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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
Inspired by BaCaPhe
sqing   2
N 4 minutes ago by lbh_qys
Source: Own
Let $ a,b,c \ge 0 $ and $ ab + bc + ca \ge 4 + abc. $ Prove that
$$  a^2  +  b^2  +  c^2-abc  \ge 4$$$$  a^2  +  b^2  +  c^2  \ge 8$$$$a^2  +  b^2  +  c^2+a^2b^2c^2\ge 8$$$$  a^2  +  b^2  +  c^2+ab+bc+ca  \ge 12$$
2 replies
1 viewing
sqing
4 hours ago
lbh_qys
4 minutes ago
Twin Prime Diophantine
awesomeming327.   18
N 5 minutes ago by EthanWYX2009
Source: CMO 2025
Determine all positive integers $a$, $b$, $c$, $p$, where $p$ and $p+2$ are odd primes and
\[2^ap^b=(p+2)^c-1.\]
18 replies
awesomeming327.
Mar 7, 2025
EthanWYX2009
5 minutes ago
Sharygin 2025 CR P12
Gengar_in_Galar   7
N 9 minutes ago by maths_enthusiast_0001
Source: Sharygin 2025
Circles $\omega_{1}$ and $\omega_{2}$ are given. Let $M$ be the midpoint of the segment joining their centers, $X$, $Y$ be arbitrary points on $\omega_{1}$, $\omega_{2}$ respectively such that $MX=MY$. Find the locus of the midpoints of segments $XY$.
Proposed by: L Shatunov
7 replies
+1 w
Gengar_in_Galar
Mar 10, 2025
maths_enthusiast_0001
9 minutes ago
Problem 1
randomusername   71
N 20 minutes ago by MihaiT
Source: IMO 2015, Problem 1
We say that a finite set $\mathcal{S}$ of points in the plane is balanced if, for any two different points $A$ and $B$ in $\mathcal{S}$, there is a point $C$ in $\mathcal{S}$ such that $AC=BC$. We say that $\mathcal{S}$ is centre-free if for any three different points $A$, $B$ and $C$ in $\mathcal{S}$, there is no points $P$ in $\mathcal{S}$ such that $PA=PB=PC$.

(a) Show that for all integers $n\ge 3$, there exists a balanced set consisting of $n$ points.

(b) Determine all integers $n\ge 3$ for which there exists a balanced centre-free set consisting of $n$ points.

Proposed by Netherlands
71 replies
randomusername
Jul 10, 2015
MihaiT
20 minutes ago
No more topics!
Most challenging inequality on the ISL 2004
orl   6
N Dec 24, 2024 by MathLuis
Source: IMO ShortList 2004, algebra problem 7
Let ${a_1,a_2,\dots,a_n}$ be positive real numbers, ${n>1}$. Denote by $g_n$ their geometric mean, and by $A_1,A_2,\dots,A_n$ the sequence of arithmetic means defined by \[ A_k=\frac{a_1+a_2+\cdots+a_k}{k},\qquad k=1,2,\dots,n. \] Let $G_n$ be the geometric mean of $A_1,A_2,\dots,A_n$. Prove the inequality \[
n \root n\of{\frac{G_n}{A_n}}+ \frac{g_n}{G_n}\le n+1 \] and establish the cases of equality.

Proposed by Finbarr Holland, Ireland
6 replies
orl
Jun 15, 2005
MathLuis
Dec 24, 2024
Most challenging inequality on the ISL 2004
G H J
Source: IMO ShortList 2004, algebra problem 7
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orl
3647 posts
#1 • 4 Y
Y by langkhach11112, rmtf1111, Adventure10, Mango247
Let ${a_1,a_2,\dots,a_n}$ be positive real numbers, ${n>1}$. Denote by $g_n$ their geometric mean, and by $A_1,A_2,\dots,A_n$ the sequence of arithmetic means defined by \[ A_k=\frac{a_1+a_2+\cdots+a_k}{k},\qquad k=1,2,\dots,n. \] Let $G_n$ be the geometric mean of $A_1,A_2,\dots,A_n$. Prove the inequality \[
n \root n\of{\frac{G_n}{A_n}}+ \frac{g_n}{G_n}\le n+1 \] and establish the cases of equality.

Proposed by Finbarr Holland, Ireland
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orl
3647 posts
#2 • 3 Y
Y by langkhach11112, Adventure10, Mango247
Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in $\LaTeX$. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions :)


Solution by Bojan Basic
This post has been edited 2 times. Last edited by orl, Jun 16, 2005, 9:19 AM
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harazi
5526 posts
#3 • 4 Y
Y by langkhach11112, Mathematicsislovely, Adventure10, Mango247
Then the quality of the inequalities proposed that year wasn't very extraordinary if this is the most challenging. Write it in the form \[ n\cdot\sqrt[n^2]{\frac{A_1...A_{n-1}}{A_n^{n-1}}}+\sqrt[n]{(2-A_1/A_2)...(n-(n-1)A_{n-1}/A_n)}\leq n+1. \] Use the AM-GM inequality to deduce that the second term of the LHS is at most \[ \frac{1+(2-A_1/A_2)+...+(n-(n-1)A_{n-1}/A_n)}{n}. \]
Thus the inequality reduces to \[  \frac{n+1}{2}+\frac{A_1/A_2+...+(n-1)A_{n-1}/A_n}{n} \geq n\cdot\sqrt[n^2]{\frac{A_1...A_{n-1}}{A_n^{n-1}}}. \] Now, apply generalized AM-GM inequality to deduce that the LHS is at least $ (n+1)/2+(n-1)/2\cdot\sqrt[(n^2-n)/2]{\frac{A_1...A_{n-1}}{A_n^{n-1}}}$. Finally, kill the last beast with another AM-GM inequality.
This post has been edited 1 time. Last edited by v_Enhance, Mar 19, 2015, 7:24 PM
Reason: Fix broken LaTeX
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k2c901_1
146 posts
#4 • 2 Y
Y by Adventure10, ehuseyinyigit
This problem was used as problem 4 of the final exam of the 3rd TST of Taiwan 2005. (Only 1 person got it right.)
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yayups
1614 posts
#5 • 1 Y
Y by samuel
Solved with Amol Rama, Raymond Feng, Samuel Wang, and William Wang.

Let $c_i=A_i/A_{i+1}$. The inequality we wish to prove reads \[n\left(c_{n-1}^{\frac{n-1}{n^2}}c_{n-2}^{\frac{n-2}{n^2}}\cdots c_1^{\frac{1}{n^2}}\right)^{1/n^2}+\left(1\cdot(2-c_1)\cdot(3-2c_2)\cdots(n-(n-1)c_{n-1})\right)^{1/n}\le n+1.\]We see that \[n\left(c_{n-1}^{\frac{n-1}{n^2}}c_{n-2}^{\frac{n-2}{n^2}}\cdots c_1^{\frac{1}{n^2}}\right)^{1/n^2}\le \frac{n-1}{n}c_{n-1}+\cdots+\frac{1}{n}c_1+\frac{n+1}{2},\]and \[\left(1\cdot(2-c_1)\cdot(3-2c_2)\cdots(n-(n-1)c_{n-1})\right)^{1/n}\le \frac{n+1}{2}-\frac{1}{n}c_1-\cdots-\frac{n-1}{n}c_{n-1},\]so adding gives the desired estimate.

Remark: The solution is motivated by the $n=2$ case, which was solved with a similar AM-GM as above.
This post has been edited 1 time. Last edited by yayups, Nov 9, 2020, 12:04 AM
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awesomeming327.
1665 posts
#6
Y by
Note that $a_k=kA_k-(k-1)A_{k-1}$. Let $b_k=\frac{A_{k-1}}{A_k}$ for all $k$. Then, we have
\begin{align*}
n \sqrt[n]{\frac{G_n}{A_n}}+ \frac{g_n}{G_n} &= n \sqrt[n^2]{\frac{A_1A_2A_3\dots A_n}{A_n^n}} + \sqrt[n]{\frac{a_1}{A_1}\cdot \frac{a_2}{A_2}\cdot \ldots \cdot \frac{a_n}{A_n}} \\
&= n\sqrt[n^2]{\left(\frac{A_1}{A_n}\right)\left(\frac{A_2}{A_n}\right)\dots \left(\frac{A_{n-1}}{A_n}\right)} + \sqrt[n]{\left(\frac{2A_2-A_1}{A_2}\right)\left(\frac{3A_2-2A_2}{A_3}\right)\dots\left(\frac{nA_n-(n-1)A_{n-1}}{A_n}\right)} \\
&=n\sqrt[n^2]{1^{\frac{n(n+1)}{2}}b_1 b_2^2b_3^3\dots b_{n-1}^{n-1}}+\sqrt[n]{(1)(2-b_1)(3-2b_2)(4-3b_3)\dots(n-(n-1)b_{n-1})} \\
&\le n\cdot \frac{\frac{n(n+1)}{2}+b_1+2b_2+\dots + (n-1)b_{n-1}}{n^2} + \frac{\frac{n(n+1)}{2}-b_1-2b_2-\dots-(b-1)b_{n-1}}{n} \\
= n+1
\end{align*}as desired.
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MathLuis
1451 posts
#7
Y by
Re-write the ineq as $n \cdot \sqrt[n^2]{\frac{A_1 \cdots A_{n-1}}{A_n^{n-1}}}+\sqrt[n]{\left(2-\frac{A_1}{A_2} \right) \cdots \left(n-\frac{(n-1)A_{n-1}}{A_n} \right)} \le n+1$.
Now from AM-GM we notice that we have that:
\[ \frac{n+1}{2}-\frac{\frac{A_1}{A_2}+\cdots+\frac{(n-1)A_{n-1}}{A_n}}{n}=\frac{1+\left(2-\frac{A_1}{A_2} \right)+ \cdots + \left(n-\frac{(n-1)A_{n-1}}{A_n} \right)}{n} \ge \sqrt[n]{\left(2-\frac{A_1}{A_2} \right) \cdots \left(n-\frac{(n-1)A_{n-1}}{A_n} \right)} \]Now now all we need to prove after replacing this, is the following:
\[ \frac{n+1}{2}+\frac{\frac{A_1}{A_2}+\cdots+\frac{(n-1)A_{n-1}}{A_n}}{n} \ge n \cdot \sqrt[n^2]{\frac{A_1 \cdots A_{n-1}}{A_n^{n-1}}} \]But from AM-GM once again one has that:
\[ \frac{\frac{n(n+1)}{2}+\frac{A_1}{A_2}+\cdots+\frac{(n-1)A_{n-1}}{A_n}}{n^2} \ge \sqrt[n^2]{\frac{1^{\frac{n(n+1)}{2}} \cdot A_1 \cdots A_{n-1}}{A_n^{n-1}}} \](And thus eq case happens when $a_1=a_2=\cdots=a_n$)
So simply rearranging this one gives the desired ineq, thus we are done :cool:.
This post has been edited 2 times. Last edited by MathLuis, Dec 24, 2024, 2:11 PM
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