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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
nice algebra
gggzul   0
a minute ago
For all real numbers $a$ find the number of ordered real triples $(x, y, z)$ such that
\[\begin{aligned}
\begin{cases}
    x+y^2+z^2=a,\\
    x^2+y+z^2=a,\\
    x^2+y^2+z=a.
\end{cases}
\end{aligned}\]
0 replies
gggzul
a minute ago
0 replies
IMO Problem 2
iandrei   47
N 7 minutes ago by asdf334
Source: IMO ShortList 2003, number theory problem 3
Determine all pairs of positive integers $(a,b)$ such that \[ \dfrac{a^2}{2ab^2-b^3+1}  \] is a positive integer.
47 replies
iandrei
Jul 14, 2003
asdf334
7 minutes ago
geometry
srnjbr   1
N 9 minutes ago by removablesingularity
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.
1 reply
srnjbr
an hour ago
removablesingularity
9 minutes ago
D1010 : How it is possible ?
Dattier   8
N 32 minutes ago by Dattier
Source: les dattes à Dattier
Is it true that$$\forall n \in \mathbb N^*, (24^n \times B \mod A) \mod 2 = 0 $$?

A=1728400904217815186787639216753921417860004366580219212750904
024377969478249664644267971025952530803647043121025959018172048
336953969062151534282052863307398281681465366665810775710867856
720572225880311472925624694183944650261079955759251769111321319
421445397848518597584590900951222557860592579005088853698315463
815905425095325508106272375728975

B=2275643401548081847207782760491442295266487354750527085289354
965376765188468052271190172787064418854789322484305145310707614
546573398182642923893780527037224143380886260467760991228567577
953725945090125797351518670892779468968705801340068681556238850
340398780828104506916965606659768601942798676554332768254089685
307970609932846902
8 replies
Dattier
Mar 10, 2025
Dattier
32 minutes ago
Linear algebra
Dynic   2
N Today at 3:32 PM by loup blanc
Let A and B be two square matrices with the same size. Prove that if AB is an invertible matrix, then A and B are also invertible matrices
2 replies
Dynic
Today at 2:48 PM
loup blanc
Today at 3:32 PM
3-dimensional matrix system
loup blanc   0
Today at 1:04 PM
Let $A=\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}$.
i) Find the matrices $B\in M_3(\mathbb{R})$ s.t. $A^TA=B^TB,AA^T=BB^T$.
ii) Show that each solution of i) is in $M_3(K)$, where $K=\mathbb{Q}[\sin^2(\dfrac{2}{3}\arctan(3\sqrt{3}))]$.
iii) Solve i) when $B\in M_3(\mathbb{C})$.
EDIT. In iii) we again consider the transpose of $B$ and not its conjugate transpose.
0 replies
loup blanc
Today at 1:04 PM
0 replies
Very hard group theory problem
mathscrazy   3
N Today at 9:50 AM by quasar_lord
Source: STEMS 2025 Category C6
Let $G$ be a finite abelian group. There is a magic box $T$. At any point, an element of $G$ may be added to the box and all elements belonging to the subgroup (of $G$) generated by the elements currently inside $T$ are moved from outside $T$ to inside (unless they are already inside). Initially $
T$ contains only the group identity, $1_G$. Alice and Bob take turns moving an element from outside $T$ to inside it. Alice moves first. Whoever cannot make a move loses. Find all $G$ for which Bob has a winning strategy.
3 replies
mathscrazy
Dec 29, 2024
quasar_lord
Today at 9:50 AM
limit of u(pi/45)
EthanWYX2009   0
Today at 7:08 AM
Source: 2025 Pi Day Challenge T5
Let \(\omega\) be a positive real number. Divide the positive real axis into intervals \([0, \omega)\), \([\omega, 2\omega)\), \([2\omega, 3\omega)\), \([3\omega, 4\omega)\), \(\ldots\), and color them alternately black and white. Consider the function \(u(x)\) satisfying the following differential equations:
\[
u''(x) + 9^2u(x) = 0, \quad \text{for } x \text{ in black intervals},
\]\[
u''(x) + 63^2u(x) = 0, \quad \text{for } x \text{ in white intervals},
\]with the initial conditions:
\[
u(0) = 1, \quad u'(0) = 1,
\]and the continuity conditions:
\[
u(x) \text{ and } u'(x) \text{ are continuous functions}.
\]Show that
\[
\lim_{\omega \to 0} u\left(\frac{\pi}{45}\right) = 0.
\]
0 replies
EthanWYX2009
Today at 7:08 AM
0 replies
Integration Bee Kaizo
Calcul8er   42
N Today at 12:57 AM by awzhang10
Hey integration fans. I decided to collate some of my favourite and most evil integrals I've written into one big integration bee problem set. I've been entering integration bees since 2017 and I've been really getting hands on with the writing side of things over the last couple of years. I hope you'll enjoy!
42 replies
Calcul8er
Mar 2, 2025
awzhang10
Today at 12:57 AM
maximum value
Tip_pay   3
N Yesterday at 9:37 PM by alexheinis
Find the value $x\in [0,4]$ at which the function $f(x)=\int_{0}^{\sqrt{x}}\ln \frac{e}{1+t^2}dt$ takes its maximum value
3 replies
Tip_pay
Yesterday at 8:48 PM
alexheinis
Yesterday at 9:37 PM
Spheres and a point source of light
mofidy   3
N Yesterday at 9:22 PM by kiyoras_2001
How many spheres are needed to shield a point source of light?
Unfortunately, I didn't find a suitable solution for it on the page below:
https://artofproblemsolving.com/community/c4h1469498p8521602
Here too, two different solutions are given:
https://math.stackexchange.com/questions/2791186
3 replies
mofidy
Mar 11, 2025
kiyoras_2001
Yesterday at 9:22 PM
Real Analysis
rljmano   4
N Yesterday at 4:38 PM by alexheinis
In [0 1], {f(t)}*{f’(t)-1} =0 and f is continuously differentiable. How do we conclude that either f is identically zero or f’(t) is identically 1 in [0 1]?
4 replies
rljmano
Yesterday at 6:32 AM
alexheinis
Yesterday at 4:38 PM
find the isomorphism
nguyenalex   14
N Yesterday at 1:49 PM by Royrik123456
I have the following exercise:

Let $E$ be an algebraic extension of $K$, and let $F$ be an algebraic closure of $K$ containing $E$. Prove that if $\sigma : E \to F$ is an embedding such that $\sigma(c) = c$ for all $c \in K$, then $\sigma$ extends to an automorphism of $F$.

My attempt:

Theorem (*): Suppose that $E$ is an algebraic extension of the field $K$, $F$ is an algebraically closed field, and $\sigma: K \to F$ is an embedding. Then, there exists an embedding $\tau: E \to F$ that extends $\sigma$. Moreover, if $E$ is an algebraic closure of $K$ and $F$ is an algebraic extension of $\sigma(K)$, then $\tau$ is an isomorphism.

Back to our main problem:

Since $K \subset E$ and $F$ is an algebraic extension of $K$, it follows that $F$ is an algebraic extension of $E$. Assume that there exists an embedding $\sigma : E \to F$ such that $\sigma(c) = c$ for all $c \in K$. By Theorem (*), there exists an embedding $\tau : F \to F$ that extends $\sigma$. Since $F$ is algebraically closed, $\tau(F)$ is also an algebraically closed field.

Furthermore, because $\sigma(c) = c$ for all $c \in K$ and $\tau$ is an extension of $\sigma$, we have
$$K = \sigma(K) \subset K \subset \sigma(E) \subset \tau(F) \subset F.$$
This implies that $F$ is an algebraic extension of $\tau(F)$. We conclude that $F = \tau(F)$, meaning that $\tau$ is an automorphism. (Finished!!)

Let choose $F = A$ be the field of algebraic numbers, $K=\mathbb{Q}$. Consider the embedding $\sigma: \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{2}) \subset A$ defined by
$$
a + b\sqrt{2} \mapsto a - b\sqrt{2}.
$$Then, according to the exercise above, $\sigma$ extends to an isomorphism
$$
\bar{\sigma}: A \to A.
$$How should we interpret $\bar{\sigma}$?
14 replies
nguyenalex
Mar 12, 2025
Royrik123456
Yesterday at 1:49 PM
find the convex sets satisfied
nguyenalex   2
N Yesterday at 9:54 AM by ILOVEMYFAMILY
In $\mathbb{R}^2$, let $B = \{(x, y) \mid x \geq 0\}$. Find all convex sets $C$ such that

\[\mathcal{E}(B \cup C) = B \cup C.\]
2 replies
nguyenalex
Yesterday at 8:57 AM
ILOVEMYFAMILY
Yesterday at 9:54 AM
Combinatorics Problem
Hopeooooo   1
N Oct 13, 2022 by M11100111001Y1R
Source: SRMC 2022 P4
In a language$,$ an alphabet with $25$ letters is used$;$ words are exactly all sequences of $($ not necessarily different $)$ letters of length $17.$ Two ends of a paper strip are glued so that the strip forms a ring$;$ the strip bears a sequence of $5^{18}$ letters$.$ Say that a word is singular if one can cut a piece bearing exactly that word from the strip$,$ but one cannot cut out two such non-overlapping pieces$.$ It is known that one can cut out $5^{16}$ non-overlapping pieces each containing the same word$.$ Determine the largest possible number of singular words$.$
(Bogdanov I.)
1 reply
Hopeooooo
May 23, 2022
M11100111001Y1R
Oct 13, 2022
Combinatorics Problem
G H J
G H BBookmark kLocked kLocked NReply
Source: SRMC 2022 P4
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Hopeooooo
819 posts
#1 • 1 Y
Y by Mango247
In a language$,$ an alphabet with $25$ letters is used$;$ words are exactly all sequences of $($ not necessarily different $)$ letters of length $17.$ Two ends of a paper strip are glued so that the strip forms a ring$;$ the strip bears a sequence of $5^{18}$ letters$.$ Say that a word is singular if one can cut a piece bearing exactly that word from the strip$,$ but one cannot cut out two such non-overlapping pieces$.$ It is known that one can cut out $5^{16}$ non-overlapping pieces each containing the same word$.$ Determine the largest possible number of singular words$.$
(Bogdanov I.)
This post has been edited 1 time. Last edited by Hopeooooo, May 23, 2022, 2:19 PM
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M11100111001Y1R
121 posts
#2 • 3 Y
Y by Hopeooooo, Mahdi.sh, a22886
I don't remember a combinatorics problem harder than this one...

Answer: $2\times5^{17}.$
Official Solution: Let the alphabet of the letters $a_1,a_2,\cdots,a_{25}.$ By a piece we always mean a piece of the strip containing exactly $17$ consecutive letters; Different piece may contain the same word. Say that a piece is singular if the word it contain is such. We start with constructing an example containing $N= 2\times5^{17}$ singular words. Define a word $W=a_1,a_2,\cdots,a_{17};$ This will be the word having $k=5^{16}$ non-overlapping copies on the strip. There exist exactly $k=25^8$ possible $8-$letter sequences, consisting of letters $a_{18},a_{19},\cdots,a_{25};$ Put them onto the strip in an arbitrary order, separating each two sequences by an instance of $W.$ Each segment of the strip containing one $8-$sequence mentioned above (and no other letters) will be referred as a part. Notice that the strip contains exactly $(8+17)k=5^{18}$ letters. Clearly, the obtained strip contains $k$ non-overlapping copies of $W.$ Now we show that any piece containing a whole part is singular moreover, that the word it contains is met on no other piece. Since a part can be situated in a piece at $10$ different positions (starting from the $1-$st, from the $2-$nd, \cdots, or from the $10-$th letter of a piece), we will get that there are at least $N=10\times 5^{16}$ singular words. Consider an arbitrary piece $p$ containing a a word $P.$ Either this piece contains a unique non empty prefix which coincides with some suffix of $W,$ or there is no such prefix only in this case we will say that such prefix is empty. Let $b$ be the length of the defined prefix. Define similarly a suffix of $P$ which coincides with a prefix of $W,$ and denote its length by $e.$ Notice that the defined prefix and suffix do not overlap whenever $P\not =W$ (if $P=W,$ we have $b=c=17$). If the piece contains no whole part, then $max\{b,e\}>9.$ If the piece contains a part then $b+e=9$ and $0\le b,e\le9.$ Thus, piece $p$ contains a part iff $max\{b,e\}\le9,$ and in this case the position of the part at $P$ (and hence the position of $p$ at the strip) is uniquely determined. Therefore, in this case $P$ is met only one piece $p,$ so this piece is singular. We have proven that the constructed example works. It remains to prove that the number of singular words cannot exceed $N.$ Enumerate the positions in the strip successively by $1,2,\cdots,5^{18}$ (the numeration is cyclic module $5^{18}$). Let $p_i$ denote the piece starting at positions $i,$ and let $P_i$ be the word on that piece. Let $n_1,\cdots,n_k$ be a positive such that the pieces $p_{n_1},p_{n_2},\cdots,p_{n_k}$ are pairwise disjoint and contain the same word $W$ (from the problem statement). Clearly, those pieces are not singular. For $i=1,2,\cdots,8$ and $1\le s \le k,$ we say that a piece $p_{n_{s+1}}$ is a rank $i$ follower, while $p_{n_{s-1}}$ is a rank $i$ predecessor. All these pieces (followers and predecessors) are distinct; Moreover. Followers of a fixed rank are pairwise disjoint, and the same holds for predecessors. We will show that $among\ 8\times 5^{16}$ followers of all ranks, at most $5^{16}$ pieces are singular (we will call this statement a quoted claim in the future); By symmetry, the same bound hold for predecessors. This will yield that there are at least $5^{16}+7 \times 5^{16}+7  \times 5^{16}=3\times 5^{17}$ non-singular pieces, which implies the desired bound. Thus, we are left to prove the claim quoted above. For any rank $i$ follower $p_{n_s+i}$ define its tail as its suffix of length $i$ (the tail consists of all letters which do not lie in $p_{n_s};$ We regard a tail as a sequence of letters). We show by induction on $m=0,1,\cdots,8$ that for every sequence $U$ consisting of $(8-m)$ letters, there are no more than $25^m$ followers whose tails contain $U$ as a prefix. The desired claim is obtained by setting $m=8.$ The base case $m=0$ is obvious; If a follower with tail $U$ is singular, then there is only one such follower. Let us perform the inductive step. If there is no singular follower whose tail is $U,$ then every singular follower's tail starting with $U$ starts in fact with some word of the form $U_{a_i}.$ For every $i=1,2,\cdots,25,$ there are at most $25^{m-1}$ such followers, by the inductive hypothesis. So the total number of such followers does not exceed $25\times 25^{m-1}=25^m,$ as desired. Finally, if there is a singular follower $p_{n_s+8-m}$ whose tail is $U,$ then such follower is unique. Therefore, all followers of larger ranks whose tails start with $U$ correspond to the same copy $p_{n_s}$ of $W.$ Then the number of such followers (including $p_{n_s+8-m }$ itself) is at most $m+1 \le 25^m,$ as desired again.
This post has been edited 2 times. Last edited by M11100111001Y1R, Oct 13, 2022, 2:10 PM
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