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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
2023 Putnam A1
giginori   28
N 37 minutes ago by pie854
For a positive integer $n$, let $f_n(x)=\cos (x) \cos (2 x) \cos (3 x) \cdots \cos (n x)$. Find the smallest $n$ such that $\left|f_n^{\prime \prime}(0)\right|>2023$.
28 replies
giginori
Dec 3, 2023
pie854
37 minutes ago
Sequence of rational numbers
mojyla222   2
N an hour ago by Assassino9931
Source: Iran 2024 3rd round number theory exam P1
Given a sequence $x_1,x_2,x_3,\cdots$ of positive integers, Ali proceed the following algorythm: In the i-th step he markes all rational numbers in the interval $[0,1]$ which have denominator equal to $x_i$. Then he write down the number $a_i$ equal to the length of the smallest interval in $[0,1]$ which both two ends of that is a marked number. Find all sequences $x_1,x_2,x_3,\cdots$ with $x_5=5$ and such that for all $n\in \mathbb N$ we have
$$
a_1+a_2+\cdots+a_n= 2-\dfrac{1}{x_n}.
$$
Proposed by Mojtaba Zare
2 replies
mojyla222
Aug 27, 2024
Assassino9931
an hour ago
Midpoint in a weird configuration
Gimbrint   1
N an hour ago by Beelzebub
Source: Own
Let $ABC$ be an acute triangle ($AB<BC$) with circumcircle $\omega$. Point $L$ is chosen on arc $AC$, not containing $B$, so that, letting $BL$ intersect $AC$ at $S$, one has $AS<CS$. Points $D$ and $E$ lie on lines $AB$ and $BC$ respectively, such that $BELD$ is a parallelogram. Point $P$ is chosen on arc $BC$, not containing $A$, such that $\angle CBP=\angle BDE$. Line $AP$ intersects $EL$ at $X$, and line $CP$ intersects $DL$ at $Y$. Line $XY$ intersects $AB$, $BC$ and $BP$ at points $M$, $N$ and $T$ respectively.

Prove that $TN=TM$.
1 reply
Gimbrint
May 23, 2025
Beelzebub
an hour ago
Another FE
M11100111001Y1R   2
N an hour ago by AblonJ
Source: Iran TST 2025 Test 2 Problem 3
Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for all $x,y>0$ we have:
$$f(f(f(xy))+x^2)=f(y)(f(x)-f(x+y))$$
2 replies
+1 w
M11100111001Y1R
Today at 8:03 AM
AblonJ
an hour ago
Shortest number theory you might've seen in your life
AlperenINAN   13
N an hour ago by lksb
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)+1$ is a perfect square, then $pq + 1$ is also a perfect square.
13 replies
AlperenINAN
May 11, 2025
lksb
an hour ago
Quadruple Binomial Coefficient Sum
P162008   1
N 2 hours ago by vmene
Source: Self made by my Elder brother
$\sum_{p=0}^{\infty} \sum_{r=0}^{\infty} \sum_{q=1}^{\infty} \sum_{s=0}^{p+q - 1} \frac{((-1)^{p+r+s+1})(2^{p+q-1}) \binom{p + q - s - 1}{p + q - 2s - 1}}{4^s(2p^2q + 2pqr + pq + qr)(2p + 2q + 2r + 3)}.$
1 reply
P162008
2 hours ago
vmene
2 hours ago
n-term Sequence
MithsApprentice   15
N 2 hours ago by Ilikeminecraft
Source: USAMO 1996, Problem 4
An $n$-term sequence $(x_1, x_2, \ldots, x_n)$ in which each term is either 0 or 1 is called a binary sequence of length $n$. Let $a_n$ be the number of binary sequences of length $n$ containing no three consecutive terms equal to 0, 1, 0 in that order. Let $b_n$ be the number of binary sequences of length $n$ that contain no four consecutive terms equal to 0, 0, 1, 1 or 1, 1, 0, 0 in that order. Prove that $b_{n+1} = 2a_n$ for all positive integers $n$.
15 replies
MithsApprentice
Oct 22, 2005
Ilikeminecraft
2 hours ago
A MATHEMATICA E BONITA
P162008   0
2 hours ago
Source: Self made by my Elder brother
Let $K = \sum_{i=0}^{\infty} \sum_{j=0}^{\infty}\sum_{m=0}^{\infty}\sum_{l=0}^{\infty} \frac{1}{(i+j+m+l)!}$ where $i,j,k$ and $l \in W.$

Now, consider the ratio $Z$ defined as
$Z = \frac{\sum_{r=0}^{\lfloor k \rfloor} \sum_{i=0}^{\lfloor k \rfloor} (-1)^r \binom{\lfloor k \rfloor}{r}(\lfloor k \rfloor - r)^i}{\sum_{r=0}^{\lfloor k \rfloor + 1}(-1)^r\binom{\lfloor k \rfloor + 1}{r}(\lfloor k \rfloor + 1 - r)^{\lfloor k \rfloor + 1}}.$

The summation function $S(n)$ is given by
$S(n) = \sum_{j=1}^{n} \left(\binom{n}{j} (j!)\left(\sum_{b=0}^{j} \frac{(-1)^b}{b!}\right)\right)$

Let $p$ denotes the number of points of intersection between the curves
$x^2 + y^2 - \tan(e^x) - \frac{|x|}{\sin y} = 0, (x\sin (a))^y + (y - x\cos(a))^x = |a|.$

Define $A(m)$ as
$A(m) = p\left(\sum_{k=0}^{m} \binom{2m + 1}{k} ((2m + 1) - 2k) (-1)^k\right).$

The value of $X$ is
$X = \lim_{n \to \infty} \frac{\sqrt{n}}{e^n} \text{exp} \left(\int_{0}^{\infty} \lfloor ne^{-x} \rfloor \text{dx}\right).$

And, $8Y =$ Number of subsets of $\left(1,2,3,\cdots,100\right)$ whose sum of elements is divisible by $5.$

Finally, compute the value of $\frac{1}{Z} +S(4) + 1 + e^{A(20)} + X\sqrt{8\pi} + Y.$
0 replies
P162008
2 hours ago
0 replies
Drawing Triangles Against Your Clone
pieater314159   19
N 2 hours ago by Ilikeminecraft
Source: 2019 ELMO Shortlist C1
Elmo and Elmo's clone are playing a game. Initially, $n\geq 3$ points are given on a circle. On a player's turn, that player must draw a triangle using three unused points as vertices, without creating any crossing edges. The first player who cannot move loses. If Elmo's clone goes first and players alternate turns, who wins? (Your answer may be in terms of $n$.)

Proposed by Milan Haiman
19 replies
pieater314159
Jun 27, 2019
Ilikeminecraft
2 hours ago
Odd digit multiplication
JuanDelPan   12
N 3 hours ago by Ilikeminecraft
Source: Pan-American Girls' Mathematical Olympiad 2021, P4
Lucía multiplies some positive one-digit numbers (not necessarily distinct) and obtains a number $n$ greater than 10. Then, she multiplies all the digits of $n$ and obtains an odd number. Find all possible values of the units digit of $n$.

$\textit{Proposed by Pablo Serrano, Ecuador}$
12 replies
JuanDelPan
Oct 6, 2021
Ilikeminecraft
3 hours ago
Cup of Combinatorics
M11100111001Y1R   7
N 3 hours ago by MathematicalArceus
Source: Iran TST 2025 Test 4 Problem 2
There are \( n \) cups labeled \( 1, 2, \dots, n \), where the \( i \)-th cup has capacity \( i \) liters. In total, there are \( n \) liters of water distributed among these cups such that each cup contains an integer amount of water. In each step, we may transfer water from one cup to another. The process continues until either the source cup becomes empty or the destination cup becomes full.

$a)$ Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most \( \frac{4n}{3} \) steps.

$b)$ Prove that in at most \( \frac{5n}{3} \) steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
7 replies
M11100111001Y1R
May 27, 2025
MathematicalArceus
3 hours ago
Inequality
knm2608   17
N 3 hours ago by Adywastaken
Source: JBMO 2016 shortlist
If the non-negative reals $x,y,z$ satisfy $x^2+y^2+z^2=x+y+z$. Prove that
$$\displaystyle\frac{x+1}{\sqrt{x^5+x+1}}+\frac{y+1}{\sqrt{y^5+y+1}}+\frac{z+1}{\sqrt{z^5+z+1}}\geq 3.$$When does the equality occur?

Proposed by Dorlir Ahmeti, Albania
17 replies
knm2608
Jun 25, 2017
Adywastaken
3 hours ago
Circumcircle of XYZ is tangent to circumcircle of ABC
mathuz   39
N 3 hours ago by zuat.e
Source: ARMO 2013 Grade 11 Day 2 P4
Let $ \omega $ be the incircle of the triangle $ABC$ and with centre $I$. Let $\Gamma $ be the circumcircle of the triangle $AIB$. Circles $ \omega $ and $ \Gamma $ intersect at the point $X$ and $Y$. Let $Z$ be the intersection of the common tangents of the circles $\omega$ and $\Gamma$. Show that the circumcircle of the triangle $XYZ$ is tangent to the circumcircle of the triangle $ABC$.
39 replies
mathuz
May 22, 2013
zuat.e
3 hours ago
Ultra-hyper saddle with logarithmic weight
randomperson1021   0
5 hours ago
Fix integers \(k\ge 3\) and \(1<r<k\), a parameter \(\lambda>0\), and a real log-exponent \(\beta\in\mathbb R\). For every real \(a\) define
$$
F_{a,\beta}^{(k,r)}(x)
  \;:=\;
  \sum_{n\ge 1}
       n^{\,a}\,(\log n)^{\beta}\,e^{\lambda n^{r}}\,x^{\,n^{k}},
  \qquad 0\le x<1.
$$
Put
$$
\Lambda_{k,r,\lambda}
   \;:=\;
   \lambda\!\left(1-\frac{r}{k}\right)
   \left(\frac{\lambda r}{k}\right)^{\!\frac{r}{\,k-r\,}},
   \qquad
   \gamma=\frac{r}{k-r}.
$$
(1) Show that there exists a real constant \(c=c(k,r)\) (independent of \(\lambda\) and of \(\beta\)) such that
$$
\lim_{x\to 1^{-}}
      F_{a,\beta}^{(k,r)}(x)\,
      e^{-\Lambda_{k,r,\lambda}\,(1-x)^{-\gamma}}
      \;=\;
      \begin{cases}
          0, & a<c,\\[6pt]
          \infty, & a>c.
      \end{cases}
$$
(2) Determine this critical value \(c\) explicitly and verify that it coincides with the classical case \(r=1\), namely \(c=-\tfrac12\).

(3) Evaluate the finite, non-zero limit that occurs at the borderline \(a=c\) (your answer may depend on \(k,r,\lambda\) but not on \(\beta\)).
0 replies
randomperson1021
5 hours ago
0 replies
maximal determinant
EthanWYX2009   4
N Apr 12, 2025 by loup blanc
Source: 2023 Aug taca-9
Let matrix
\[A=\begin{bmatrix} 1&1&1&1&1\\1&-1&1&-1&1\\?&?&?&?&?\\?&?&?&?&?\\?&?&?&?&?\end{bmatrix}\in\mathbb R^{5\times 5}\]satisfy $\text{tr} (AA^T)=28.$ Determine the maximum value of $\det A.$
4 replies
EthanWYX2009
Apr 9, 2025
loup blanc
Apr 12, 2025
maximal determinant
G H J
G H BBookmark kLocked kLocked NReply
Source: 2023 Aug taca-9
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EthanWYX2009
872 posts
#1 • 1 Y
Y by Euler_Gauss
Let matrix
\[A=\begin{bmatrix} 1&1&1&1&1\\1&-1&1&-1&1\\?&?&?&?&?\\?&?&?&?&?\\?&?&?&?&?\end{bmatrix}\in\mathbb R^{5\times 5}\]satisfy $\text{tr} (AA^T)=28.$ Determine the maximum value of $\det A.$
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loup blanc
3601 posts
#2
Y by
With a computer, I find $72$.
Assuming that the last three rows of $A$ are real.
This post has been edited 1 time. Last edited by loup blanc, Apr 9, 2025, 6:01 PM
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loup blanc
3601 posts
#3
Y by
$\bullet$ Let $A_{i,j}$ be the matrix obtained from $A$ removing the $i^{th}$ row and the $j^{th}$ column.
If $i\geq 3$, then $\dfrac{\partial \det(A)}{\partial a_{i,j}}=(-1)^{i+j}\det(A_{i,j})$.
The imposed condition is $\sum_{i\geq 3,j\geq 1}{a_{i,j}}^2=18$; this set is a compact of $\mathbb{R}^{15}$,
what implies that the required maximum is reached and -easy to see- is $>0$.
Then an extremum satisfies (Kuhn-Tucker) there is $k$ s.t. for every $i\geq 3,j\geq 1$, $(-1)^{i+j}\det(A_{i,j})=ka_{i,j}$.
Thus $\det(A)=k\sum_j{a_{3,j}}^2=k\sum_j{a_{4,j}}^2=k\sum_j{a_{5,j}}^2$.
Then $\sum_j{a_{3,j}}^2=\sum_j{a_{4,j}}^2=\sum_j{a_{5,j}}^2=6$ and $\det(A)=6k$.
$\bullet$ let $A=\begin{pmatrix}U_{2,5}\\V_{3,5}\end{pmatrix}$, where $U$ is known and $V$ unknown. $AA^T=\begin{pmatrix}UU^T&UV^T\\VU^T&VV^T\end{pmatrix}$ is symmetric $>0$ and $tr(VV^T)=18$ (*).
$\det(A)^2=\det(UU^T)\det(VV^T-VU^T(UU^T)^{-1}UV^T)=24\det(VV^T-VU^T(UU^T)^{-1}UV^T)$.
Note that $VU^T(UU^T)^{-1}UV^T\in M_3$ is symmetric $\geq 0$.
If I'm not mistaken, $Z=VV^T-VU^T(UU^T)^{-1}UV^T$ is -as $AA^T$- symmetric $> 0$ and $VV^T$ is invertible.
We seek the maximum of $\det(Z)$. Under the condition (*), $\max(\det(VV^T))=6^3$.
Finally, if we find $V$ s.t. $VV^T=6I_3$ and $VU^T=0$, then we obtain $\max\det(A)^2=24\det(6 I_3)=24\times 6^3$
and $\max(\det(A))=72$ and $k=12$.
$\textbf{Conclusion.}$ It remains to show that there is $V$ s.t. $VV^T=6I_3$ and $VU^T=0_{3,2}$.
This post has been edited 2 times. Last edited by loup blanc, Apr 9, 2025, 11:36 PM
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solyaris
651 posts
#4 • 1 Y
Y by loup blanc
Nice problem! What is taca-9?

A somewhat different approach: The trace condition means that $g(A) = \sum_{i,j} a_{i,j}^2 = 28$. By compactness the maximum is attained (as pointed out in the post above).

Consider the determinant as a function of the remaining row vectors $u,v,w$ of $A$. Since the determinant is linear as a function of every row, $\det(A)$ can not be maximal if one of the rows is the $o$-vector. Also $\det(A)$ can be maximal only if the Euclidean norm of $u,v,w$ are equal: If $|u| > |v| > 0$, then choose $c > 1$ such that $\frac 1 c |u| =  c |v|$ and replace $u$ by $\frac 1 c u$ and $v$ by $cv$. This does not change the value of the determinant, but it is easy to see that it decreases the value of $g(A)$. (If you write this out, this boils down to the AM-GM-inequality.) In conclusion the three remaining rows have norm $\sqrt 6$ (and this takes care of the $g(A) = 0$ condition).

Finally: For $u,v,w$ with fixed lengths the (absolute value of the) determinant is maximal iff $u,v,w$ are orthogonal to each other and to the first two row vectors. (This follows geometrically by interpreting the deterimant as the volume of the parallelepiped spanned by the row vectors. (It is clear that $u,v,w$ with this property exists and by swapping two rows we can make the determinant positive.) Calculating the area of the parallelogramm spanned by the first two row vectors gives $\sqrt{5^2-1^2} = \sqrt {24}$. Thus the maximal value of the deterimant is $\sqrt{24} \sqrt 6^3 = 72$ (which matches the answer given above). Note that there are many matrices such that equality occurs.
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loup blanc
3601 posts
#5
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The end of my post #3.
That follows is a matrix that satisfies the relations required in the conclusion.
$V=\begin{pmatrix}-\sqrt{2}&-1&\sqrt{2}&1&0\\-\sqrt{2}&1&0&-1&\sqrt{2}\\0&-1&-\sqrt{2}&1&\sqrt{2}\end{pmatrix}$.
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a