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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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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
(a-1)(b-1)(c-1) is a divisor of abc-1
ehsan2004   22
N 36 minutes ago by reni_wee
Source: IMO 1992, Day 1, Problem 1
Find all integers $\,a,b,c\,$ with $\,1<a<b<c\,$ such that \[ (a-1)(b-1)(c-1)  \] is a divisor of $abc-1.$
22 replies
ehsan2004
Jan 22, 2005
reni_wee
36 minutes ago
Balkan MO 2025 p1
Mamadi   0
an hour ago
Source: Balkan MO 2025
An integer \( n > 1 \) is called good if there exists a permutation \( a_1, a_2, \dots, a_n \) of the numbers \( 1, 2, 3, \dots, n \), such that:

\( a_i \) and \( a_{i+1} \) have different parities for every \( 1 \le i \le n - 1 \)

the sum \( a_1 + a_2 + \dots + a_k \) is a quadratic residue modulo \( n \) for every \( 1 \le k \le n \)

Prove that there exist infinitely many good numbers, as well as infinitely many positive integers which are not good.

Remark: Here an integer \( z \) is considered a quadratic residue modulo \( n \) if there exists an integer \( y \) such that \( y^2 \equiv z \pmod{n} \).
0 replies
+2 w
Mamadi
an hour ago
0 replies
Number theory
MathsII-enjoy   3
N an hour ago by KevinYang2.71
Prove that when $x^p+y^p$ | $(p^2-1)^n$ with $x,y$ are positive integers and $p$ is prime ($p>3$), we get: $x=y$
3 replies
MathsII-enjoy
Yesterday at 3:22 PM
KevinYang2.71
an hour ago
Inspired by Bet667
sqing   1
N an hour ago by ytChen
Source: Own
Let $ a,b $ be a real numbers such that $a^2+kab+b^2\ge a^3+b^3.$Prove that$$a+b\leq k+2$$Where $ k\geq 0. $
1 reply
sqing
5 hours ago
ytChen
an hour ago
4-var inequality
sqing   1
N an hour ago by arqady
Source: SXTB (4)2025 Q2837
Let $ a,b,c,d> 0  $. Prove that
$$   \frac{1}{(3a+1)^4}+ \frac{1}{(3b+1)^4}+\frac{1}{(3c+1)^4}+\frac{1}{(3d+1)^4} \geq \frac{1}{16(3abcd+1)}$$
1 reply
sqing
5 hours ago
arqady
an hour ago
Extremaly hard inequality
blug   1
N an hour ago by arqady
Source: Polish Math Olympiad Training Camp 2024
Let $a, b, c$ be non-negative real numbers. Prove that
$$a+b+c+\sqrt{a^2+b^2+c^2-ab-bc-ca}\geq \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+c^2}.$$Looking for an algebraic solution!
1 reply
blug
2 hours ago
arqady
an hour ago
Common external tangents of two circles
a1267ab   56
N 2 hours ago by wassupevery1
Source: USA Winter TST for IMO 2020, Problem 2, by Merlijn Staps
Two circles $\Gamma_1$ and $\Gamma_2$ have common external tangents $\ell_1$ and $\ell_2$ meeting at $T$. Suppose $\ell_1$ touches $\Gamma_1$ at $A$ and $\ell_2$ touches $\Gamma_2$ at $B$. A circle $\Omega$ through $A$ and $B$ intersects $\Gamma_1$ again at $C$ and $\Gamma_2$ again at $D$, such that quadrilateral $ABCD$ is convex.

Suppose lines $AC$ and $BD$ meet at point $X$, while lines $AD$ and $BC$ meet at point $Y$. Show that $T$, $X$, $Y$ are collinear.

Merlijn Staps
56 replies
a1267ab
Dec 16, 2019
wassupevery1
2 hours ago
That's Vietnamese geo!
wassupevery1   8
N 2 hours ago by cj13609517288
Source: 2025 Vietnam National Olympiad - Problem 3
Let $ABC$ be an acute, scalene triangle with circumcenter $O$, circumcircle $(O)$, orthocenter $H$. Line $AH$ meets $(O)$ again at $D \neq A$. Let $E, F$ be the midpoint of segments $AB, AC$ respectively. The line through $H$ and perpendicular to $HF$ meets line $BC$ at $K$.
a) Line $DK$ meets $(O)$ again at $Y \neq D$. Prove that the intersection of line $BY$ and the perpendicular bisector of $BK$ lies on the circumcircle of triangle $OFY$.
b) The line through $H$ and perpendicular to $HE$ meets line $BC$ at $L$. Line $DL$ meets $(O)$ again at $Z \neq D$. Let $M$ be the intersection of lines $BZ, OE$; $N$ be the intersection of lines $CY, OF$; $P$ be the intersection of lines $BY, CZ$. Let $T$ be the intersection of lines $YZ, MN$ and $d$ be the line through $T$ and perpendicular to $OA$. Prove that $d$ bisects $AP$.
8 replies
wassupevery1
Dec 25, 2024
cj13609517288
2 hours ago
Equation has no integer solution.
Learner94   33
N 2 hours ago by bjump
Source: INMO 2013
Let $a,b,c,d \in \mathbb{N}$ such that $a \ge b \ge c \ge d $. Show that the equation $x^4 - ax^3 - bx^2 - cx -d = 0$ has no integer solution.
33 replies
Learner94
Feb 3, 2013
bjump
2 hours ago
Problem 1 — Symmetric Squares, Symmetric Products
RockmanEX3   8
N 2 hours ago by Baimukh
Source: 46th Austrian Mathematical Olympiad National Competition Part 1 Problem 1
Let $a$, $b$, $c$, $d$ be positive numbers. Prove that

$$(a^2 + b^2 + c^2 + d^2)^2 \ge (a+b)(b+c)(c+d)(d+a)$$
When does equality hold?

(Georg Anegg)
8 replies
RockmanEX3
Jul 14, 2018
Baimukh
2 hours ago
Math solution
Techno0-8   0
2 hours ago
Solution
0 replies
Techno0-8
2 hours ago
0 replies
D1027 : Super Schoof
Dattier   0
2 hours ago
Source: les dattes à Dattier
Let $p>11$ a prime number with $a=\text{card}\{(x,y) \in \mathbb Z/ p \mathbb Z: y^2=x^3+1\}$ and $b=\dfrac 1 {((p-1)/2)! \times ((p-1)/3)! \times ((p-1)/6)!} \mod p$ when $p \mod 3=1$.



Is it true that if $p \mod 3=1$ then $a \in \{b,p-b, \min\{b,p-b\}+p\}$ else $A=p$.
0 replies
Dattier
2 hours ago
0 replies
Interesting inequality
sealight2107   0
3 hours ago
Source: Own
Let $a,b,c>0$ such that $a+b+c=3$. Find the minimum value of:
$Q=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc}$
0 replies
sealight2107
3 hours ago
0 replies
Bosnia and Herzegovina 2022 IMO TST P1
Steve12345   3
N 3 hours ago by waterbottle432
Let $ABC$ be a triangle such that $AB=AC$ and $\angle BAC$ is obtuse. Point $O$ is the circumcenter of triangle $ABC$, and $M$ is the reflection of $A$ in $BC$. Let $D$ be an arbitrary point on line $BC$, such that $B$ is in between $D$ and $C$. Line $DM$ cuts the circumcircle of $ABC$ in $E,F$. Circumcircles of triangles $ADE$ and $ADF$ cut $BC$ in $P,Q$ respectively. Prove that $DA$ is tangent to the circumcircle of triangle $OPQ$.
3 replies
Steve12345
May 22, 2022
waterbottle432
3 hours ago
Kiepert triangle problem
VUThanhTung   5
N May 5, 2015 by VUThanhTung
Source: Own
Consider a triangle $ABC$. Let $A_\alpha B_\alpha C_\alpha $ be the points satisfying
$ \angle A_\alpha BC=\angle A_\alpha CB =\angle B_\alpha CA=\angle B_\alpha AC=\angle C_\alpha AB=\angle C_\alpha BA=\alpha $ ($\alpha=0$ to $2\pi$).
Let $ K_\alpha=AA_\alpha\cap BB_\alpha\cap CC_\alpha $ and $L_\alpha$ be the isogonal conjugate of $  K_\alpha $ WRT $\triangle ABC$. $\theta$ is fixed and $\beta$ is varying.
1. Prove that $L_{-\theta}$, $K_\beta$, $K_{\theta-\beta}$ are collinear.
2. Prove that $K_\beta K_{\theta+\beta}$ is tangent to a fixed conic $c_\theta$.
5 replies
VUThanhTung
Mar 28, 2015
VUThanhTung
May 5, 2015
Kiepert triangle problem
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G H BBookmark kLocked kLocked NReply
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VUThanhTung
67 posts
#1 • 2 Y
Y by Adventure10, Mango247
Consider a triangle $ABC$. Let $A_\alpha B_\alpha C_\alpha $ be the points satisfying
$ \angle A_\alpha BC=\angle A_\alpha CB =\angle B_\alpha CA=\angle B_\alpha AC=\angle C_\alpha AB=\angle C_\alpha BA=\alpha $ ($\alpha=0$ to $2\pi$).
Let $ K_\alpha=AA_\alpha\cap BB_\alpha\cap CC_\alpha $ and $L_\alpha$ be the isogonal conjugate of $  K_\alpha $ WRT $\triangle ABC$. $\theta$ is fixed and $\beta$ is varying.
1. Prove that $L_{-\theta}$, $K_\beta$, $K_{\theta-\beta}$ are collinear.
2. Prove that $K_\beta K_{\theta+\beta}$ is tangent to a fixed conic $c_\theta$.
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TelvCohl
2312 posts
#4 • 4 Y
Y by VUThanhTung, Leooooo, Adventure10, Mango247
1.

My solution:

Lemma 1 :

$ K_{\phi} L_{-\phi} $ pass through the Centroid $ G $ of $ \triangle ABC $

Proof of lemma 1:

From post #3 (remark) and post #4 in Property of Kariya point $ \Longrightarrow K_{\phi}K_{-\phi} $ pass through the Symmedian point of $ \triangle ABC $ ,
so from the lemma mentioned at post #3 in the topic Angle related to Fermat points and Isodynamic points we get $  K_{\phi} L_{-\phi} $ pass through the isogonal conjugate of the Symmedian point of $ \triangle ABC $ . i.e. $K_{\phi} L_{-\phi} $ pass through the Centroid $ G $ of $ \triangle ABC $
________________________________________
Lemma 2 :

Let $ A_{\phi}^* $ be the isogonal conjugate of $ A_{\phi} $ WRT $ \triangle ABC $ .

Then $ A_{\phi}^* A_{90^{\circ}-\phi} $ pass through the orthocenter $ H $ of $ \triangle ABC $

Proof of lemma 2 :

Since $ \angle A_{\phi}^*BA=\angle A_{\phi}^*CA=180^{\circ}-\phi, \angle A_{90^{\circ}-\phi} BC=\angle A_{90^{\circ}-\phi} CB=90^{\circ}-\phi $ ,
so we get $ \angle A_{\phi}^*BA_{90^{\circ}-\phi}=90^{\circ}-\angle CBA=\angle HCB, \angle A_{\phi}^*CA_{90^{\circ}-\phi}=90^{\circ}-\angle ACB=\angle HBC $ ,
hence from the problem A useful collinearity we get $ H, A_{90^{\circ}-\phi}, A_{\phi}^* $ are collinear .
________________________________________
Lemma 3 :

$ K_{\phi} L_{90^{\circ}-\phi} $ pass through $ H $

Proof of lemma 3 :

Easy to see $ A \in B_{90^{\circ}-\phi}C_{90^{\circ}-\phi}^*, A \in C_{90^{\circ}-\phi}B_{90^{\circ}-\phi}^*, B \in A_{90^{\circ}-\phi}C_{90^{\circ}-\phi}^*, C \in A_{90^{\circ}-\phi}B_{90^{\circ}-\phi}^* $ .

Since $ CB_{\phi} $ is the isogonal conjugate of $ CA_{\phi} $ WRT $ \angle ACB $ ,
so we get $ A_{\phi}^* \in CB_{\phi} $ (Similarly we can prove $ A_{\phi}^* \in BC_{\phi} $) .

From lemma 2 we get $ H \in A_{\phi}^*A_{90^{\circ}-\phi} $ .
Similarly we can prove $ H $ lie on $ B_{\phi}B_{90^{\circ}-\phi}^* $ and $ C_{\phi}C_{90^{\circ}-\phi}^* $ ,
so from Desargue theorem ( for $ \triangle BC_{\phi}C_{90^{\circ}-\phi}^* $ and $ \triangle CB_{\phi}B_{90^{\circ}-\phi}^* $ ) we get $ BC, B_{\phi}C_{\phi}, B_{90^{\circ}-\phi}^*C_{90^{\circ}-\phi}^* $ are concurrent ,
hence from Desargue theorem ( for $ \triangle BB_{\phi}B_{90^{\circ}-\phi}^* $ and $ \triangle CC_{\phi}C_{90^{\circ}-\phi}^* $ ) we get $ K_{\phi}, L_{90^{\circ}-\phi},  H $ are collinear .
( notice that $ K_{\phi} \equiv BB_{\phi} \cap CC_{\phi}, L_{90^{\circ}-\phi} \equiv BB_{90^{\circ}-\phi}^* \cap CC_{90^{\circ}-\phi}^*, H \equiv B_{\phi}B_{90^{\circ}-\phi}^* \cap C_{\phi}C_{90^{\circ}-\phi}^* $ )
________________________________________
Back to the main problem :

Since $ K_{\beta} \mapsto K_{\theta-\beta} $ form an involution on Kiepert hyperbola ,
so $ K_{\beta}  K_{\theta-\beta} $ pass through a fixed point $ T $ (the pole of the involution) when $ \beta $ varies .

When $ \beta=0^{\circ} $, from lemma 1 we get $ K_{\beta}  K_{\theta-\beta} \equiv  G  K_{\theta} $ pass through $ L_{-\theta} $ . ... $ (1) $
When $ \beta=90^{\circ} $ , from lemma 3 we get $ K_{\beta}  K_{\theta-\beta} \equiv H K_{90^{\circ}+\theta} $ pass through $ L_{-\theta} $ . ... $ (2) $
From $ (1), (2) $ we get $ T \equiv L_{-\theta} $ . i.e. $  K_{\beta}  K_{\theta-\beta} $ always pass through $ L_{-\theta} $ when $ \beta $ varies

Q.E.D
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TelvCohl
2312 posts
#5 • 4 Y
Y by VUThanhTung, Leooooo, Adventure10, Mango247
2.

Since $ K_{\beta} \mapsto K_{\theta+\beta} $ is a homography on Kiepert hyperbola ,
so from the problem Homography on a Conic we get $ K_{\beta} K_{\theta+\beta} $ is tangent to a fixed conic .

Q.E.D
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VUThanhTung
67 posts
#6 • 1 Y
Y by Adventure10
wow, thank you for the solution :)
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VUThanhTung
67 posts
#7 • 1 Y
Y by Adventure10
My solution for the part 1 is using Trilinear coordinates:
The trilinear coordinates of $K_\beta$ is $(\csc(A+\beta),\csc(B+\beta),\csc(C+\beta))$
The trilinear coordinates of $K_{\theta-\beta}$ is $(\csc(A+\theta-\beta),\csc(B+\theta-\beta),\csc(C+\theta-\beta))$
The trilinear coordinates of $K_{-\theta}$ is $(\csc(A-\theta),\csc(B-\theta),\csc(C-\theta))$ so the trilinear coordinates of $L_{-\theta}$ is $(1/(\csc(A-\theta),1/\csc(B-\theta),1/\csc(C-\theta))=(\sin(A-\theta),\sin(B-\theta),\sin(C-\theta))$.

$L_{-\theta}$, $K_\beta$, $K_{\theta-\beta}$ are collinear if and only if $\det \begin{vmatrix}\csc(A+\beta)&\csc(B+\beta)&\csc(C+\beta)\\
\csc(A+\theta-\beta)&\csc(B+\theta-\beta)&\csc(C+\theta-\beta)\\\sin(A-\theta)&\sin(B-\theta)&\sin(C-\theta)\end{vmatrix}=0$.
Although the computation of $\det ...$ has not been done yet but we all know that it must be equal to $0$ and we are done :D
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VUThanhTung
67 posts
#8 • 2 Y
Y by Adventure10, Mango247
Here is an interesting collorary of 1. : $K_{\beta}L_{-2\beta}$ is tangent to the Kiepert hyperbola of $\triangle ABC$ :)
This post has been edited 1 time. Last edited by VUThanhTung, May 5, 2015, 11:54 AM
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