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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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0 replies
jlacosta
Apr 2, 2025
0 replies
FE inequality from Iran
mojyla222   2
N a few seconds ago by lucas3617
Source: Iran 2025 second round P5
Find all functions $f:\mathbb{R}^+ \to \mathbb{R}$ such that for all $x,y,z>0$
$$
3(x^3+y^3+z^3)\geq f(x+y+z)\cdot f(xy+yz+xz) \geq (x+y+z)(xy+yz+xz).
$$
2 replies
mojyla222
4 hours ago
lucas3617
a few seconds ago
Similar triangles formed by angular condition
Mahdi_Mashayekhi   3
N 7 minutes ago by mathuz
Source: Iran 2025 second round P3
Point $P$ lies inside of scalene triangle $ABC$ with incenter $I$ such that $:$
$$ 2\angle ABP = \angle BCA , 2\angle ACP = \angle CBA $$Lines $PB$ and $PC$ intersect line $AI$ respectively at $B'$ and $C'$. Line through $B'$ parallel to $AB$ intersects $BI$ at $X$ and line through $C'$ parallel to $AC$ intersects $CI$ at $Y$. Prove that triangles $PXY$ and $ABC$ are similar.
3 replies
+1 w
Mahdi_Mashayekhi
2 hours ago
mathuz
7 minutes ago
This year's Diophantine equation
GreekIdiot   1
N 20 minutes ago by Math_Physico
Source: own
Let $x,y,z \in \mathbb {Z}$ such that $5^x-y^2=z^3+2025$. Find all such $(x,y,z)$.
1 reply
1 viewing
GreekIdiot
an hour ago
Math_Physico
20 minutes ago
Mildly interesting
GreekIdiot   0
27 minutes ago
Source: my teacher
Let numbers $a_1,a_2,a_3,\cdots,a_n \in \mathbb{Z_+}$ such that $\forall \: 1 \leq i \leq n, a_i<1000$ and $\forall \: i \neq j, \: lcm(a_i,a_j)>1000$. Prove that $\sum_{i=1}^{n} \dfrac{1}{a_i}<3/2$.
0 replies
GreekIdiot
27 minutes ago
0 replies
Differential equations , Matrix theory
c00lb0y   1
N 3 hours ago by loup blanc
Source: RUDN MATH OLYMP 2024 problem 4
Any idea?? Diff equational system combined with Matrix theory.
Consider the equation dX/dt=X^2, where X(t) is an n×n matrix satisfying the condition detX=0. It is known that there are no solutions of this equation defined on a bounded interval, but there exist non-continuable solutions defined on unbounded intervals of the form (t ,+∞) and (−∞,t). Find n.
1 reply
c00lb0y
Apr 17, 2025
loup blanc
3 hours ago
OMOUS-2025 (Team Competition) P7
enter16180   1
N 3 hours ago by RobertRogo
Source: Open Mathematical Olympiad for University Students (OMOUS-2025)
Let $R$ be a ring not assumed to have an identity, with the following properties:
(i) There is an element of $R$ that is not nilpotent.
(ii) If $x_{1}, \ldots, x_{2024}$ are nonzero elements of $R$, then $\sum_{j=1}^{2024} x_{j}^{2025}=0$.

Show that $R$ is a division ring, that is, the nonzero elements of R form a group under multiplication.
1 reply
enter16180
Yesterday at 12:04 PM
RobertRogo
3 hours ago
Question
Riptide1901   1
N 4 hours ago by Mathzeus1024
Isn't their a left hand side of the graph that they're leaving out? I'm confused why they only consider $x>\sqrt{15}/2$ and not $x<-\sqrt{15}/2.$
1 reply
Riptide1901
Jan 29, 2025
Mathzeus1024
4 hours ago
Putnam 2019 A5
hoeij   17
N Today at 5:34 AM by Ilikeminecraft
Let $p$ be an odd prime number, and let $\mathbb{F}_p$ denote the field of integers modulo $p$. Let $\mathbb{F}_p[x]$ be the ring of polynomials over $\mathbb{F}_p$, and let $q(x) \in \mathbb{F}_p[x]$ be given by $q(x) = \sum_{k=1}^{p-1} a_k x^k$ where $a_k = k^{(p-1)/2}$ mod $p$. Find the greatest nonnegative integer $n$ such that $(x-1)^n$ divides $q(x)$ in $\mathbb{F}_p[x]$.
17 replies
hoeij
Dec 10, 2019
Ilikeminecraft
Today at 5:34 AM
Projection of angle into planes
geekmath-31   0
Today at 5:14 AM
Question:
consider the angle formed by 2 half lines in the three dimensional space. Prove that the average of the projection of the angle into all of the planes is equal to the angle

The answer is in the attachments.

Please could anyone prove the answer to me in detail.
0 replies
geekmath-31
Today at 5:14 AM
0 replies
Projection of angle into planes
geekmath-31   0
Today at 4:31 AM
Question:
consider the angle formed by 2 half lines in the three dimensional space. Prove that the average of the projection of the angle into all of the planes is equal to the angle

The answer is in the attachments.

Please could anyone prove the answer to me in detail.
0 replies
geekmath-31
Today at 4:31 AM
0 replies
Soviet Union University Mathematical Contest
geekmath-31   0
Today at 3:40 AM
Given a n*n matrix A, prove that there exists a matrix B such that ABA = A

Solution: I have submitted the attachment

The answer is too symbol dense for me to understand the answer.
What I have undertood:

There is use of direct product in the orthogonal decomposition. The decomposition is made with kernel and some T (which the author didn't mention) but as per orthogonal decomposition it must be its orthogonal complement.

Can anyone explain the answer in much much more detail with less use of symbols ( you can also use symbols but clearly define it).

Also what is phi | T ?
0 replies
geekmath-31
Today at 3:40 AM
0 replies
Dimension of a Linear Space
EthanWYX2009   0
Today at 2:50 AM
Source: 2024 May taca-10
Let \( V \) be a $10$-dimensional inner product space of column vectors, where for \( v = (v_1, v_2, \dots, v_{10})^T \) and \( w = (w_1, w_2, \dots, w_{10})^T \), the inner product of \( v \) and \( w \) is defined as \[\langle v, w \rangle = \sum_{i=1}^{10} v_i w_i.\]For \( u \in V \), define a linear transformation \( P_u \) on \( V \) as follows:
\[ P_u : V \to V, \quad x \mapsto x - \frac{2\langle x, u \rangle u}{\langle u, u \rangle} \]Given \( v, w \in V \) satisfying
\[ 0 < \langle v, w \rangle < \sqrt{\langle v, v \rangle \langle w, w \rangle} \]let \( Q = P_v \circ P_w \). Then the dimension of the linear space formed by all linear transformations \( P : V \to V \) satisfying \( P \circ Q = Q \circ P \) is $\underline{\quad\quad}.$
0 replies
EthanWYX2009
Today at 2:50 AM
0 replies
Matrices and Determinants
Saucepan_man02   5
N Today at 1:23 AM by Saucepan_man02
Hello

Can anyone kindly share some problems/handouts on matrices & determinants (problems like Putnam 2004 A3, which are simple to state and doesnt involve heavy theory)?

Thank you..
5 replies
Saucepan_man02
Apr 4, 2025
Saucepan_man02
Today at 1:23 AM
Putnam 1960 B1
sqrtX   4
N Yesterday at 11:26 PM by KAME06
Source: Putnam 1960
Find all integer solutions $(m,n)$ to $m^{n}=n^{m}.$
4 replies
sqrtX
Jun 18, 2022
KAME06
Yesterday at 11:26 PM
Quadrilateral and Symmetric Points
MathPanda1   3
N Aug 25, 2015 by Rice12
Source: Singapore 2014 #1
The quadrilateral $ABCD$ is inscribed in a circle which has diameter $BD$. Points $A'$ and $B'$ are symmetric to $A$ and $B$ with respect to the lines $BD$ and $AC$ respectively. If the lines $A'C$ and $BD$ intersect at $P$, and the lines $AC$ and $B'D$ intersect at $Q$, prove that $PQ$ is perpendicular to $AC$.
3 replies
MathPanda1
Aug 24, 2015
Rice12
Aug 25, 2015
Quadrilateral and Symmetric Points
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G H BBookmark kLocked kLocked NReply
Source: Singapore 2014 #1
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MathPanda1
1135 posts
#1 • 2 Y
Y by Adventure10, Mango247
The quadrilateral $ABCD$ is inscribed in a circle which has diameter $BD$. Points $A'$ and $B'$ are symmetric to $A$ and $B$ with respect to the lines $BD$ and $AC$ respectively. If the lines $A'C$ and $BD$ intersect at $P$, and the lines $AC$ and $B'D$ intersect at $Q$, prove that $PQ$ is perpendicular to $AC$.
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gavrilos
233 posts
#2 • 2 Y
Y by Adventure10, Mango247
Hello!

Suppose that $E\equiv BD\cap AC$.

Obviously $A'$ lies on $c$ (i.e. the circumcircle of $ABCD$).Since $BD$ is the perpendicular bisector of $AA'$ we get

$\angle{BPA}=\angle{BPC} \ (1)$.Also $\angle{DBA}=\angle{DBA'}$ thus

$\angle{PCD}=\angle{A'CD}=180^{\circ}-\angle{DBA'}=180^{\circ}-\angle{DBA}=\angle{ABP} \ (2)$.

$(1),(2)$ imply that $\triangle{ABP}\simeq \triangle{PCD}$ thus $\angle{BAP}=\angle{PDC}=\angle{BDC}=\angle{BAC}=\angle{BAE}$,that is,

$BA$ is the internal bisector of $\angle{EAP}$.Since $BD$ is a diameter,we have $AD\perp BA$ thus $AD$ is the

external bisector of $\angle{EAP}$ whence we get that the pencil $A.(P,B,E,D)$ is harmonic.

Hence $(P,E/B,D)$ is a harmonic quadruple.Thus,$Q.(P,B,E,D)$ is a harmonic pencil.

Also $AC$ is the perpendicular bisector of $BB'$.

Since $Q$ lies on $AC$,we get $\angle{BQE}=\angle{DQE}$.Since $Q.(P,B,E,D)$ is harmonic,$QE$ bisects $\angle{BQD}$

$PQ$ is necessarily the external bisector of $\angle{BQD}$ which implies that $PQ\perp QE\Leftrightarrow PQ\perp AC$

which is the desired result.
Attachments:
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LREDK
22 posts
#3 • 2 Y
Y by Adventure10, Mango247
gavrilos wrote:
Hello!

Suppose that $E\equiv BD\cap AC$.

Obviously $A'$ lies on $c$ (i.e. the circumcircle of $ABCD$).Since $BD$ is the perpendicular bisector of $AA'$ we get

$\angle{BPA}=\angle{BPC} \ (1)$.Also $\angle{DBA}=\angle{DBA'}$ thus

$\angle{PCD}=\angle{A'CD}=180^{\circ}-\angle{DBA'}=180^{\circ}-\angle{DBA}=\angle{ABP} \ (2)$.

$(1),(2)$ imply that $\triangle{ABP}\simeq \triangle{PCD}$ thus $\angle{BAP}=\angle{PDC}=\angle{BDC}=\angle{BAC}=\angle{BAE}$,that is,

$BA$ is the internal bisector of $\angle{EAP}$.Since $BD$ is a diameter,we have $AD\perp BA$ thus $AD$ is the

external bisector of $\angle{EAP}$ whence we get that the pencil $A.(P,B,E,D)$ is harmonic.

Hence $(P,E/B,D)$ is a harmonic quadruple.Thus,$Q.(P,B,E,D)$ is a harmonic pencil.

Also $AC$ is the perpendicular bisector of $BB'$.

Since $Q$ lies on $AC$,we get $\angle{BQE}=\angle{DQE}$.Since $Q.(P,B,E,D)$ is harmonic,$QE$ bisects $\angle{BQD}$

$PQ$ is necessarily the external bisector of $\angle{BQD}$ which implies that $PQ\perp QE\Leftrightarrow PQ\perp AC$

which is the desired result.

do you have a way that which not use 【Projective geometry】?
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Rice12
1 post
#4 • 1 Y
Y by Adventure10
There is another one.
Points $C'$ and $D'$ are symmetric to $C$ and $D$ with respect to $BD$ and $AC$ respectively. $E=BD\bigcap AC$ , $H=AA'\bigcap BD$, $Z=BB'\bigcap AC$.

We need to prove that $PBHA$ is circumscribed. Than we need to prove that $\frac{EQ}{EP} = \frac{EH}{EA}$. As we know $\frac{EH}{EA} = \frac{EZ}{EB}$.

$P$ is a center of homothety converting $AA'$ to $CC'$ and $Q$ is a center of homothety converting $BB'$ to $DD'$. Then it's easy to find $\frac{EP}{EH}$ and $\frac{EZ}{EQ}$. After some counting we will only need to prove that $\frac{ \mid EB - ED \mid}{ \mid EA - EC \mid} = \frac{EH}{EA}$. That's also an easy statement.
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