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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
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0 replies
jlacosta
Mar 2, 2025
0 replies
D1010 : How it is possible ?
Dattier   8
N 9 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
9 minutes ago
GOTEEM #5: Circumcircle passes through fixed point
tworigami   21
N 33 minutes ago by Ilikeminecraft
Source: GOTEEM: Mock Geometry Contest
Let $ABC$ be a triangle and let $B_1$ and $C_1$ be variable points on sides $\overline{BA}$ and $\overline{CA}$, respectively, such that $BB_1 = CC_1$. Let $B_2 \neq B_1$ denote the point on $\odot(ACB_1)$ such that $BC_1$ is parallel to $B_1B_2$, and let $C_2 \neq C_1$ denote the point on $\odot(ABC_1)$ such that $CB_1$ is parallel to $C_1C_2$. Prove that as $B_1, C_1$ vary, the circumcircle of $\triangle AB_2C_2$ passes through a fixed point, other than $A$.

Proposed by tworigami
21 replies
tworigami
Jan 2, 2020
Ilikeminecraft
33 minutes ago
Strike the inequality
giangtruong13   1
N 36 minutes ago by arqady
Source: Idk
Let $a,b,c \geq 0$ satisfy that $a+b+c=3$. Prove that $$\sum a\sqrt{b^3+1} \leq 5$$
1 reply
giangtruong13
5 hours ago
arqady
36 minutes ago
Calculus rather than inequalities
darij grinberg   12
N 39 minutes ago by asdf334
Source: German TST, IMO ShortList 2003, algebra problem 3
Consider pairs of the sequences of positive real numbers \[a_1\geq a_2\geq a_3\geq\cdots,\qquad b_1\geq b_2\geq b_3\geq\cdots\]and the sums \[A_n = a_1 + \cdots + a_n,\quad B_n = b_1 + \cdots + b_n;\qquad n = 1,2,\ldots.\]For any pair define $c_n = \min\{a_i,b_i\}$ and $C_n = c_1 + \cdots + c_n$, $n=1,2,\ldots$.


(1) Does there exist a pair $(a_i)_{i\geq 1}$, $(b_i)_{i\geq 1}$ such that the sequences $(A_n)_{n\geq 1}$ and $(B_n)_{n\geq 1}$ are unbounded while the sequence $(C_n)_{n\geq 1}$ is bounded?

(2) Does the answer to question (1) change by assuming additionally that $b_i = 1/i$, $i=1,2,\ldots$?

Justify your answer.
12 replies
darij grinberg
Jul 15, 2004
asdf334
39 minutes ago
No more topics!
f(y+f(x))=f(y)+x
janlaw   6
N Nov 24, 2024 by AshAuktober
Source: problem 1081. from http://genesiszero617.blogspot.com
For all $x,y$ real find continuous $f(x)$ from real numbers to real numbers satisfying the following equation:

$f(y+f(x))=f(y)+x$

EDITED

my solution is in the attachment.
6 replies
janlaw
Feb 24, 2018
AshAuktober
Nov 24, 2024
f(y+f(x))=f(y)+x
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G H BBookmark kLocked kLocked NReply
Source: problem 1081. from http://genesiszero617.blogspot.com
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janlaw
3 posts
#1 • 1 Y
Y by Adventure10
For all $x,y$ real find continuous $f(x)$ from real numbers to real numbers satisfying the following equation:

$f(y+f(x))=f(y)+x$

EDITED

my solution is in the attachment.
Attachments:
This post has been edited 1 time. Last edited by janlaw, Feb 24, 2018, 11:15 AM
Reason: continuity added
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Vrangr
1600 posts
#2 • 2 Y
Y by Adventure10, Mango247
You can't assume continuity.

This is my 666th post. :diablo:
This post has been edited 1 time. Last edited by Vrangr, Feb 24, 2018, 11:14 AM
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TheDarkPrince
3042 posts
#3 • 2 Y
Y by Maths_Guy, Adventure10
janlaw wrote:
For all $x,y$ real find continuous $f(x)$ from real numbers to real numbers satisfying the following equation:

$f(y+f(x))=f(y)+x$

Let $P(x,y)$ be the assertion.
Claim: $f$ is injective.
Proof: Suppose $f(a)=f(b)$, $P(a,y)$ and $P(b,y)$ give injectivity.

$P(0,0)$ gives $f(f(0))=f(0)\Rightarrow f(0)=0$. $P(x,0)$ gives $f(f(x))=x$. $P(f(x),y)$ gives $f(x+y)=f(x)+f(y)$. $f$ is continuous, so $f(x)=cx$. Plugging this in the original equation, $f(x)=x$ and $f(x)=-x$ are the only solutions.
This post has been edited 1 time. Last edited by TheDarkPrince, Feb 24, 2018, 1:30 PM
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WizardMath
2487 posts
#4 • 1 Y
Y by Adventure10
@above, $f(x) = -x$ also works.
This post has been edited 1 time. Last edited by WizardMath, Feb 24, 2018, 11:36 AM
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janlaw
3 posts
#5 • 2 Y
Y by Adventure10, Mango247
Once we know that the solution has the form $f(x)=cx$ then returning back to the original equation we obtain two solutions $f(x)=x$ or $f(x)=-x$.
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nguyenhuybao_06
38 posts
#6
Y by
For all $x,y,z\in\mathbb{R}$ we have $$f(f(x)+y+z)=x+f(y+z).$$We also have $$f(f(x)+y+z)=f(f(f(y)+x)+z)=f(y)+x+f(z),\forall x,y,z\in\mathbb{R}.$$Now we have $$f(y+z)=f(y)+f(z),\forall x,y,z\in\mathbb{R}.$$From $f$ is continuous we claim that $f(x)=cx$ for all $x\in\mathbb{R}$ and $c$ is constant.
We have $c^2x+cy=x+cy$ so that $c=1$ or $c=-1.$
So that $f(x)\equiv x$ or $f(x)\equiv -x$ are solutions.
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AshAuktober
912 posts
#7
Y by
$f$ is bijective as $P(x, 0) $ shows.
In particular $f^2(x) = x+c$
Now replace $x$ with $f(x) $ and turn it into cauchy to see $f$ is linear, from where we can check for solutions
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