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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
Different Paths Probability
Qebehsenuef   7
N 3 hours ago by GreenKeeper
Source: OBM
A mouse initially occupies cage A and is trained to change cages by going through a tunnel whenever an alarm sounds. Each time the alarm sounds, the mouse chooses any of the tunnels adjacent to its cage with equal probability and without being affected by previous choices. What is the probability that after the alarm sounds 23 times the mouse occupies cage B?
7 replies
Qebehsenuef
Apr 28, 2025
GreenKeeper
3 hours ago
A sharp one with 3 var
mihaig   10
N 3 hours ago by mihaig
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$ab+bc+ca+abc\geq4.$$
10 replies
mihaig
May 13, 2025
mihaig
3 hours ago
\int_{0}^{\pi/4} \frac{1}{\cos 2\phi} \cdot 2 \ln(\cot \phi) \cdot 2 \, d\phi
Martin.s   0
3 hours ago
\[
I = 2 \int_{0}^{\pi/4} \frac{1}{\cos 2\phi} \cdot 2 \ln(\cot \phi) \cdot 2 \, d\phi
\]
0 replies
Martin.s
3 hours ago
0 replies
Another right angled triangle
ariopro1387   1
N 3 hours ago by lolsamo
Source: Iran Team selection test 2025 - P7
Let $ABC$ be a right angled triangle with $\angle A=90$. Point $M$ is the midpoint of side $BC$ And $P$ be an arbitrary point on $AM$. The reflection of $BP$ over $AB$ intersects lines $AC$ and $AM$ at $T$ and $Q$, respectively. The circumcircles of $BPQ$ and $ABC$ intersect again at $F$. Prove that the center of the circumcircle of $CFT$ lies on $BQ$.
1 reply
ariopro1387
Yesterday at 4:13 PM
lolsamo
3 hours ago
nice integral
Martin.s   1
N 4 hours ago by aiops
$$\int_0^{\pi/2} \Bigl[
\log\bigl(\sqrt{5}-\sin\theta+1\bigr)
+\log\bigl(\sqrt{5}+\sin\theta-1\bigr)
-\log\bigl(\sqrt{5}-\sin\theta-1\bigr)
-\log\bigl(\sqrt{5}+\sin\theta+1\bigr)
\Bigr]\,d\theta.
$$
1 reply
Martin.s
4 hours ago
aiops
4 hours ago
four points lie on a circle
pohoatza   78
N 4 hours ago by ezpotd
Source: IMO Shortlist 2006, Geometry 2, AIMO 2007, TST 1, P2
Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic.

Proposed by Vyacheslev Yasinskiy, Ukraine
78 replies
pohoatza
Jun 28, 2007
ezpotd
4 hours ago
JBMO TST Bosnia and Herzegovina 2023 P4
FishkoBiH   2
N 4 hours ago by Stear14
Source: JBMO TST Bosnia and Herzegovina 2023 P4
Let $n$ be a positive integer. A board with a format $n*n$ is divided in $n*n$ equal squares.Determine all integers $n$3 such that the board can be covered in $2*1$ (or $1*2$) pieces so that there is exactly one empty square in each row and each column.
2 replies
FishkoBiH
Yesterday at 1:38 PM
Stear14
4 hours ago
Does there exist 2011 numbers?
cyshine   8
N 4 hours ago by TheBaiano
Source: Brazil MO, Problem 4
Do there exist $2011$ positive integers $a_1 < a_2 < \ldots < a_{2011}$ such that $\gcd(a_i,a_j) = a_j - a_i$ for any $i$, $j$ such that $1 \le i < j \le 2011$?
8 replies
cyshine
Oct 20, 2011
TheBaiano
4 hours ago
D1036 : Composition of polynomials
Dattier   1
N 4 hours ago by Dattier
Source: les dattes à Dattier
Find all $A \in \mathbb Q[x]$ with $\exists Q \in \mathbb Q[x], Q(A(x))= x^{2025!+2}+x^2+x+1$ and $\deg(A)>1$.
1 reply
Dattier
Saturday at 1:52 PM
Dattier
4 hours ago
number sequence contains every large number
mathematics2003   3
N 4 hours ago by sttsmet
Source: 2021ChinaTST test3 day1 P2
Given distinct positive integer $ a_1,a_2,…,a_{2020} $. For $ n \ge 2021 $, $a_n$ is the smallest number different from $a_1,a_2,…,a_{n-1}$ which doesn't divide $a_{n-2020}...a_{n-2}a_{n-1}$. Proof that every number large enough appears in the sequence.
3 replies
mathematics2003
Apr 13, 2021
sttsmet
4 hours ago
IMO ShortList 2002, geometry problem 2
orl   28
N 4 hours ago by ezpotd
Source: IMO ShortList 2002, geometry problem 2
Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \]
28 replies
orl
Sep 28, 2004
ezpotd
4 hours ago
Arc Midpoints Form Cyclic Quadrilateral
ike.chen   56
N 4 hours ago by ezpotd
Source: ISL 2022/G2
In the acute-angled triangle $ABC$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $AF$. The lines through $P$ parallel to $AC$ and $AB$ meet $BC$ at $D$ and $E$, respectively. Points $X \ne A$ and $Y \ne A$ lie on the circles $ABD$ and $ACE$, respectively, such that $DA = DX$ and $EA = EY$.
Prove that $B, C, X,$ and $Y$ are concyclic.
56 replies
ike.chen
Jul 9, 2023
ezpotd
4 hours ago
Non-linear Recursive Sequence
amogususususus   4
N 5 hours ago by GreekIdiot
Given $a_1=1$ and the recursive relation
$$a_{i+1}=a_i+\frac{1}{a_i}$$for all natural number $i$. Find the general form of $a_n$.

Is there any way to solve this problem and similar ones?
4 replies
amogususususus
Jan 24, 2025
GreekIdiot
5 hours ago
nice ecuation
MihaiT   1
N 5 hours ago by Hello_Kitty
Find real values $m$ , s.t. ecuation: $x+1=me^{|x-1|}$ have 2 real solutions .
1 reply
MihaiT
Yesterday at 2:03 PM
Hello_Kitty
5 hours ago
Putnam 2006 B1
Kent Merryfield   54
N Apr 25, 2025 by Ilikeminecraft
Show that the curve $x^{3}+3xy+y^{3}=1$ contains only one set of three distinct points, $A,B,$ and $C,$ which are the vertices of an equilateral triangle.
54 replies
Kent Merryfield
Dec 4, 2006
Ilikeminecraft
Apr 25, 2025
Putnam 2006 B1
G H J
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jj_ca888
2726 posts
#41 • 2 Y
Y by Williamgolly, HWenslawski
Good Laugh.

We use the key identity that $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$. Let $a = x$, $b = y$, $c = -1$, which gives us our problem statement. Hence, it factors to\[(x + y - 1)(x^2 + y^2 + 1 - xy + x + y) = 0\]so either $x + y = 1$ or $(x+1)^2 + (y+1)^2 + (x - y)^2 = 0$. Hence the locus of all points satisfying the problem is the line $x + y = 1$ and the point $(-1, -1)$. The other two vertices of this triangle are clearly unique so the triangle is unique.

In fact, the distance from $(-1, -1)$ to $x + y = 1$ can be calculated to be $\tfrac{3}{\sqrt{2}}$, which also happens to be the height of the unique equilateral triangle. Hence, the equilateral triangle has area $\tfrac{h^2}{\sqrt{3}} = \tfrac{3\sqrt{3}}{2}$.
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Zorger74
760 posts
#42 • 1 Y
Y by HWenslawski
Solved with andyxpandy99.

Solution
This post has been edited 1 time. Last edited by Zorger74, Apr 25, 2022, 1:43 PM
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IAmTheHazard
5003 posts
#43 • 1 Y
Y by HWenslawski
oh titu...

We have
\begin{align*}
x^3+3xy+y^3&=1\\
x^3+y^3+(-1)^3-3xy(-1)&=0\\
(x+y-1)((x-y)^2+(x+1)^2+(y+1)^2)&=0,
\end{align*}so the graph of the curve is the union of $x+y=1$ and $(x,y)=(-1,-1)$. From here the result is immediate. $\blacksquare$
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loup blanc
3600 posts
#44
Y by
Consider the behavior of the curves $C_b$: $f(x,y)=x^3+y^3+bxy-1=0$, when $b\in (-3-\epsilon,3+\epsilon)$ in a neighborhood of the point $A=(-1,-1)$.
The system $\dfrac{\partial f}{\partial x}=0,\dfrac{\partial f}{\partial y}=0,f(x,y)=0$ give the singular points of $C_b$.
For every $b\not= 3$, there are no singular points; for $b=3$, we obtain the sole point $A$ which is an isolated point of $C_3$.
When $b=3-h,h>0$ small , $C_b$ has no points close to $A$.
When $b=3+h$, the curve $C_b$ presents a kind of small ellipse close to $A$.

In fact, when $b<3$ the complex set $C_b$ contains two conjugate curves that don't intersect; yet, they intersect -when $b=3$- in the isolated singular real point $A$.
A simpler example is as follows:
Consider the curves $C_a$: $x^2+y^2-a=0$, where $a\leq0$. If $a<0$, then $C_a$ contains the conjugate complex curves
$\{(x,y);x\in\mathbb{R},y=i\sqrt{x^2-a}\}\},\{(x,y);x\in\mathbb{R},y=-i\sqrt{x^2-a}\}$ that don't intersect.
When $a=0$, these $2$ curves intersect in the isolated point $(0,0)$.
Of course, when $a>0$ small, we obtain a little circle close to $(0,0)$.
This post has been edited 2 times. Last edited by loup blanc, Apr 16, 2022, 1:42 PM
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ZETA_in_olympiad
2211 posts
#45
Y by
The A1 of Putnam 2006 is more cool.
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ZETA_in_olympiad
2211 posts
#46
Y by
The curve can be factored to $.5((x+1)^2+(y+1)^2+(x-y)^2).$ It follows that the curve consists of a point $(-1,-1)$ and the line $x+y=1.$ We need one of the vertices to be $(-1,-1)$ for the triangle with three points on this curve to be formed. Moreover, the distance from $(-1,-1)$ to $(.5,.5)$ is $3\sqrt{2}/2$, in fact this is the altitude from $(-1,-1).$ So the area of the triangle is $\boxed{3\sqrt{3}/2}.$
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loup blanc
3600 posts
#47 • 1 Y
Y by ZETA_in_olympiad
Zeta, read my post.
I don't want to solve the problem; moreover the same solution has already been written at least 4 times above (5 times with yours).
I explain how an isolated point of an algebraic curve can be dynamically created.
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ZETA_in_olympiad
2211 posts
#48
Y by
loup blanc wrote:
Zeta, read my post.
I don't want to solve the problem; moreover the same solution has already been written at least 4 times above (5 times with yours).
I explain how an isolated point of an algebraic curve can be dynamically created.

I see. I guess my solution happens to be "the most obvious one."
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OronSH
1748 posts
#49 • 1 Y
Y by alsk
We see $x^3+3xy+y^3-1$ factors as $(x+y-1)((x+1)^2+(y+1)^2-(x+1)(y+1))=0$ and we may notice that the second factor is zero only at $x=y=-1$ since it is the average of $(x+1)^2+(y+1)^2$ and $(x-y)^2.$ Since the curve is a point and a line we see that one of $A,B,C$ must be $(-1,-1)$ and the other two are on $x+y=1.$ Their midpoint must be the foot from $(-1,-1)$ to $x+y=1,$ and now it is easy to finish. The triangle has height $\frac{3\sqrt2}2$ so it has area $\frac{3\sqrt3}2.$
This post has been edited 1 time. Last edited by OronSH, Jan 10, 2024, 9:49 PM
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RedFireTruck
4243 posts
#50
Y by
Notice that $$x^3+y^3+3xy-1=(x+y)^3-3x^2y-3xy^2+3xy-1=(x+y)^3-3xy((x+y)-1)-1$$so the given equation is equivalent to $(x+y)^3-1=3xy((x+y)-1)$. Notice that this implies that $x+y=1$ is a solution. Since this is a line, an equilateral triangle can be made from any point such that $x+y\ne 1$ and $2$ points on $x+y=1$. Therefore, it suffices to prove that there is only one solution to $(x+y)^3-1=3xy((x+y)-1)$ such that $x+y\ne1$. Since we've already considered $x+y=1$, we can divide by $x+y-1$ to get $(x+y)^2+(x+y)+1=3xy$ or $x^2-xy+y^2+x+y+1=0$. We can rewrite this as $(x+1)^2+(y+1)^2-(x+1)(y+1)=0$. Clearly, $(-1, -1)$ is a solution to this. Plugging in $x=-1+a$ and $y=-1-a$ into the original equation gives $a=0$, and we have already considered $x=y=-1$, so we multiply both sides by $x+y+2=(x+1)+(y+1)$ to get that $(x+1)^3+(y+1)^3=0$ so $x+y+2=0$, which we've already shown only has solution $(-1, -1)$. Now, we calculate the area of this unique equilateral triangle to be $\frac{3\sqrt2}{2}\cdot\frac{3\sqrt\frac23}{2}=\frac{3\sqrt3}{2}$.
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chakrabortyahan
385 posts
#51 • 1 Y
Y by D_S
The problem is veryyyy easyyy.The phrasing of the problem is bit confusing though $x^3+y^3+3xy-1 = (x+y-1)\frac{1}{2}((x-y)^2+(x+1)^2+(y+1)^2)$
Now Drop perpendicular from $(-1,-1)$ to the line and there are two unique points on the line making $30^{\circ}$ with the perp and hence we are done $\blacksquare\smiley$
@below ole baba cho cutee
This post has been edited 2 times. Last edited by chakrabortyahan, May 4, 2024, 7:47 PM
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sanyalarnab
947 posts
#52
Y by
I wasn't even a month old when this problem was posted :-D
Attachments:
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naonaoaz
334 posts
#53
Y by
Factor into
\[(x+y-1)(x^2+y^2+x+y-xy+1) = 0\]So either $x+y = 1$ or
\[x^2+y^2+x+y-xy+1 = 0\]Since $x$ is real, taking the discriminant gives
\[-3(y+1)^2 \ge 0 \implies y = -1 \implies x = -1\]Thus we have $x+y = 1$ or $(x,y) = (-1,-1)$. Now simple geometry gives an answer of
\[\boxed{\frac{3\sqrt{3}}{2}}\]
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Sagnik123Biswas
421 posts
#54
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It is a very standard factorization that $a^3+b^3+c^3 = \frac{1}{2}(a+b+c)((a-b)^2 + (a-c)^2 + (b-c)^2)$. Now it suffices to twiddle with the given equation.

It can be re-arranged to $x^3+y^3+(-1)^3 - 3(x)(y)(-1) = 0 \implies (x+y-1)((x-y)^2+(x+1)^2+(y+1)^2)=0$. So either $(x, y) = (-1, -1)$ or $x+y=1$. So the three vertices of the equilateral triangle are $(-1, -1), (\frac{1}{2} - \frac{\sqrt{3}}{2}, \frac{1}{2} + \frac{\sqrt{3}}{2} ), (\frac{1}{2} + \frac{\sqrt{3}}{2}, \frac{1}{2} - \frac{\sqrt{3}}{2}) $
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Ilikeminecraft
658 posts
#55
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\begin{align*}
    x^3 + y^3 + 3xy - 1 & = x^3 + y^3 - 1 - 3xy(-1) \\
    & = (x + y - 1)(x^2 + y^2 - xy + x + y + 1) \\
    & = (x + y - 1)((x + 1)^2 + (y+ 1)^2 - (x + 1)(y + 1))
\end{align*}It can be shown that the 2nd factor is never 0 unless $x = y = -1.$ From here, the problem is done, since one vertex must be at $(-1, -1),$ and the other two points lie on the line.
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