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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
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
Problem 2
blug   1
N 8 minutes ago by aidan0626
Source: Polish Junior Math Olympiad Finals 2025
A party is attended by boys and girls. Each person attending the party knows exactly 3 boys and exactly 7 girls among the other people. Prove that the number of all the people attending the party is divisible by 20.
1 reply
blug
35 minutes ago
aidan0626
8 minutes ago
Polynomial produces perfect powers
TheUltimate123   21
N 10 minutes ago by pi271828
Source: ELMO 2023/1
Let \(m\) be a positive integer. Find, in terms of \(m\), all polynomials \(P(x)\) with integer coefficients such that for every integer \(n\), there exists an integer \(k\) such that \(P(k)=n^m\).

Proposed by Raymond Feng
21 replies
TheUltimate123
Jun 26, 2023
pi271828
10 minutes ago
geometry
srnjbr   0
13 minutes ago
the points f,n,o, t a lie in the plane such that the triangles tfo ton are similar, preserving direction and order, and fano is a parallelogram. show that of×on=oa×ot.
0 replies
srnjbr
13 minutes ago
0 replies
Problem 5
blug   0
30 minutes ago
Source: Polish Junior Math Olympiad Finals 2025
Each square on a 5×5 board contains an arrow pointing up, down, left, or right. Show that it is possible to remove exactly 20 arrows from this board so that no two of the remaining five arrows point to the same square.
0 replies
blug
30 minutes ago
0 replies
Diophantine equation with large moduli
Assassino9931   2
N 31 minutes ago by Assassino9931
Source: Bulgaria, Concours Generale Minko Balkanski 2024
Solve in positive integers $2^x - 23^y = 9$.
2 replies
Assassino9931
4 hours ago
Assassino9931
31 minutes ago
Problem 4
blug   0
31 minutes ago
Source: Polish Junior Math Olympiad Finals 2025
In a rhombus $ABCD$, angle $\angle ABC=100^{\circ}$. Point $P$ lies on $CD$ such that $\angle PBC=20^{\circ}$. Line parallel to $AD$ passing trough $P$ intersects $AC$ at $Q$. Prove that $BP=AQ$.
0 replies
blug
31 minutes ago
0 replies
Problem 3
blug   0
33 minutes ago
Source: Polish Junior Math Olympiad Finals 2025
Find all primes $(p, q, r)$ such that
$$pq+4=r^4.$$
0 replies
blug
33 minutes ago
0 replies
Problem 1
blug   0
35 minutes ago
Source: Polish Junior Math Olympiad Finals 2025
Do there exists a tetrahedron, in which the lenghts of the edges are six different integers such that their sum is 25?
0 replies
blug
35 minutes ago
0 replies
A two-variable & non-homogenous inequality that seems hard to me
MyLifeMyChoice   3
N 44 minutes ago by Radin_
Source: Developing from a larger, three-variable one
For $a,b>0$, prove/disprove the following claim: :maybe:

$a^3b^3+\frac{1}{a^3}+\frac{1}{b^3}+3\stackrel{?}{\ge}a^2b+b^2a+\frac{1}{a^2b}+\frac{1}{b^2a}+\frac{a}{b}+\frac{b}{a}$
3 replies
MyLifeMyChoice
Mar 13, 2025
Radin_
44 minutes ago
exponential diophantine with factorials
skellyrah   4
N an hour ago by InftyByond
find all non negative integers (x,y) such that $$ x! + y! = 2025^x + xy$$
4 replies
skellyrah
Feb 24, 2025
InftyByond
an hour ago
Point satisfies triple property
62861   35
N an hour ago by Sanjana42
Source: USA Winter Team Selection Test #2 for IMO 2018, Problem 2
Let $ABCD$ be a convex cyclic quadrilateral which is not a kite, but whose diagonals are perpendicular and meet at $H$. Denote by $M$ and $N$ the midpoints of $\overline{BC}$ and $\overline{CD}$. Rays $MH$ and $NH$ meet $\overline{AD}$ and $\overline{AB}$ at $S$ and $T$, respectively. Prove that there exists a point $E$, lying outside quadrilateral $ABCD$, such that
[list]
[*] ray $EH$ bisects both angles $\angle BES$, $\angle TED$, and
[*] $\angle BEN = \angle MED$.
[/list]

Proposed by Evan Chen
35 replies
62861
Jan 22, 2018
Sanjana42
an hour ago
Prove concyclic and tangency
syk0526   40
N an hour ago by Ilikeminecraft
Source: Japan Olympiad Finals 2014, #4
Let $ \Gamma $ be the circumcircle of triangle $ABC$, and let $l$ be the tangent line of $\Gamma $ passing $A$. Let $ D, E $ be the points each on side $AB, AC$ such that $ BD : DA= AE : EC $. Line $ DE $ meets $\Gamma $ at points $ F, G $. The line parallel to $AC$ passing $ D $ meets $l$ at $H$, the line parallel to $AB$ passing $E$ meets $l$ at $I$. Prove that there exists a circle passing four points $ F, G, H, I $ and tangent to line $ BC$.
40 replies
syk0526
May 17, 2014
Ilikeminecraft
an hour ago
p^2+3*p*q+q^2
mathbetter   0
2 hours ago
\[
\text{Find all prime numbers } (p, q) \text{ such that } p^2 + 3pq + q^2 \text{ is a fifth power of an integer.}
\]
0 replies
mathbetter
2 hours ago
0 replies
two sequences of positive integers and inequalities
rmtf1111   49
N 2 hours ago by dolphinday
Source: EGMO 2019 P5
Let $n\ge 2$ be an integer, and let $a_1, a_2, \cdots , a_n$ be positive integers. Show that there exist positive integers $b_1, b_2, \cdots, b_n$ satisfying the following three conditions:

$\text{(A)} \ a_i\le b_i$ for $i=1, 2, \cdots , n;$

$\text{(B)} \ $ the remainders of $b_1, b_2, \cdots, b_n$ on division by $n$ are pairwise different; and

$\text{(C)} \ $ $b_1+b_2+\cdots b_n \le n\left(\frac{n-1}{2}+\left\lfloor \frac{a_1+a_2+\cdots a_n}{n}\right \rfloor \right)$

(Here, $\lfloor x \rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)
49 replies
rmtf1111
Apr 10, 2019
dolphinday
2 hours ago
Equal angles (a very old problem)
April   54
N Today at 12:49 AM by joshualiu315
Source: ISL 2007, G3, VAIMO 2008, P5
The diagonals of a trapezoid $ ABCD$ intersect at point $ P$. Point $ Q$ lies between the parallel lines $ BC$ and $ AD$ such that $ \angle AQD = \angle CQB$, and line $ CD$ separates points $ P$ and $ Q$. Prove that $ \angle BQP = \angle DAQ$.

Author: Vyacheslav Yasinskiy, Ukraine
54 replies
April
Jul 13, 2008
joshualiu315
Today at 12:49 AM
Equal angles (a very old problem)
G H J
Source: ISL 2007, G3, VAIMO 2008, P5
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huashiliao2020
1292 posts
#46
Y by
This is a boring problem :unamused: it's one construction which trivializes, nothing to learn from it.
Let the line through D parallel to BQ intersect PQ at R. It's obvious by sim triangle ratio of sides that ADR and CBQ are homothetic at P (RP/RQ=RD/RB=RA/RC). Then ARD=CQB=AQD, meaning R lies on the circumcircle of ADQ. Then DAQ=DRQ=BQP, which finishes. $\blacksquare$
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OronSH
1716 posts
#47 • 2 Y
Y by megarnie, ihatemath123
Cute!

Suppose the homothety at $P$ sending $BC$ to $DA$ takes $Q$ to point $R.$ Then we have $BQ \parallel DR,$ and $\angle ARD=\angle CQB=\angle AQD,$ so $ARQD$ is cyclic. Then, we have $\angle BQP=\angle BQR=\angle DRQ=\angle DAQ$ and we are done.
This post has been edited 1 time. Last edited by OronSH, Aug 30, 2023, 1:01 AM
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ihatemath123
3426 posts
#48 • 1 Y
Y by OronSH
Let $X$ be the second intersection of line $PQ$ with $(BQC)$. Clearly $P$ is the center of the homothety from $(BQC)$ to $(AQD)$. This homothety sends arc $BX$ in $(BQC)$ to arc $QD$ in $(AQD)$, hence $\angle BQX = \angle DAQ$.
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Shreyasharma
666 posts
#49
Y by
Extend $QP$ to meet the circumcircle of $ADQ$ at a point $R$. Then it is easy to see from the angle condition that there exists a homothety taking $\triangle BCQ \mapsto \triangle DAR$ so we have $RD \parallel BQ$ from which we finish as $\angle QAD = \angle QRD = \angle BQP$.
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trk08
614 posts
#50
Y by
Take a homothety centered at $P$ that maps $C$ to $A$ and $B$ to $D$, because they are similar triangles. Let us say this homothety sends $\triangle{BCQ}$ to $\triangle{DAT}$. Then:
\[\angle DTA=\angle BQC=\angle AQD,\]so $ADQT$ is cyclic.

We then angle chase to see:
\[\angle BQP=\angle DTQ=\angle DAQ,\]as desired $\square$
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asdf334
7574 posts
#52 • 1 Y
Y by OronSH
ok heres th e dumb solution first i guess

By $DDIT$ involution swaps $(QA,QB),(QC,QD),(QP,Q\infty_{AD})$ so we're done by angle chase.
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shendrew7
787 posts
#53
Y by
Homothety at $P$ sending $AD$ to $CB$ will also send $Q$ to a point $R$ such that $AQ \parallel CR$ and $BQ \parallel DR$. Thus
\[\angle DRA = \angle BQC = \angle DQA \implies ADQR \text{ cyclic} \implies \angle BQP = \angle QRD = \angle QAD. \quad \blacksquare\]
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EpicBird08
1730 posts
#54
Y by
Assume WLOG that $\frac{AD}{BC} = r \le 1,$ and let $Q'$ be the image of $Q$ under a homothety $h$ centered at $P$ with scale $-r.$ Then $h(A) = C$ and $h(D) = B,$ so the given condition is equivalent to $ADQQ'$ being cyclic. This means that $\angle DAQ = \angle DQ'Q.$ Therefore, we'd like to show that $\angle BQQ' = \angle DQ'Q,$ or $DQ' \parallel BQ.$ This is evident since $Q \to Q'$ and $D \to B.$
This post has been edited 1 time. Last edited by EpicBird08, Jan 13, 2024, 3:46 AM
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dolphinday
1310 posts
#55
Y by
Let $\mathcal{H}$ be the homothety sending $AD \to BC$, centered at $P$. Then let $\mathcal{H}(Q) = X$. Notice that $\triangle{BCQ} \sim \triangle DAX$, by homothety. It follows that $\angle{AQD} = \angle BQC = \angle{DXA}$, so $ADQX$ is cyclic. From our homothety, we also know that $BQ \parallel XD \implies \angle DAQ = \angle QXD = \angle BQP$, so we are done.
Nice problem. I think the solution makes it look easy but it's a bit hard to spot the homothety centered at $P$.
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IAmTheHazard
4999 posts
#56 • 1 Y
Y by ihatemath123
If $Q'$ is the image of $Q$ under the homothety (centered at $P$) sending $\overline{AD}$ to $\overline{CB}$ then the angle condition implies $BCQQ'$ cyclic and $\measuredangle BQP=\measuredangle BQQ'=\measuredangle BCQ'=\measuredangle DAQ$. $\blacksquare$


Remark: I think this problem becomes 10x (not an exaggeration) easier if you make $ABCD$ a self-intersecting trapezoid, i.e. such that $ADBC$ is convex instead of $ABCD$. If we translate the given angle conditions to the appropriate directed angle equalities (by drawing the diagram for when $ABCD$ is convex and inspecting orientations), the problem should still be true when $ADBC$ is convex. It's generally much easier to spot positive homotheties than negative ones, and in this case it becomes much easier (maybe even obvious) to notice that $\overline{AD} \mapsto \overline{CB}$ gives you a cyclic quadrilateral. In fact, when $P$ goes to infinity in this case I think the resulting problem is quite well-known (and older than this).
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Aiden-1089
277 posts
#57
Y by
Using directed angles, the condition is $\measuredangle AQD = \measuredangle CQB$, and we need to show that $\measuredangle BQP = \measuredangle DAQ$.
Take a pole-polar duality at $Q$ with arbitrary radius. Denote by $l_P$ the polar of $P$, $P_l$ the pole of $l$.
Since $BC//AD$, $P_{BC}, Q, P_{AD}$ are collinear. Also, $l_P=P_{AC}P_{BD}$.
$\measuredangle AQD = \measuredangle CQB \implies \measuredangle(l_A,l_D)=\measuredangle(l_C,l_B) \implies \measuredangle P_{AC}P_{AD}P_{BD} = \measuredangle P_{AC}P_{BC}P_{BD} \implies P_{AC}, P_{AD}, P_{BC}, P_{BD}$ are concyclic.
Thus, $\measuredangle P_{BC}P_{BD}P_{AC} = \measuredangle P_{BC}P_{AD}P_{AC} = \measuredangle  QP_{AD}P_{AC} \implies \measuredangle(l_B,l_P) = \measuredangle(\overline{AD},l_A) \implies \measuredangle BQP = \measuredangle DAQ$.
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john0512
4170 posts
#58
Y by
wow homothety is so nice but imagine being able to find that :(

The condition is essentially saying that $QB$ and $QA$ are isogonal in $\triangle QCD$.

The key idea is to that since we need to understand how $QP$ divides up the angle $\angle BQA$, we use Trig Ceva on $\triangle QAB$. We have by Trig Ceva that $$\frac{\sin\angle BQP}{\sin\angle AQP}=\frac{\sin\angle QBD}{\sin\angle ABD}\cdot \frac{\sin\angle CAB}{\sin\angle QAC}$$.

However, we have by law of sines on $\triangle BQD$ and $\triangle BAD$ that $$\frac{\sin\angle QBD}{\sin\angle ABD}=\frac{QD}{AD}\cdot\frac{\sin\angle BQD}{\sin\angle BAD}$$and similarly $$\frac{\sin\angle CAB}{\sin\angle QAC}=\frac{BC}{QC}\cdot\frac{\sin\angle ABC}{\sin\angle CQA}.$$However note that $\sin\angle ABC=\sin\angle BAD$ because $BC\parallel AD$, and $\angle BQD=\angle CQA$ by the isogonality of $QB$ and $QA$. Thus, we simply have $$\frac{\sin\angle BQP}{\sin\angle AQP}=\frac{QD}{AD}\cdot \frac{BC}{QC}=\frac{\sin\angle QAD}{\sin\angle AQD}\cdot\frac{\sin\angle CQB}{\sin\angle CBQ}=\frac{\sin\angle QAD}{\sin\angle CBQ}.$$
Note that $$\angle BQA=180-\angle QBA-\angle QAB=\angle CBQ+\angle DAQ,$$so we also have $\angle BQP+\angle AQP=\angle QAD+\angle CBQ$, and because of $$\frac{\sin\angle BQP}{\sin\angle AQP}=\frac{\sin\angle QAD}{\sin\angle CBQ},$$we have $\angle BQP=\angle QAD$ and $\angle AQP=\angle CBQ$, as desired.
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ihategeo_1969
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#59 • 1 Y
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Apply DDIT from $Q$ to $ADBC$ to get involution pairs $(\overline{QP},\overline{Q\infty})$; $(\overline{QA},\overline{QB})$; $(\overline{QC},\overline{QD})$. The last two pairs tell us the involution is an isogonality and just for convenience define $T=\overline{Q\infty} \cap \overline{AB}$, just look at this simple angle chase \[\angle BQP=\angle TQA=\angle DAQ\]and we are done.
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AshAuktober
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#60
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By ILL< we have $QP, Q\infty_{BC}$ are isogonal in triangle $QCD$, from where it's trivial.
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joshualiu315
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#61
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WLOG assume that $BC<AD$, and let $\overline{PQ}$ intersect $(AQD)$ again at a point $R$.

There exists a homothety $\mathcal{H}$ centered at $P$ that maps $C$ to $A$ and $B$ to $D$. Note that $\overline{QR}$ passes through the center of $\mathcal{H}$ and

\[\angle BQC = \angle AQD = \angle ARD,\]
which implies $\mathcal{H}$ maps $R$ to $Q$ as well. Hence, $\triangle PQB \sim \triangle PRD$, so

\[\angle PQB = \angle PRD = \angle QRD = \angle QAD,\]
as desired. $\blacksquare$
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