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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=b
k2c901_1   89
N 3 minutes ago by reni_wee
Source: Taiwan 1st TST 2006, 1st day, problem 3
Let $a$, $b$ be positive integers such that $b^n+n$ is a multiple of $a^n+n$ for all positive integers $n$. Prove that $a=b$.

Proposed by Mohsen Jamali, Iran
89 replies
k2c901_1
Mar 29, 2006
reni_wee
3 minutes ago
Strange angle condition and concyclic points
lminsl   129
N 7 minutes ago by Aiden-1089
Source: IMO 2019 Problem 2
In triangle $ABC$, point $A_1$ lies on side $BC$ and point $B_1$ lies on side $AC$. Let $P$ and $Q$ be points on segments $AA_1$ and $BB_1$, respectively, such that $PQ$ is parallel to $AB$. Let $P_1$ be a point on line $PB_1$, such that $B_1$ lies strictly between $P$ and $P_1$, and $\angle PP_1C=\angle BAC$. Similarly, let $Q_1$ be the point on line $QA_1$, such that $A_1$ lies strictly between $Q$ and $Q_1$, and $\angle CQ_1Q=\angle CBA$.

Prove that points $P,Q,P_1$, and $Q_1$ are concyclic.

Proposed by Anton Trygub, Ukraine
129 replies
lminsl
Jul 16, 2019
Aiden-1089
7 minutes ago
Simple inequality
sqing   12
N 17 minutes ago by Rayvhs
Source: MEMO 2018 T1
Let $a,b$ and $c$ be positive real numbers satisfying $abc=1.$ Prove that$$\frac{a^2-b^2}{a+bc}+\frac{b^2-c^2}{b+ca}+\frac{c^2-a^2}{c+ab}\leq a+b+c-3.$$
12 replies
sqing
Sep 2, 2018
Rayvhs
17 minutes ago
Random concyclicity in a square config
Maths_VC   2
N 19 minutes ago by Maths_VC
Source: Serbia JBMO TST 2025, Problem 1
Let $M$ be a random point on the smaller arc $AB$ of the circumcircle of square $ABCD$, and let $N$ be the intersection point of segments $AC$ and $DM$. The feet of the tangents from point $D$ to the circumcircle of the triangle $OMN$ are $P$ and $Q$ , where $O$ is the center of the square. Prove that points $A$, $C$, $P$ and $Q$ lie on a single circle.
2 replies
Maths_VC
an hour ago
Maths_VC
19 minutes ago
Serbian selection contest for the IMO 2025 - P3
OgnjenTesic   3
N 22 minutes ago by atdaotlohbh
Source: Serbian selection contest for the IMO 2025
Find all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that:
- $f$ is strictly increasing,
- there exists $M \in \mathbb{N}$ such that $f(x+1) - f(x) < M$ for all $x \in \mathbb{N}$,
- for every $x \in \mathbb{Z}$, there exists $y \in \mathbb{Z}$ such that
\[
            f(y) = \frac{f(x) + f(x + 2024)}{2}.
        \]Proposed by Pavle Martinović
3 replies
OgnjenTesic
May 22, 2025
atdaotlohbh
22 minutes ago
Easy P4 combi game with nt flavour
Maths_VC   0
33 minutes ago
Source: Serbia JBMO TST 2025, Problem 4
Two players, Alice and Bob, play the following game, taking turns. In the beginning, the number $1$ is written on the board. A move consists of adding either $1$, $2$ or $3$ to the number written on the board, but only if the chosen number is coprime with the current number (for example, if the current number is $10$, then in a move a player can't choose the number $2$, but he can choose either $1$ or $3$). The player who first writes a perfect square on the board loses. Prove that one of the players has a winning strategy and determine who wins in the game.
0 replies
Maths_VC
33 minutes ago
0 replies
USAMO 2003 Problem 1
MithsApprentice   70
N 34 minutes ago by endless_abyss
Prove that for every positive integer $n$ there exists an $n$-digit number divisible by $5^n$ all of whose digits are odd.
70 replies
MithsApprentice
Sep 27, 2005
endless_abyss
34 minutes ago
Basic ideas in junior diophantine equations
Maths_VC   0
40 minutes ago
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
0 replies
Maths_VC
40 minutes ago
0 replies
Two permutations
Nima Ahmadi Pour   13
N 43 minutes ago by awesomeming327.
Source: Iran prepration exam
Suppose that $ a_1$, $ a_2$, $ \ldots$, $ a_n$ are integers such that $ n\mid a_1 + a_2 + \ldots + a_n$.
Prove that there exist two permutations $ \left(b_1,b_2,\ldots,b_n\right)$ and $ \left(c_1,c_2,\ldots,c_n\right)$ of $ \left(1,2,\ldots,n\right)$ such that for each integer $ i$ with $ 1\leq i\leq n$, we have
\[ n\mid a_i - b_i - c_i
\]

Proposed by Ricky Liu & Zuming Feng, USA
13 replies
Nima Ahmadi Pour
Apr 24, 2006
awesomeming327.
43 minutes ago
Easy but unusual junior ineq
Maths_VC   0
an hour ago
Source: Serbia JBMO TST 2025, Problem 2
Real numbers $x, y$ $\ge$ $0$ satisfy $1$ $\le$ $x^2 + y^2$ $\le$ $5$. Determine the minimal and the maximal value of the expression $2x + y$
0 replies
Maths_VC
an hour ago
0 replies
p is an odd prime congruent to 2 mod 3
Valentin Vornicu   9
N an hour ago by reni_wee
Source: Balkan MO 1999, Problem 2
Let $p$ be an odd prime congruent to 2 modulo 3. Prove that at most $p-1$ members of the set $\{m^2 - n^3 - 1 \mid 0 < m,\ n < p\}$ are divisible by $p$.
9 replies
Valentin Vornicu
Apr 24, 2006
reni_wee
an hour ago
Circle is tangent to circumcircle and incircle
ABCDE   74
N an hour ago by zuat.e
Source: 2016 ELMO Problem 6
Elmo is now learning olympiad geometry. In triangle $ABC$ with $AB\neq AC$, let its incircle be tangent to sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. The internal angle bisector of $\angle BAC$ intersects lines $DE$ and $DF$ at $X$ and $Y$, respectively. Let $S$ and $T$ be distinct points on side $BC$ such that $\angle XSY=\angle XTY=90^\circ$. Finally, let $\gamma$ be the circumcircle of $\triangle AST$.

(a) Help Elmo show that $\gamma$ is tangent to the circumcircle of $\triangle ABC$.

(b) Help Elmo show that $\gamma$ is tangent to the incircle of $\triangle ABC$.

James Lin
74 replies
ABCDE
Jun 24, 2016
zuat.e
an hour ago
Division on 1989
mistakesinsolutions   3
N an hour ago by reni_wee
Prove that for positive integer $n$ greater than $3,$ $n^{n^{n^n}} - n^{n^n}$ is divisible by $1989.$
3 replies
mistakesinsolutions
Jun 14, 2023
reni_wee
an hour ago
exponential diophantine in integers
skellyrah   0
2 hours ago
find all integers x,y,z such that $$ 45^x = 5^y + 2000^z $$
0 replies
skellyrah
2 hours ago
0 replies
Prove excircle is tangent to circumcircle
sarjinius   8
N Apr 24, 2025 by Lyzstudent
Source: Philippine Mathematical Olympiad 2025 P4
Let $ABC$ be a triangle with incenter $I$, and let $D$ be a point on side $BC$. Points $X$ and $Y$ are chosen on lines $BI$ and $CI$ respectively such that $DXIY$ is a parallelogram. Points $E$ and $F$ are chosen on side $BC$ such that $AX$ and $AY$ are the angle bisectors of angles $\angle BAE$ and $\angle CAF$ respectively. Let $\omega$ be the circle tangent to segment $EF$, the extension of $AE$ past $E$, and the extension of $AF$ past $F$. Prove that $\omega$ is tangent to the circumcircle of triangle $ABC$.
8 replies
sarjinius
Mar 9, 2025
Lyzstudent
Apr 24, 2025
Prove excircle is tangent to circumcircle
G H J
Source: Philippine Mathematical Olympiad 2025 P4
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sarjinius
249 posts
#1 • 4 Y
Y by MathLuis, mpcnotnpc, JollyEggsBanana, Rounak_iitr
Let $ABC$ be a triangle with incenter $I$, and let $D$ be a point on side $BC$. Points $X$ and $Y$ are chosen on lines $BI$ and $CI$ respectively such that $DXIY$ is a parallelogram. Points $E$ and $F$ are chosen on side $BC$ such that $AX$ and $AY$ are the angle bisectors of angles $\angle BAE$ and $\angle CAF$ respectively. Let $\omega$ be the circle tangent to segment $EF$, the extension of $AE$ past $E$, and the extension of $AF$ past $F$. Prove that $\omega$ is tangent to the circumcircle of triangle $ABC$.
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ilovemath0402
188 posts
#2
Y by
bump bump this problem is so nice
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sarjinius
249 posts
#3
Y by
ilovemath0402 wrote:
bump bump this problem is so nice

Thanks, I proposed this problem :)
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SimplisticFormulas
123 posts
#4
Y by
what’s the solution? I am completely stuck
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MathLuis
1556 posts
#5 • 4 Y
Y by drago.7437, sarjinius, Mysteriouxxx, radian_51
Well this geo is really amazing I have to say...solved in around 30 mins but I think this could even be around 30-35 MOHS because the way to find things on this problem requires deep intuition.
Let $BI \cap (ABC)=M_B$ and $CI \cap (ABC)=M_C$, also let $N_A$ be midpoint of arc $BAC$ on $(ABC)$, now let reflections of $D$ over $EX, FY, BI, CI, Y \infty_{\perp CI}, X \infty_{\perp BI}$ be $D_B, D_C, L', K', K, L$ respectively now let reflection of $D_B$ over $AX$ be $L_1$ and reflection of $D_C$ over $AY$ be $K_1$.
Using the paralelogram we can easly see from the direction of the reflections that $KK'$ and $LL'$ are diameters on $(Y, YD), (X, XD)$ respectively, now let $I_B, I_C$ be the $B,C$ excenters of $\triangle ABC$ then notice we have $\measuredangle II_CA=\measuredangle CBI=\measuredangle CDY=\measuredangle YK'C$ which implies $I_CAK'Y$ cyclic and similarily $I_BAL'X$ is cyclic however since $\measuredangle CDY=\measuredangle YD_CF$ we also get that $I_CAK'YD_C$ is cyclic and similarily $I_BAL'XD_B$ is cyclic, however it doesn't end here...
Now notice that $YK'=YD_C$ so $Y$ is midpoint of arc $K'D_C$ on $(I_CAK')$ however $D, K'$ are symetric in $CI$ which means both $I_CD, I_CD'$ are reflections of $I_CK'$ over $CI$ and thus $I_C, D, D_C$ are colinear, and similarily $I_B, D, D_B$ are colinear.
Now $\measuredangle CDY=\measuredangle YD_CA=\measuredangle AK_1Y$ which means $CK_1YD$ is cyclic and similarily we have $L_1BXD$ cyclic, but also note that $\measuredangle L_1DL=\measuredangle L_1L'L=\measuredangle AI_BX=\measuredangle ACI=\measuredangle K_1DY$ which means that $L_1, D, K_1$ are colinear.
Now from here notice that $\measuredangle DL_1A=\measuredangle DXI=\measuredangle IYD=\measuredangle AK_1D$ which does in fact show that $\triangle L_1AK_1$ is isosceles and therefore $AK_1=AL_1$, and from reflections this gives $AD_B=AD_C$, but notice from other reflections we have $D_BG=DG=D_CG$ where $EX \cap FY=G$ (clearly then $G$ is A-excenter of $\triangle EAF$), but now also note that we have $\measuredangle AD_BG=\measuredangle GDE=\measuredangle AD_CG$ which means that $AD_BGD_C$ is cyclic but by summing arcs we end up realising $AG$ is diameter and in fact now this means $(D_BDD_C)$ is $\omega$ from the tangencies.
To finish let $J$ be the miquelpoint of $L_1BCK_1$ then $J$ lies on $(ABC)$ but also from Reim's we get $N_A, D, J$ colinear and then Reim's twice gives $M_CX \cap M_BY=J$ and from double Reim's once again we have that $(AL'X) \cap (AK'Y)=J$ and this is excellent news because now we can note that $\measuredangle D_CJD_B=\measuredangle D_CJA+\measuredangle AJD_B=\measuredangle D_CI_CA+\measuredangle AI_BD_B=\measuredangle D_CDD_B$ which shows that $J$ lies on $\omega$ as well, but since $N_A, D, J$ are colinear from the converse of Archiemedes Lemma (or just shooting Lemma/homothety) we have that $\omega, (ABC)$ are tangent at $J$ as desired thus we are done :cool:
This post has been edited 1 time. Last edited by MathLuis, Mar 13, 2025, 8:16 PM
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AndreiVila
210 posts
#6 • 1 Y
Y by Lyzstudent
Notice that $X$ and $Y$ are the incenters of $\triangle ABE$ and $\triangle ACF$. Let $X'$ and $Y'$ be the projections of $X$ and $Y$ onto $BC$. Let $T$ and $S$ be the projections of $X$ onto $AE$ and $AB$ respectively, and let $K$ be the tangency point of $\omega$ with $AE$.

Claim 1. The $A$-excircle of $\triangle AEF$ is tangent to $EF$ in $D$.
Proof: Since $IXDY$ is a parallelogram, by projecting onto $BC$ we get that $BX'+BY'=BT+BD$. This is equivalent to $$BE+AB-AE+2BF+FC+AF-AC=BA+BC-AC+2BD.$$Simplifying yields $AE+ED=AF+FD$, which is equivalent to $D$ being the tangency point of the excircle.

Claim 2. Circle $\omega$ is tangent to $(ABC)$.
Proof: By Casey's Theorem, we need to prove that $$b\cdot BD + c\cdot CD = a\cdot AK.$$But $$AK=AT+TK=AT+X'D=AT-BX'+BD=AS-BS+BD=c-2BS+BD.$$With Thales' Theorem, $\frac{BS}{p-b}=\frac{BX}{BI}=\frac{BD}{a},$ so $BS=\frac{BD(p-b)}{a},$ thus getting $AK=\frac{ac+BD(b-c)}{a},$ and the conclusion follows.
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SimplisticFormulas
123 posts
#7
Y by
I found that $X,Y$ are in centres, $XE$ meets $YF$ in $Z=$$A$- excentre of $AEF$ and that$A$ appears to be Miquel point of $IXYZ$.
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markam
4 posts
#10
Y by
sarjinius, what solution did you have in mind at first, when you proposed this problem?
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Lyzstudent
1 post
#11
Y by
AndreiVila wrote:
Notice that $X$ and $Y$ are the incenters of $\triangle ABE$ and $\triangle ACF$. Let $X'$ and $Y'$ be the projections of $X$ and $Y$ onto $BC$. Let $T$ and $S$ be the projections of $X$ onto $AE$ and $AB$ respectively, and let $K$ be the tangency point of $\omega$ with $AE$.

Claim 1. The $A$-excircle of $\triangle AEF$ is tangent to $EF$ in $D$.
Proof: Since $IXDY$ is a parallelogram, by projecting onto $BC$ we get that $BX'+BY'=BT+BD$. This is equivalent to $$BE+AB-AE+2BF+FC+AF-AC=BA+BC-AC+2BD.$$Simplifying yields $AE+ED=AF+FD$, which is equivalent to $D$ being the tangency point of the excircle.

Claim 2. Circle $\omega$ is tangent to $(ABC)$.
Proof: By Casey's Theorem, we need to prove that $$b\cdot BD + c\cdot CD = a\cdot AK.$$But $$AK=AT+TK=AT+X'D=AT-BX'+BD=AS-BS+BD=c-2BS+BD.$$With Thales' Theorem, $\frac{BS}{p-b}=\frac{BX}{BI}=\frac{BD}{a},$ so $BS=\frac{BD(p-b)}{a},$ thus getting $AK=\frac{ac+BD(b-c)}{a},$ and the conclusion follows.
Excellent!!!Much better than the solution above.
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