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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)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Jul 1, 2025
0 replies
CMO qualifying repechange 2025
Zeroin   5
N 2 hours ago by Zeroin
In scalene triangle $ABC$, the circumcentre and incentre are respectively $O$ and $I$.
Let $AD$ be the altitude to line $BC$, with $D$ lying on line $BC$. Given that the radius of the
circumcircle and A-excircle are equal, prove that the points $O$, $I$, and $D$ are collinear.
5 replies
Zeroin
Today at 12:24 PM
Zeroin
2 hours ago
Korea csat problem, so-called “Killer problem”..
darrime627   1
N 3 hours ago by Zennggg
Find the values of \( a \) and \( b \) such that the function \( f(x) \), which has a second derivative for all \( x \in \mathbb{R} \), satisfies the following conditions:

\[
(\gamma) \quad [f(x)]^5 + [f(x)]^3 + ax + b = \ln \left( x^2 + x + \frac{5}{2} \right)
\]
\[
(\delta) \quad f(-3) f(3) < 0, \quad f'(2) > 0
\]
1 reply
darrime627
Jul 14, 2025
Zennggg
3 hours ago
Probably a Cool Optimization Problem
PhysicsIsChad   4
N 4 hours ago by aaravdodhia
What i am trying to do is finding a general solution to this earlier problem (https://artofproblemsolving.com/community/u1056796h3610029p35343126) . Here i am trying to generalise such a solution for n points . Simply stated find the min value of \[
f(x, y) = \sum_{i=1}^{n} \sqrt{(x - a_i)^2 + (y - b_i)^2}
\]. I would love to hear your insights on this problem .
My insights :-
\[
\text{Let } (x_1, y_1), (x_2, y_2), \dots, (x_n, y_n) \text{ be the points}
\]\[
\text{Minimize } f(x, y) = \sqrt{(x - x_1)^2 + (y - y_1)^2} + \sqrt{(x - x_2)^2 + (y - y_2)^2} + \cdots + \sqrt{(x - x_n)^2 + (y - y_n)^2}
\]\[
\text{Set } \frac{\partial f}{\partial x} = 0 \quad \text{and} \quad \frac{\partial f}{\partial y} = 0
\]\[
\frac{\partial f}{\partial x} = \frac{x - x_1}{\sqrt{(x - x_1)^2 + (y - y_1)^2}} + \frac{x - x_2}{\sqrt{(x - x_2)^2 + (y - y_2)^2}} + \cdots + \frac{x - x_n}{\sqrt{(x - x_n)^2 + (y - y_n)^2}} = 0
\]\[
\frac{\partial f}{\partial y} = \frac{y - y_1}{\sqrt{(x - x_1)^2 + (y - y_1)^2}} + \frac{y - y_2}{\sqrt{(x - x_2)^2 + (y - y_2)^2}} + \cdots + \frac{y - y_n}{\sqrt{(x - x_n)^2 + (y - y_n)^2}} = 0
\]I am not able to move ahead but probably we will find a constraint here through some inequality which will help us reach the solution .


4 replies
PhysicsIsChad
5 hours ago
aaravdodhia
4 hours ago
NT By Probabilistic Method
EthanWYX2009   2
N 4 hours ago by Manivs
Source: 2024 March 谜之竞赛-6
Given a positive integer \( k \) and a positive real number \( \varepsilon \), prove that there exist infinitely many positive integers \( n \) for which we can find pairwise coprime integers \( n_1, n_2, \cdots, n_k \) less than \( n \) satisfying
\[\text{gcd}(\varphi(n_1), \varphi(n_2), \cdots, \varphi(n_k)) \geq n^{1-\varepsilon}.\]Proposed by Cheng Jiang from Tsinghua University
2 replies
EthanWYX2009
Jul 14, 2025
Manivs
4 hours ago
Concurrency of Lines Involving Altitudes and Circumcenters in a Triangle
JackMinhHieu   2
N 6 hours ago by JackMinhHieu
Hi everyone,
I recently came across an interesting geometry problem that I'd like to share. It involves a triangle inscribed in a circle, altitudes, points on arcs, and a surprising concurrency involving circumcenters. Here's the problem:
Problem:
Let ABC be an acute triangle inscribed in a circle (O). Let the altitudes AD, BE, CF intersect at the orthocenter H.
Points M and N lie on the minor arcs AB and AC of circle (O), respectively, such that MN // BC.
Let I be the circumcenter of triangle NEC, and J be the circumcenter of triangle MFB.
Prove that the lines OD, BI, and CJ are concurrent.

I find the configuration quite elegant, and I'm looking for different ways to approach the problem — whether it's synthetic, coordinate, vector-based, or inversion.
Any ideas, hints, or full solutions are appreciated. Thank you!
2 replies
JackMinhHieu
Jul 14, 2025
JackMinhHieu
6 hours ago
Find the min value of function f(x,y)
PhysicsIsChad   4
N 6 hours ago by PhysicsIsChad
Let
\[
f(x, y) = \sqrt{x^2 + y^2} + \sqrt{x^2 + y^2 - 2x + 1} + \sqrt{x^2 + y^2 - 2y + 1} + \sqrt{x^2 + y^2 - 6x - 8y + 25}
\]for all real \( x, y \in \mathbb{R} \).
Find the min value of this function . What are your insights on this problem ?
My insights :
All these quadratics under the square root can be converted into \[
f(x, y) = \sqrt{(x - 0)^2 + (y - 0)^2} 
+ \sqrt{(x - 1)^2 + (y - 0)^2} 
+ \sqrt{(x - 0)^2 + (y - 1)^2} 
+ \sqrt{(x - 3)^2 + (y - 4)^2}
\]This function can be interpreted as the sum of the distances of points (0,0) , (1,0) , (0,1) , (3,4) from (x,y) .
We can prove that for minimising this distance (x,y) must be the mean of the x and y coordinates of the points respectively . Thus we get (x,y) = (1,1.25) . \[
\begin{aligned}
f(x, y) &= \sqrt{(x - 0)^2 + (y - 0)^2} 
+ \sqrt{(x - 1)^2 + (y - 0)^2} \\
&\quad + \sqrt{(x - 0)^2 + (y - 1)^2} 
+ \sqrt{(x - 3)^2 + (y - 4)^2} \\
\\
f(1, 1.25) &= \sqrt{1^2 + 1.25^2} 
+ \sqrt{(1 - 1)^2 + 1.25^2} \\
&\quad + \sqrt{1^2 + (1.25 - 1)^2} 
+ \sqrt{(1 - 3)^2 + (1.25 - 4)^2} \\
\\
&= \sqrt{2.5625} + \sqrt{1.5625} + \sqrt{1.0625} + \sqrt{11.5625} \\
\\
&\approx 1.6 + 1.25 + 1.03078 + 3.40184 \\
\\
&= \boxed{7.2826}
\end{aligned}
\]But apparently this is wrong . I will post the proof for the intermediate i used in the replies .





4 replies
PhysicsIsChad
Yesterday at 4:36 PM
PhysicsIsChad
6 hours ago
AM-GM Problem
arcticfox009   15
N 6 hours ago by DAVROS
Let $x, y$ be positive real numbers such that $xy \geq 1$. Find the minimum value of the expression

\[ \frac{(x^2 + y)(x + y^2)}{x + y}. \]
answer confirmation
15 replies
arcticfox009
Jul 11, 2025
DAVROS
6 hours ago
Grade 10 Inequality
BomboaneMentos316   2
N 6 hours ago by BomboaneMentos316
Prove that for any x,y,z real and positive the following is true:
(PS: It is written correctly)

\begin{align*}
\frac{x^3}{y^2 + y z + z^2}
\;+\;
\frac{y^3}{2 z^2 + y z x}
\;+\;
\frac{z^3}{x^2 + x y + y^2}
\;\ge\;
\frac{x + y + z}{3}.
\end{align*}
2 replies
BomboaneMentos316
Yesterday at 6:41 PM
BomboaneMentos316
6 hours ago
Help me please
Hahahsafdk3l   1
N Today at 1:20 PM by littleduckysteve
Let \( \triangle ABC \) be a triangle with incircle \( (I) \) touching \( BC, CA, AB \) at \( D, E, F \), respectively. Points \( H \) and \( K \) are such that \( BHEC \) and \( BFKC \) are parallelograms. Let \( G \) be the intersection of \( BE \) and \( CF \). Prove that \( GH = GK \).
1 reply
Hahahsafdk3l
Today at 1:02 PM
littleduckysteve
Today at 1:20 PM
[PMO26 Qualis] II.2 Coin sides except it's not probability
aops-g5-gethsemanea2   1
N Today at 1:12 PM by Siopao_Enjoyer
There are 2024 coins laid out in a row. For some positive integer \( x \), at least $\frac{2}{3}$ of the first \( x \) coins are heads, and at least $\frac{4}{5}$ of the last \( x \) coins are tails. What is the maximum possible value of \( x \)?

\(\text{(a) }1012\qquad\text{(b) }1079\qquad\text{(c) }1288\qquad\text{(d) }1380\)

Answer confirmation
1 reply
aops-g5-gethsemanea2
Feb 15, 2025
Siopao_Enjoyer
Today at 1:12 PM
combination
Doanh   2
N Today at 1:05 PM by Doanh
\textbf{Problem 14.} There are \( m \) girls and \( n \) boys. It is known that for any two boys and any two girls, there exists at least one boy--girl pair who do not know each other (mutually). Prove that the total number of boy--girl pairs who know each other (regardless of order) does not exceed
\[
m + \frac{n(n - 1)}{2}.
\]
2 replies
Doanh
Today at 6:20 AM
Doanh
Today at 1:05 PM
Inequalities
sqing   13
N Today at 1:02 PM by sqing
Let $ a,b,c $ be real numbers . Prove that
$$- \frac{64(9+2\sqrt{21})}{9} \leq \frac {(ab-4)(bc-4)(ca-4) } {(a^2+a +1)(b^2+b +1)(c^2+c +1)}\leq \frac{16}{9}$$$$- \frac{8(436+79\sqrt{31})}{27} \leq  \frac {(ab-5)(bc-5)(ca-5) } {(a^2+a +1)(b^2+b +1)(c^2+c +1)}\leq \frac{25}{12}$$
13 replies
sqing
Jul 6, 2025
sqing
Today at 1:02 PM
Subset Ordered Pairs of {1, 2, ..., 10}
ahaanomegas   12
N Today at 3:08 AM by mudkip42
Source: Putnam 1990 A6
If $X$ is a finite set, let $X$ denote the number of elements in $X$. Call an ordered pair $(S,T)$ of subsets of $ \{ 1, 2, \cdots, n \} $ $ \emph {admissible} $ if $ s > |T| $ for each $ s \in S $, and $ t > |S| $ for each $ t \in T $. How many admissible ordered pairs of subsets $ \{ 1, 2, \cdots, 10 \} $ are there? Prove your answer.
12 replies
ahaanomegas
Jul 12, 2013
mudkip42
Today at 3:08 AM
Partial sum of Taylor series of e^x
NamelyOrange   2
N Today at 2:09 AM by genius_007
Source: 2022 National Taiwan University STEM Development Program Admissions Test, P1
For positive integer $n$ and positive real $x$, define $e_n(x) = \sum_{k=0}^{n}\frac{x^k}{k!}$.

(a) Prove that $\frac{1}{n!}\le \frac{m^{m-n-1}}{(m-1)!}$ for all integer $n\ge m >1$.

(b) Use (a) to prove that $e_m(x)\le e_n(x)\le e_{m-1}(x)+\frac{x^m}{(m-1)!(m-x)}$ for all integer $n\ge m >1$ and real $0<x<m$.

(c) Use (b) to prove that for all positive real $x$, there exists some positive $L$ such that $e_n(x)<L$ for all positive integer $n$.

(d) Prove that for positive integer $m<n$ and positive real $x,y$ such that $x+y<m$, we have $0\le e_n(x+y)-e_n(x)e_n(y)\le \frac{m^{m-n-1}(x+y)^{n+1}}{(m-1)!(m-x-y)}$.
2 replies
NamelyOrange
Yesterday at 3:54 AM
genius_007
Today at 2:09 AM
triangle
123...   1
N Apr 23, 2021 by 277546
Supose $a, b , c$ are the vertices of closed equilateral triangle in the plane,
find the maximum value of $|z-a||z-b||z-c|$ where $z$ range over this triangle.
1 reply
123...
May 29, 2010
277546
Apr 23, 2021
triangle
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#1 • 1 Y
Y by Adventure10
Supose $a, b , c$ are the vertices of closed equilateral triangle in the plane,
find the maximum value of $|z-a||z-b||z-c|$ where $z$ range over this triangle.
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277546
1607 posts
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Y by
Let $\Omega$ be the interior of the closed triangle $\Delta$. Clearly $\Omega$ is a bounded region. Now $f(z) := |z-a||z-b||z-c|$ is a continuous, nonconstant function on $\bar{\Omega}=\Delta$ so by the Maximum Modulus Principle we have $$|f(z)|<\left\|{f}\right\|_{\partial\Omega}=\left\|{f}\right\|_{\partial\Delta}$$for all $z \in \Omega$. Thus, $$\max_{z \in \Delta}f(z)=\max_{z \in \partial\Delta}f(z)$$That is to say, if $z \in \Delta$ is a point that maximizes $f(z)=|z-a||z-b||z-c|$, then $z$ lies on the boundary of the triangle.

It is now a matter of elementary geometry to maximize $f$. Let $z \in \{z \in \partial\Delta| f(z)=\max_{z \in \Delta}f(z)\}$ (the latter set is nonempty). WLOG suppose $z$ lies on $[a,b]$. To simplify notation, let $s=|a-b|=|b-c|=|c-a$. Then, let $x=|z-a|$. We then have $|b-z|=s-x$. We then get by the Pythagorean Theorem that $$|z-c|=\sqrt{\left(x-\frac{s}{2}\right)^2+\left(\frac{s\sqrt{3}}{2}\right)^2}=\sqrt{x^2-sx+s^2}=x(s-x)\sqrt{x^2-s(x-s)}$$Thus, $$|z-a||z-b||z-c|=x(s-x)\sqrt{x^2-sx+s^2}$$We now maximize this quantity as $x$ ranges from $[0,s]$. Taking the derivative with respect to $s$, we have by the first derivative test that
$$(s-2x)\sqrt{x^2-sx+s^2}+(xs-x^2)\frac{1}{2}(2x-s)\frac{1}{\sqrt{x^2-sx+s^2}}=0$$so rearranging, we get $$(s-2x)(\frac{3}{2}x^2-\frac{3}{2}xs+s^2)=0$$Remark that $$\frac{3}{2}x^2-\frac{3}{2}xs+s^2=\frac{3}{2}(x-\frac{s}{2})^2+\frac{5s^2}{8}>0$$for all $x \in [0,s]$ so we must have $s-2x=0$, that is $x=\frac{s}{2}$. In this case, we have $$|z-a||z-b||z-c|=\frac{s}{2}(\frac{s}{2})\sqrt{(\frac{s}{2})^2-s(\frac{s}{2})+s^2}=\frac{\sqrt{3}s^3}{8}$$Thus, $\max(|z-a||z-b||z-c|)=\boxed{\frac{\sqrt{3}s^3}{8}}$ where $s=|a-b|=|b-c|=|c-a|$.

Addendum: There is also an elementary proof using the Erdos-Mordell Inequality. Lets see if someone here is smart enough to think of it.
This post has been edited 2 times. Last edited by 277546, Apr 23, 2021, 7:33 AM
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