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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
Very hard
steven_zhang123   0
3 minutes ago
Source: An article
Given a positive integer \(n\) and a positive real number \(m\), non-negative real numbers \(a_1, a_2, \cdots, a_n\) satisfy \(\sum_{i=1}^n a_i = m\). Define \(N\) as the number of elements in the following collection of subsets:

$$
\left\{ I \subseteq \{1, 2, \cdots, n\} : \prod_{i \in I} a_i \geq 1 \right\}.
$$
Find the maximum possible value of \(N\).
0 replies
steven_zhang123
3 minutes ago
0 replies
Interesting Inequality
lbh_qys   1
N 26 minutes ago by nexu
Let $ a, b, c$ be real numbers such that $ (3a-2b-c)(3b-2c-a)(3c-2a-b)\neq 0 $ and $ a + b + c = 3 . $ Prove that
$$ \left( \frac{1}{3a - 2b - c} + \frac{1}{3b - 2c - a} + \frac{1}{3c - 2a - b} \right)^2 + a^2 + b^2 + c^2 \geq 3 + 3\sqrt{\frac 27}$$
1 reply
lbh_qys
2 hours ago
nexu
26 minutes ago
Goofy geometry
giangtruong13   1
N 29 minutes ago by nabodorbuco2
Source: A Specialized School's Math Entrance Exam
Given the circle $(O)$, from $A$ outside the circle, draw tangents $AE,AF$ ($E,F$ are tangential points) and secant $ABC$ ($B,C$ lie on circle $O$, $B$ is between $A$ and $C$). $OA$ intersects $EF$ at $H$; $I$ is midpoint of $BC$. The line crossing $I$, paralleling with $CE$, intersects $EF$ at $D$. $CD$ intersects $AE$ at $K$. Let $N$ lie inside the triangle $FBC$ such that: $AF$=$AN$. From $N$ draw chords $BQ$, $RC$, $FP$ on circle $(O)$. Prove that: $PRQ$ is a isosceles triangle
1 reply
giangtruong13
Yesterday at 4:23 PM
nabodorbuco2
29 minutes ago
Rainbow vertices
goodar2006   3
N 38 minutes ago by Sina_Sa
Source: Iran 3rd round 2012-Combinatorics exam-P1
We've colored edges of $K_n$ with $n-1$ colors. We call a vertex rainbow if it's connected to all of the colors. At most how many rainbows can exist?

Proposed by Morteza Saghafian
3 replies
goodar2006
Sep 20, 2012
Sina_Sa
38 minutes ago
Geometry hard problem.
noneofyou34   1
N 42 minutes ago by User21837561
In a circle of radius R, three chords of length R are given. Their ends are joined with segments to
obtain a hexagon inscribed in the circle. Show that the midpoints of the new chords are the vertices of
an equilateral triang
1 reply
noneofyou34
2 hours ago
User21837561
42 minutes ago
Prove that two different boards can be obtained
hectorleo123   2
N 43 minutes ago by alpha31415
Source: 2014 Peru Ibero TST P2
Let $n\ge 4$ be an integer. You have two $n\times n$ boards. Each board contains the numbers $1$ to $n^2$ inclusive, one number per square, arbitrarily arranged on each board. A move consists of exchanging two rows or two columns on the first board (no moves can be made on the second board). Show that it is possible to make a sequence of moves such that for all $1 \le i \le n$ and $1 \le j \le n$, the number that is in the $i-th$ row and $j-th$ column of the first board is different from the number that is in the $i-th$ row and $j-th$ column of the second board.
2 replies
hectorleo123
Sep 15, 2023
alpha31415
43 minutes ago
Inspired by lbh_qys
sqing   2
N 2 hours ago by JARP091
Source: Own
Let $ a, b $ be real numbers such that $ (a-3)(b-3)(a-b)\neq 0 $ and $ a + b =6 . $ Prove that
$$ \left( \frac{a}{b - 3} + \frac{b}{3 - a} + \frac{3}{a - b} \right)^2 + 2(a^2 + b^2 )\geq54$$Equality holds when $ (a,b)=\left(\frac{3}{2},\frac{9}{2}\right)$
$$\left( \frac{a+1}{b - 3} + \frac{b+1}{3 - a} + \frac{4}{a - b} \right)^2 + 2(a^2 + b^2 )\geq 60$$Equality holds when $ (a,b)=\left(3-\sqrt 3,3+\sqrt 3\right)$
$$  \left( \frac{a+3}{b - 3} + \frac{b+3}{3 - a} + \frac{6}{a - b} \right)^2 + 2\left(a^2 + b^2\right)\geq 72$$Equality holds when $ (a,b)=\left(3-\frac{3}{\sqrt 2},3+\frac{3}{\sqrt 2}\right)$
2 replies
sqing
2 hours ago
JARP091
2 hours ago
Interesting Inequality
lbh_qys   4
N 2 hours ago by lbh_qys
Given that \( a, b, c \) are pairwise distinct $\mathbf{real}$ numbers and \( a + b + c = 9 \), find the minimum value of

\[
\left( \frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b} \right)^2  + a^2 + b^2 + c^2.
\]
4 replies
2 viewing
lbh_qys
6 hours ago
lbh_qys
2 hours ago
Sum of complex fractions is an integer
Miquel-point   2
N 2 hours ago by cazanova19921
Source: Romanian IMO TST 1981, Day 3 P4
Let $n\geqslant 3$ be a fixed integer and $\omega=\cos\dfrac{2\pi}n+i\sin\dfrac{2\pi}n$.
Show that for every $a\in\mathbb{C}$ and $r>0$, the number
\[\sum\limits_{k=1}^n \dfrac{|a-r\omega^k|^2}{|a|^2+r^2}\]is an integer. Interpet this result geometrically.

Octavian Stănășilă
2 replies
1 viewing
Miquel-point
Apr 6, 2025
cazanova19921
2 hours ago
mexico 1999
cuenca   1
N 3 hours ago by FrancoGiosefAG
A polygon has each side integral and each pair of adjacent sides perpendicular (it is not necessarily convex). Show that if it can be covered by non-overlapping $2 x 1$ dominos, then at least one of its sides has even length.
1 reply
cuenca
Nov 18, 2006
FrancoGiosefAG
3 hours ago
segment of projections is half as sidelength, right triangle inscribed in right
parmenides51   4
N 3 hours ago by sunken rock
Source: 2020 Austrian Federal Competition For Advanced Students, Part 1, p2
Let $ABC$ be a right triangle with a right angle in $C$ and a circumcenter $U$. On the sides $AC$ and $BC$, the points $D$ and $E$ lie in such a way that $\angle EUD = 90 ^o$. Let $F$ and $G$ be the projection of $D$ and $E$ on $AB$, respectively. Prove that $FG$ is half as long as $AB$.

(Walther Janous)
4 replies
parmenides51
Nov 22, 2020
sunken rock
3 hours ago
Hard Inequality
danilorj   2
N 3 hours ago by sqing
Let $a, b, c > 0$ with $a + b + c = 1$. Prove that:
\[
\sqrt{a + (b - c)^2} + \sqrt{b + (c - a)^2} + \sqrt{c + (a - b)^2} \geq \sqrt{3},
\]with equality if and only if $a = b = c = \frac{1}{3}$.
2 replies
1 viewing
danilorj
Today at 5:17 AM
sqing
3 hours ago
Interesting inequalities
sqing   6
N 3 hours ago by sqing
Source: Own
Let $a,b,c \geq 0 $ and $ab+bc+ca- abc =3.$ Show that
$$a+k(b+c)\geq 2\sqrt{3 k}$$Where $ k\geq 1. $
Let $a,b,c \geq 0 $ and $2(ab+bc+ca)- abc =31.$ Show that
$$a+k(b+c)\geq \sqrt{62k}$$Where $ k\geq 1. $
6 replies
sqing
May 16, 2025
sqing
3 hours ago
hexagon area = 1/3 triangle area, plus concurrency question
parmenides51   1
N 3 hours ago by FrancoGiosefAG
Source: Mexican Mathematical Olympiad 1999 OMM P3
A point $P$ is given inside a triangle $ABC$. Let $D,E,F$ be the midpoints of $AP,BP,CP$, and let $L,M,N$ be the intersection points of $ BF$ and $CE, AF$ and $CD, AE$ and $BD$, respectively.
(a) Prove that the area of hexagon $DNELFM$ is equal to one third of the area of triangle $ABC$.
(b) Prove that $DL,EM$, and $FN$ are concurrent.
1 reply
parmenides51
Jul 28, 2018
FrancoGiosefAG
3 hours ago
computational, right triangle inscribed related
parmenides51   1
N Nov 27, 2020 by rafaello
Source: 2019 VU MIF Olympiad XI, XI p3 - Vilnius University, Lithuania
The point $C \ne A, B$ belongs to a circle of diameter $AB$. Point $D$ divides the shorter arc $BC$ in half. Lines $AD$ and $BC$ intersects at point $E$. Find the length of the segment $AB$ if $CE = 3$, $BD = 2\sqrt5$ .
1 reply
parmenides51
Oct 4, 2020
rafaello
Nov 27, 2020
computational, right triangle inscribed related
G H J
G H BBookmark kLocked kLocked NReply
Source: 2019 VU MIF Olympiad XI, XI p3 - Vilnius University, Lithuania
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parmenides51
30653 posts
#1
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The point $C \ne A, B$ belongs to a circle of diameter $AB$. Point $D$ divides the shorter arc $BC$ in half. Lines $AD$ and $BC$ intersects at point $E$. Find the length of the segment $AB$ if $CE = 3$, $BD = 2\sqrt5$ .
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rafaello
1079 posts
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Oh, d*mn, I wonder is there an easier solution. I actually hoped that you do not have to use trig, so I looked like dumb trying angle-bisector and Ptolemy and stuff.

$\sin{\angle DAB}=\frac{DB}{2R}$
$\sin{(90^\circ-2\angle DAB)}=\frac{AC}{2R}$
$\tan{(90^\circ-\angle DAB)}=\frac{AC}{CE}$
$\sin{(90^\circ-2\angle DAB)}=\frac{\tan{(90^\circ-\angle DAB)}CE}{2R}$
$$\sin{(90^\circ-2\angle DAB)}=\frac{\sin{(90^\circ-\angle DAB)}CE}{\cos{(90^\circ-\angle DAB)}2R}=\frac{\sin{(90^\circ-\angle DAB)}CE}{\sin{\angle DAB}\cdot 2R}$$$$\cos{2\angle DAB}=\frac{\cos{\angle DAB} \cdot CE}{DB}$$$2\sqrt5(2\cos^2{\angle DAB}-1)=3\cos{\angle DAB}$
$4\sqrt5 \cos^2{\angle DAB}-3\cos{\angle DAB}-2\sqrt5=0$
$\cos{\angle DAB}=\frac{3\pm\sqrt{9-4\cdot (-2\sqrt5)4\sqrt5  }}{2\cdot 4\sqrt5}=\frac{3\pm13}{2\cdot 4\sqrt5}=\frac{3\pm13}{2\cdot 4\sqrt5}$.
Negative value is false, since then $\angle DAB>90^\circ$, which means that $\angle CAB>180^\circ$, which is false since max $\angle CAB$ value can be $\to 90^\circ$ (approaching to it) .
Hence, $\cos{\angle DAB}=\frac{2\sqrt5}{5}$
Let $F=BC\cap OD$, where $O$ is the centre of that circle.
Thus, $\frac{BF}{BD}=\cos{\angle DAB}=\frac{2\sqrt5}{5}\implies BF=4$.
Thus, $BC=8$.
And we have that $\cos{\angle CAB}=\cos{2\angle DAB}=2\cos^2{\angle DAB}-1=\frac{8\cdot 5}{25}-1=\frac{15}{25}=\frac{3}{5}$.
$\frac{AC}{AB}=\cos{\angle CAB}=\frac{3}{5}\implies AC=\frac{3\cdot AB}{5}$
By Pythagorean theorem, $AC^2+BC^2=AB^2$, we get that $$\frac{9\cdot AB^2}{25}+64=AB^2\implies \frac{16\cdot AB^2}{25}=64\implies AB^2=100\implies \boxed{AB=10}.$$
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