1, x_2, \ldots$ defined by $x_0 = 1$ and \[ x_{n+1} = \frac{x_n p(x_n)}{q(x_n)} \] for $n \geq 0$. Find all $n$ such that $x_n = 1995$.
18 replies
orl
Aug 10, 2008
happypi31415
an hour ago
IMO 2018 Problem 5
orthocentre   80
N an hour ago by OronSH
Source: IMO 2018
Let $a_1$, $a_2$, $\ldots$ be an infinite sequence of positive integers. Suppose that there is an integer $N > 1$ such that, for each $n \geq N$, the number
$$\frac{a_1}{a_2} + \frac{a_2}{a_3} + \cdots + \frac{a_{n-1}}{a_n} + \frac{a_n}{a_1}$$is an integer. Prove that there is a positive integer $M$ such that $a_m = a_{m+1}$ for all $m \geq M$.

Proposed by Bayarmagnai Gombodorj, Mongolia
80 replies
orthocentre
Jul 10, 2018
OronSH
an hour ago
Line passes through fixed point, as point varies
Jalil_Huseynov   60
N 2 hours ago by Rayvhs
Source: APMO 2022 P2
Let $ABC$ be a right triangle with $\angle B=90^{\circ}$. Point $D$ lies on the line $CB$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $AD$ and let $F$ be the seconf intersection point of the circumcircle of $\triangle ACD$ and the circumcircle of $\triangle BDE$. Prove that as $D$ varies, the line $EF$ passes through a fixed point.
60 replies
Jalil_Huseynov
May 17, 2022
Rayvhs
2 hours ago
Tangent to two circles
Mamadi   2
N 2 hours ago by A22-
Source: Own
Two circles \( w_1 \) and \( w_2 \) intersect each other at \( M \) and \( N \). The common tangent to two circles nearer to \( M \) touch \( w_1 \) and \( w_2 \) at \( A \) and \( B \) respectively. Let \( C \) and \( D \) be the reflection of \( A \) and \( B \) respectively with respect to \( M \). The circumcircle of the triangle \( DCM \) intersect circles \( w_1 \) and \( w_2 \) respectively at points \( E \) and \( F \) (both distinct from \( M \)). Show that the line \( EF \) is the second tangent to \( w_1 \) and \( w_2 \).
2 replies
Mamadi
May 2, 2025
A22-
2 hours ago
Deduction card battle
anantmudgal09   55
N 3 hours ago by deduck
Source: INMO 2021 Problem 4
A Magician and a Detective play a game. The Magician lays down cards numbered from $1$ to $52$ face-down on a table. On each move, the Detective can point to two cards and inquire if the numbers on them are consecutive. The Magician replies truthfully. After a finite number of moves, the Detective points to two cards. She wins if the numbers on these two cards are consecutive, and loses otherwise.

Prove that the Detective can guarantee a win if and only if she is allowed to ask at least $50$ questions.

Proposed by Anant Mudgal
55 replies
1 viewing
anantmudgal09
Mar 7, 2021
deduck
3 hours ago
Geometry
Lukariman   7
N 3 hours ago by vanstraelen
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that $\angle HDM$ = 2∠AMP.
7 replies
Lukariman
Yesterday at 12:43 PM
vanstraelen
3 hours ago
perpendicularity involving ex and incenter
Erken   20
N 4 hours ago by Baimukh
Source: Kazakhstan NO 2008 problem 2
Suppose that $ B_1$ is the midpoint of the arc $ AC$, containing $ B$, in the circumcircle of $ \triangle ABC$, and let $ I_b$ be the $ B$-excircle's center. Assume that the external angle bisector of $ \angle ABC$ intersects $ AC$ at $ B_2$. Prove that $ B_2I$ is perpendicular to $ B_1I_B$, where $ I$ is the incenter of $ \triangle ABC$.
20 replies
Erken
Dec 24, 2008
Baimukh
4 hours ago
Isosceles Triangle Geo
oVlad   4
N 4 hours ago by Double07
Source: Romania Junior TST 2025 Day 1 P2
Consider the isosceles triangle $ABC$ with $\angle A>90^\circ$ and the circle $\omega$ of radius $AC$ centered at $A.$ Let $M$ be the midpoint of $AC.$ The line $BM$ intersects $\omega$ a second time at $D.$ Let $E$ be a point on $\omega$ such that $BE\perp AC.$ Let $N$ be the intersection of $DE$ and $AC.$ Prove that $AN=2\cdot AB.$
4 replies
oVlad
Apr 12, 2025
Double07
4 hours ago
Geometry
Lukariman   1
N 4 hours ago by Primeniyazidayi
Given acute triangle ABC ,AB=b,AC=c . M is a variable point on side AB. The circle circumscribing triangle BCM intersects AC at N.

a)Let I be the center of the circle circumscribing triangle AMN. Prove that I always lies on a fixed line.

b)Let J be the center of the circle circumscribing triangle MBC. Prove that line segment IJ has a constant length.
1 reply
Lukariman
Today at 4:02 PM
Primeniyazidayi
4 hours ago
Kingdom of Anisotropy
v_Enhance   24
N 4 hours ago by deduck
Source: IMO Shortlist 2021 C4
The kingdom of Anisotropy consists of $n$ cities. For every two cities there exists exactly one direct one-way road between them. We say that a path from $X$ to $Y$ is a sequence of roads such that one can move from $X$ to $Y$ along this sequence without returning to an already visited city. A collection of paths is called diverse if no road belongs to two or more paths in the collection.

Let $A$ and $B$ be two distinct cities in Anisotropy. Let $N_{AB}$ denote the maximal number of paths in a diverse collection of paths from $A$ to $B$. Similarly, let $N_{BA}$ denote the maximal number of paths in a diverse collection of paths from $B$ to $A$. Prove that the equality $N_{AB} = N_{BA}$ holds if and only if the number of roads going out from $A$ is the same as the number of roads going out from $B$.

Proposed by Warut Suksompong, Thailand
24 replies
v_Enhance
Jul 12, 2022
deduck
4 hours ago
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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shendrew7
795 posts
#83
Y by
We first prove the case $n=3$. Assume FTSOC we have $(t_1,t_2,t_3) = (a,b,a+b+k)$, where $k \ge 0$. Then
\begin{align*}
3^2+1 &> (t_1+t_2+t_3)\left(\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3}\right) \\
&\ge 5 + \frac{t_3}{t_1} + \frac{t_3}{t_2} + \frac{t_1}{t_2} + \frac{t_1}{t_3} \\
&= 7 + \frac{c+k}{b} + \frac{b+k}{c} + \frac{b+c}{b+c+k} \\
&\ge 10 + k\left(\frac 1b + \frac 1c - \frac{1}{b+c+k}\right) \\
&\ge 10,
\end{align*}
contradiction. We now show the case $n \ge 4$. Assume again FTSOC $t_1, t_2, t_3$ do not form the sides of a triangle. Then
\begin{align*}
n^2+1 &> \left(t_1 + t_2 + \ldots + t_n\right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_n} \right) \\
&= 10 + (n-3)^2 + \left(t_1+\ldots+t_3\right)\left(\frac{1}{t_4}+\ldots+\frac{1}{t_n}\right) + \left(t_4+\ldots+t_n\right) \left(\frac{1}{t_1}+\ldots+\frac{1}{t_3}\right) \\
&\ge 10 + (n-3)^2 + 2\sqrt{10(n-3)^2} \\
&= \left(n+\sqrt{10}-3\right)^2 \\
&\ge n^2+1,
\end{align*}
contradiction. $\blacksquare$
Z K Y
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RedFireTruck
4223 posts
#84
Y by
Assume WLOG that $t_1\le t_2\le \dots\le t_n$.

Claim: $a\ge b+c$ for $a,b,c>0$ means that $\frac{b+c}{a}+\frac{a}{b}+\frac{a}{c}\ge 5$.

Proof: Let $a=b+c+d$. Then, we wish to prove that $(1-\frac{d}{b+c+d})+2+\frac{b}{c}+\frac{c}{b}+\frac{d}{b}+\frac{d}{c}\ge 5$. By AM-GM, $\frac{b}{c}+\frac{c}{b}\ge 2$ so it suffices to prove that $\frac{d}{b}+\frac{d}{c}\ge \frac{d}{b+c+d}$, which is true because $\frac1b+\frac1c\ge \frac1{b+c}$.

Notice that for $n=3$, we just need to prove that $t_3\ge t_1+t_2$ implies that $(t_1+t_2+t_3)(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3})\ge 10$. Expanding gives $$(t_1+t_2+t_3)(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3})=3+\frac{t_1}{t_2}+\frac{t_2}{t_1}+\frac{t_1+t_2}{t_3}+\frac{t_3}{t_1}+\frac{t_3}{t_2}\ge 3+2+5=10$$as desired.

Assume that for some $n-1\ge 3$, the assertion is true. We will now prove that the assertion is true for $n$.

If there exists $1\le i<j<k\le n-1$ such that $t_i$, $t_j$, $t_k$ don't form the side lengths of a triangle, then $$(t_1+t_2+\dots+t_{n-1})(\frac{1}{t_1}+\frac{1}{t_2}+\dots+\frac{1}{t_{n-1}})\ge (n-1)^2+1$$and $$t_n(\frac1{t_1}+\frac1{t_2}+\dots+\frac1{t_{n-1}})+(t_1+t_2+\dots+t_{n-1})\frac{1}{t_n}+1\ge 2n-1$$by AM-GM so $$(t_1+t_2+\dots+t_{n})(\frac{1}{t_1}+\frac{1}{t_2}+\dots+\frac{1}{t_{n}})\ge n^2+1$$as desired.

Otherwise, assume that $t_n\ge t_1+t_2$. By AM-GM and our claim, $$(t_1+t_2+\dots+t_{n})(\frac{1}{t_1}+\frac{1}{t_2}+\dots+\frac{1}{t_{n}})\ge n+2\binom{n}{2}-4+\frac{t_1+t_2}{t_n}+\frac{t_n}{t_1}+\frac{t_n}{t_2}\ge n^2+1$$as desired.

We are done by induction.
Z K Y
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blueprimes
353 posts
#86
Y by
For the sake of contradiction, assume that three of the $t_i$ are equal to $a$, $b$, $a + b + x$ where $a$, $b$ are positive reals and $x$ is nonnegative. By Cauchy-Schwarz, we have
$$(t_1 + t_2 + \dots + t_n) \left( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} \right) \ge \left(n - 3 + \sqrt{(2a + 2b + x) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{a + b + x} \right)} \right)^2.$$Now
$$\sqrt{(2a + 2b + x) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{a + b + x} \right)} = \sqrt{6 + \frac{2a}{b} + \frac{2b}{a} + x \left( \frac{1}{a} + \frac{1}{b} - \frac{1}{a + b + x} \right) } \ge \sqrt{10}$$by AM-GM. To reach contradiction it is now sufficient to show $(n - 3 + \sqrt{10})^2 \ge n^2 + 1 \iff n \ge 3$ which is true. We are done.
This post has been edited 2 times. Last edited by blueprimes, Jun 24, 2024, 1:53 PM
Z K Y
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dudade
139 posts
#88
Y by
FTSOC, suppose $t_1 > t_2 + t_3$. Therefore, we want to show
\begin{align*}
n^2 + 1 &> \sum_{1 \leq i<j \leq n} \left(\dfrac{t_i}{t_j} + \dfrac{t_j}{t_i}\right) + n \\
&= \left(\left(\dfrac{t_1}{t_2}+\dfrac{t_2}{t_1}\right) + \left(\dfrac{t_2}{t_3}+\dfrac{t_3}{t_2}\right) + \left(\dfrac{t_3}{t_1}+\dfrac{t_1}{t_3}\right)\right) + \sum_{4\leq i<j \leq n} \left(\dfrac{t_i}{t_j} + \dfrac{t_j}{t_i}\right)+n \\
&\geq \left(\left(\dfrac{t_1}{t_2}+\dfrac{t_2}{t_1}\right) + \left(\dfrac{t_2}{t_3}+\dfrac{t_3}{t_2}\right) + \left(\dfrac{t_3}{t_1}+\dfrac{t_1}{t_3}\right)\right) + \left(n^2 - n - 6\right) + n.
\end{align*}Thus, we want
\begin{align*}
7 > \left(\dfrac{t_1}{t_2}+\dfrac{t_2}{t_1}\right) + \left(\dfrac{t_2}{t_3}+\dfrac{t_3}{t_2}\right) + \left(\dfrac{t_3}{t_1}+\dfrac{t_1}{t_3}\right)
\end{align*}if $t_1 > t_2 + t_3$. By AM-GM-HM, note that
\begin{align*}
\left(\dfrac{t_1}{t_2}+\dfrac{t_2}{t_1}\right) + \left(\dfrac{t_2}{t_3}+\dfrac{t_3}{t_2}\right) + \left(\dfrac{t_3}{t_1}+\dfrac{t_1}{t_3}\right) &= \left(\dfrac{t_2}{t_1} + \dfrac{t_3}{t_1}\right) + \left(\dfrac{t_1}{t_2} + \dfrac{t_1}{t_3}\right) + 2 \\
&\geq \left(\dfrac{t_2}{t_1} + \dfrac{t_3}{t_1}\right)+ \dfrac{4}{\left(\dfrac{t_1}{t_2} + \dfrac{t_1}{t_3}\right)} + 2 = k + \tfrac{4}{k} + 2.
\end{align*}Since $k < 1$, then we must have that $k + \tfrac{4}{k} + 2 \leq 7$ which is clearly a contradiction. Thus, for all $i$, $j$, and $k$, there exists a triangle with side lengths $t_i$, $t_j$, and $t_k$, as desired.
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jolynefag
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#89
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Elegant problem which is still being solved by people these days. So beautiful!
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SomeonesPenguin
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#90 • 1 Y
Y by zzSpartan
storage
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numbertheory97
42 posts
#91
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Really nice problem! I think I accidentally ended up proving the stronger version.

Solution. Without loss of generality, suppose to the contrary that $t_1 \geq t_2 + t_3$. Then \[t_1 + t_2 + t_3 = \frac25 t_1 + \frac35 t_1 + (t_2 + t_3) \geq \frac25 t_1 + \frac85(t_2 + t_3)\]and \[\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} \geq \frac{1}{t_1} + \frac{4}{t_2 + t_3}\]by Cauchy-Schwarz, so we have \[(t_1 + t_2 + \dots + t_n)\left(\frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n}\right)\]\[ \geq \left(\frac25 t_1 + \frac85(t_2 + t_3) + t_4 + \dots + t_n\right)\left(\frac{1}{t_1} + \frac{4}{t_2 + t_3} + \frac{1}{t_4} + \dots + \frac{1}{t_n}\right)\]\[ \overset{\text{C-S}}{\geq} \left(\sqrt{\frac25} + \sqrt{\frac{32}{5}} + (n - 3)\right)^2 = \left(n + \sqrt{10} - 3\right)^2.\]But $f(x) = \left(x + \sqrt{10} - 3\right)^2 - x^2$ is strictly increasing, so $f(x) \geq f(3) = 1$, a contradiction. $\square$

Remark. (Motivation) Why $2/5$ and $3/5$? I originally tried using $1/2$'s, which was very close but broke slightly for $n = 3$. But if we use arbitrary coefficients $x$ and $1 - x$, it suffices to maximize the function \[f(x) = \sqrt x + 2\sqrt{2 - x},\]which by C-S or similar methods is greatest when $x = 2/5$. This fits perfectly when $n = 3$.
This post has been edited 1 time. Last edited by numbertheory97, Aug 23, 2024, 11:51 PM
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ezpotd
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#92
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Sort $t_n$. Assume to the contrary that there exists some tuple $(t_1 \cdots t_n)$ such that the inequality is satisfied and $t_1 + t_2 \le t_n$. We show that if $n \ge 4$ we can form a similar tuple with $n -1$ elements. Of course, just remove $t_3$, the decrease on the left is $2n - 1$, the decrease on the right is $t_3 \frac{1}{t_3} + \sum_{i \neq 3} \frac{t_i}{t_3} + \frac{t_3}{t_i} \ge 2n - 1$ by AM-GM, so the inequality is preserved. Now we can just reduce to the $n = 3$ case, where we MUST have $10 > 3 + \sum_{sym} \frac{t_i}{t_j}$. Since $\frac{t_3}{t_1} + \frac{t_1}{t_3}, \frac{t_3}{t_2} + \frac{t_2}{t_3}$ are increasing for $t_3 > t_1, t_2$, if the inequality holds true for $t_3 > t_1 + t_2$, it must hold for $t_3 = t_1 + t_2$. Now plugging this value of $t_3$ gives $7 > 2(\frac{t_1}{t_2} + \frac{t_2}{t_1}) + 3$, which is NEVER true by AM-GM, so the original inequality can never hold in the conditions described.
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GeorgeRP
130 posts
#93 • 1 Y
Y by Bumfuzzle
We will solve the problem by induction on n.

Base case: n=3. We have that:
$$10>(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \Leftrightarrow 7abc>\sum{a^2b}$$WLOG we can assume $a$ is the largest. Thus we need to show $a<b+c$. If we rewrite as a quadratic function in respect to $a$ (considering $b$ and $c$ as parameters) we have:
$$0>a^2(b+c)+a(b^2+c^2-7bc)+(b^2c+c^2b)=f(a)$$We however know:
$$ b+c>a \Leftrightarrow \begin{cases}
  f(b+c)\geq0 \\
  b+c>\frac{7bc-b^2-c^2}{2(b+c)} 
\end{cases} \Leftrightarrow \begin{cases}
  2(b+c)^3-8(b+c)bc\geq0\\
  3b^2-3bc+3c^2>0 
\end{cases} \text{ which are both true, finishing the base}$$
IH: for n-1

IS: for n. Let $A=t_1+t_2+\cdots+t_{n-1}, B=\frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_{n-1}}$.WLOG we can assume $t_1\leq t_2 \leq \cdots \leq t_n$. We know that $n^2+1>(A+t_n)(B+\frac{1}{t_n}) \Leftrightarrow n^2-Bt_n-\frac{A}{t_n}>AB$. We will first show that $Bt_n+\frac{A}{t_n}\geq2n-2$. This follows from the fact:
$$Bt_n+\frac{A}{t_n}=\sum_{1\leq i \leq n-1}{\frac{t_n}{t_i}+\frac{t_i}{t_n}} \overset{AM-GM}{\geq} 2n-2$$From this it directly follows that $(n-1)^2+1>AB$, implying that $t_i, t_j, t_k$ are the sides of a triangle for $1\leq i<j<k\leq n-1$. We are now left to show that $t_n<t_1+t_2$ from which the desired will follow (because $t_1, t_2$ are the smallest). For simplicity let $t_1=x, t_2=y$. FTSOC let $t_n=x+y+k$ for some $k\geq0$. We know that by Cauchy-Schwarz $AB\geq(n-1)^2$. We will now show that $t_nB+\frac{A}{t_n}\geq 2n-1$.
$$t_nB+\frac{A}{t_n}= \frac{t_n}{t_1}+\frac{t_1}{t_n}+\frac{t_2}{t_n}+\frac{t_n}{t_2}+(\frac{t_3}{t_n}+\cdots+\frac{t_{n-1}}{t_n})+(\frac{t_n}{t_3}+\cdots+\frac{t_n}{t_{n-1}}) \overset{\text{AM-GM}}{\geq} $$$$ \geq \frac{a+b+k}{a}+\frac{a}{a+b+k}+\frac{a+b+k}{b}+\frac{b}{a+b+k}+2n-6 \geq \frac{a+b}{a}+\frac{a}{a+b}+\frac{a+b}{b}+\frac{b}{a+b}+2n-6=$$$$=\frac{(a+b)^2}{ab}+2n-5\overset{\text{AM-GM}}{\geq}2n-1$$From here it follows that:
$$n^2> AB+t_nB+\frac{A}{t_n}\geq n^2 \text{ \Large{ \lightning }} \text{ with which we are done}$$
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N3bula
269 posts
#94
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Expanding gives us:
\[\left(t_1 + t_2 + \dots + t_n\right)
  \left( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} \right) = n+\sum_{i, j \in \{1, 2, \dots, n\}, i\neq j} \frac{t_i}{t_j} +\frac{t_j}{t_i}\]Thus we get if for any $i$, $j$ and $k$ that $\sum_{sym}\frac{t_i}{t_j}\geq 7$ we get a contradiction,
thus it suffices to prove that if we have $a+b\leq c$ that $\sum_{sym}\frac{a}{b}\geq 7$, thus I will
prove that that sum is minimised when $a+b=c$, we get that proving that is equivalent to proving
\[\frac{1}{a}+\frac{1}{b}-\frac{a+b}{c(c+k)}\geq 0\]When $a+b=c$ and $k$ is a positive real. Thus the above is equivalent to:
\[\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\geq 0\]\[\frac{b+a}{ab}-\frac{1}{a+b}\]The above is minimised when $a=b$, thus its equivalent to:
\[\frac{2a}{a^2}-\frac{1}{2a}\geq 0\]Which is clearly true.
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eg4334
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#95
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WLOG $t_1 \leq t_2 \leq \dots \leq t_n$, and FTSOC let $t_i + t_j < t_k$. The condition then gives by Cauchy that $$n^2 +1 \geq \left( n -3 + \sqrt{3+\sum \frac{t_i}{t_j} } \right)^2$$where the summation is taken over all ordered pairs in $t_i, t_j, t_k$. Now notice that $n^2 +1 \geq (n - 3 + \sqrt{10})^2$ has no integer solutions for $n \geq 3$, so we just need to prove that $$\frac{t_i}{t_j} + \frac{t_i}{t_k} + \frac{t_j}{t_i} + \frac{t_j}{t_k} + \frac{t_k}{t_i} + \frac{t_k}{t_j} \geq 7 $$$$\frac{t_i}{t_k} + \frac{t_j}{t_k} + \frac{t_k}{t_i} + \frac{t_k}{t_j} \geq5$$by AMGM. Also, $(\frac{1}{t_i} + \frac{1}{t_j})(t_i + t_j) \geq 4$ by Cauchy so this reduces into $$\frac{t_i+t_j}{t_k} + \frac{4t_k}{t_i+t_i} \geq 5$$which is obvious.
This post has been edited 2 times. Last edited by eg4334, Nov 29, 2024, 8:56 PM
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Maximilian113
575 posts
#96
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We show by induction on $n$ that if there exists at least one triple $(t_i, t_j, t_k)$ that are not the sides of a triangle, then $$n^2+1 \leq (t_1+t_2+\cdots+t_n)\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_n} \right).$$$\text{Base Case: }$ For $n=3,$ we must show that if $t_1, t_2, t_3$ cannot be the sides of a triangle, then $$10 \leq (t_1+t_2+t_3)\left( \frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3} \right).$$WLOG, assume that $t_1 \leq t_2 \leq t_3,$ then we have that $t_3 > t_1+t_2.$ Also, because our inequality is homogenous, assume that $t_3=1.$ Letting $x=t_1+t_2,$ it suffices to show that $$7 \leq \frac{t_1}{t_2}+\frac{t_2}{t_1}+x+\frac{1}{t_1}+\frac{1}{t_2}.$$But by AM-GM, $\frac{t_1}{t_2}+\frac{t_2}{t_1} \geq 2,$ so it suffices to show that $$x+\frac{1}{t_1}+\frac{1}{t_2} \geq 5.$$But, $x \leq t_3=1 \implies (x-1)(x-4) \geq 0 \implies 5 \leq x+\frac{4}{x} \leq x+\frac{1}{t_1t_2} \leq x+\frac{1}{t_1}+\frac{1}{t_2}$ by $2$ applications of AM-GM at the end. Therefore, our base case is proven.

$\text{Inductive Step: }$ Now, assume that for $n=k-1 \geq 3$ our proposition holds. Consider the sequence $t_1, t_2, \cdots t_k$ of positive real numbers. Then if there is at least one triple, say WLOG $(t_1, t_2, t_3),$ which are not the sides of a triangle then we have by our inductive hypothesis that
\begin{align*}
(t_1+t_2+\cdots+t_k)\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_k} \right) &=(t_1+t_2+\cdots+t_{k-1})\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_{k-1}} \right) + \sum_{i=1}^{k-1} \left( \frac{t_i}{t_k}+\frac{t_k}{t_i} \right)+1 \\
&\geq \left((k-1)^2+1\right)+(k-1)\cdot 2+1 \text{ by AM-GM} \\
&= k^2+1.
\end{align*}Therefore, our induction is complete. QED

Now, for the sake of a contradiction assume that there exists $t_i, t_j, t_k$ which are not the sides of a triangle, and $$n^2+1 > (t_1+t_2+\cdots+t_n)\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_n} \right).$$But by our induction above, we have that $$n^2+1 \leq (t_1+t_2+\cdots+t_n)\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_n} \right),$$a contradiction. Therefore, all $t_i, t_j, t_k$ are the sides of a triangle. QED
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Marcus_Zhang
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#97
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IMO P4
This post has been edited 1 time. Last edited by Marcus_Zhang, Mar 23, 2025, 3:46 AM
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cubres
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#98
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Yapping
Storage - grinding IMO problems
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Ilikeminecraft
617 posts
#99
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Assume there exists a sequence $t_i$ such that $t_1 \geq t_2 + t_3.$ Then, we have that:
\begin{align*}
  & (t_1 + t_2 + \cdots + t_n)\left(\frac1{t_1} + \frac1{t_2} + \cdots + \frac1{t_n}\right) \\
  & \geq (t_1 + t_2 + t_3)\left(\frac1{t_1} + \frac1{t_2} + \frac1{t_3}\right) + (t_1 + t_2 + t_3)\left(\frac1{t_4} + \cdots + \frac1{t_n}\right)\\
  & \qquad + (t_4 + \cdots + t_n)\left(\frac1{t_1} + \frac1{t_2} + \frac1{t_3}\right) + (n - 3)^2 \\
  & \geq (n - 3)^2 + 2(3n - 9) + 3 + (t_1 + t_2 + t_3)\left(\frac1{t_1} + \frac1{t_2} + \frac1{t_3}\right)
\end{align*}Thus, all that is left is to prove that $$(t_1 + t_2 + t_3)\left(\frac1{t_1} + \frac1{t_2} + \frac1{t_3}\right) \geq 10$$Expanding, we have that $$3 + \frac{t_1}{t_2} + \frac{t_3}{t_2} + \frac{t_2 + t_3}{t_1} + \frac{t_1}{t_3} + \frac{t_2}{t_3}$$Now, differentiate w.r.t. $t_1.$ We get $f(t_1) = \frac1{t_2} + \frac1{t_3} - (t_2 + t_3)\frac1{t_1^2}.$ We claim that $t_1^2 \geq t_2t_3.$ However, this is clearly true because $t_1^2 \geq t_2^2 + t_3^2 + 2t_2t_3,$ while $t_2^2 + t_3^2 + t_2 t_3 \geq 0.$ Thus, $f(t_1)$ is increasing as $t_1$ increases. Thus, we can assume that $t_1 = t_2 + t_3.$ This clearly implies that $f(t_1) \geq 10.$ Hence, we are done.
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