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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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0 replies
jlacosta
May 1, 2025
0 replies
abc = 1 Inequality generalisation
CHESSR1DER   4
N 12 minutes ago by CHESSR1DER
Source: Own
Let $a,b,c > 0$, $abc=1$.
Find min $ \frac{1}{a^m(bx+cy)^n} + \frac{1}{b^m(cx+ay)^n} + \frac{1}{c^m(cx+ay)^n}$.
$1)$ $m,n,x,y$ are fixed positive integers.
$2)$ $m,n,x,y$ are fixed positive real numbers.
4 replies
CHESSR1DER
an hour ago
CHESSR1DER
12 minutes ago
Fond all functions in M with a) f(1)=5/2, b) f(1)=√3
Amir Hossein   4
N 26 minutes ago by mkultra42
Source: IMO LongList 1982 - P34
Let $M$ be the set of all functions $f$ with the following properties:

(i) $f$ is defined for all real numbers and takes only real values.

(ii) For all $x, y \in \mathbb R$ the following equality holds: $f(x)f(y) = f(x + y) + f(x - y).$

(iii) $f(0) \neq 0.$

Determine all functions $f \in M$ such that

(a) $f(1)=\frac 52$,

(b) $f(1)= \sqrt 3$.
4 replies
Amir Hossein
Mar 18, 2011
mkultra42
26 minutes ago
help me please
thuanz123   6
N an hour ago by pavel kozlov
find all $a,b \in \mathbb{Z}$ such that:
a) $3a^2-2b^2=1$
b) $a^2-6b^2=1$
6 replies
thuanz123
Jan 17, 2016
pavel kozlov
an hour ago
Problem 5 (Second Day)
darij grinberg   78
N an hour ago by cj13609517288
Source: IMO 2004 Athens
In a convex quadrilateral $ABCD$, the diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$. The point $P$ lies inside $ABCD$ and satisfies \[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\] Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP=CP$.
78 replies
darij grinberg
Jul 13, 2004
cj13609517288
an hour ago
Find solution of IVP
neerajbhauryal   2
N 6 hours ago by Mathzeus1024
Show that the initial value problem \[y''+by'+cy=g(t)\] with $y(t_o)=0=y'(t_o)$, where $b,c$ are constants has the form \[y(t)=\int^{t}_{t_0}K(t-s)g(s)ds\,\]

What I did
2 replies
neerajbhauryal
Sep 23, 2014
Mathzeus1024
6 hours ago
fourier series?
keroro902   2
N Today at 12:54 PM by Mathzeus1024
f(x)=$\sum _{n=0}^{\infty } \text{cos}(nx)/2^{n}$
f(x) = ?
thanks
2 replies
keroro902
May 14, 2010
Mathzeus1024
Today at 12:54 PM
integrals
FFA21   1
N Today at 11:53 AM by alexheinis
Source: OSSM Comp'25 P1 (HSE IMC qualification)
Find all continuous functions $f:[1,8]\to R$ that:
$\int_1^2f(t^3)^2dt+2\int_1^2sin(t)f(t^3)dt=\frac{2}{3}\int_1^8f(t)dt-\int_1^2(t^2-sin(t))^2dt$
1 reply
FFA21
Yesterday at 8:05 PM
alexheinis
Today at 11:53 AM
Sets on which a continuous function exists
Creativename27   1
N Today at 10:49 AM by alexheinis
Source: My head
Find all $X\subseteq R$ that exist function $f:R\to R$ such $f$ continuous on $X$ and discontinuous on $R/X$
1 reply
Creativename27
Today at 9:50 AM
alexheinis
Today at 10:49 AM
Square on Cf
GreekIdiot   1
N Today at 9:04 AM by alexheinis
Let $f$ be a continuous function defined on $[0,1]$ with $f(0)=f(1)=0$ and $f(t)>0 \: \forall \: t \in (0,1)$. We define the point $X'$ to be the projection of point $X$ on the x-axis. Prove that there exist points $A, B \in C_f$ such that $ABB'A'$ is a square.
1 reply
GreekIdiot
May 12, 2025
alexheinis
Today at 9:04 AM
Japanese Olympiad
parkjungmin   6
N Today at 5:01 AM by mathNcheese_aops
It's about the Japanese Olympiad

I can't solve it no matter how much I think about it.

If there are people who are good at math

Please help me.
6 replies
parkjungmin
May 10, 2025
mathNcheese_aops
Today at 5:01 AM
sum of trace matrices
FFA21   1
N Yesterday at 8:50 PM by Etkan
Source: OSSM Comp'25 P4 (HSE IMC qualification)
$G$ is a finite group of $n \times n$ matrices with respect to multiplication. Prove that:
if $\sum_{M\in G}tr(M)=0$ that $\sum_{M\in G}M=0_{n\times n}$
1 reply
FFA21
Yesterday at 8:27 PM
Etkan
Yesterday at 8:50 PM
lines intersecting motions of an ellipse
FFA21   0
Yesterday at 8:33 PM
Source: OSSM Comp'25 P5 (HSE IMC qualification)
Let $E$ be an infinite set of translated copies (i.e., obtained by parallel translation) of a given ellipse $e$ in the plane, and let $r$ be a fixed straight line. It is known that every straight line parallel to $r$ intersects at least one ellipse from $E$. Prove that there exist infinitely many triples of ellipses from $E$ such that there exists a straight line intersecting all three ellipses in the triple.
0 replies
FFA21
Yesterday at 8:33 PM
0 replies
fibonacci number theory
FFA21   0
Yesterday at 8:21 PM
Source: OSSM Comp'25 P3 (HSE IMC qualification)
$F_n$ fibonacci numbers ($F_1=1, F_2=1$) find all n such that:
$\forall i\in Z$ and $0\leq i\leq F_n$
$C^i_{F_n}\equiv (-1)^i\pmod{F_n+1}$
0 replies
FFA21
Yesterday at 8:21 PM
0 replies
strong polinom
FFA21   0
Yesterday at 8:13 PM
Source: OSSM Comp'25 P2 (HSE IMC qualification)
A polynomial will be called 'strong' if it can be represented as a product of two non-constant polynomials with real non-negative coefficients.
Prove that:
$\exists n$ that $p(x^n)$ 'strong' and $deg(p)>1$ $\implies$ $p(x)$ 'strong'
0 replies
FFA21
Yesterday at 8:13 PM
0 replies
Max and min of Sum of d_k^2
Kunihiko_Chikaya   1
N Apr 24, 2025 by Mathzeus1024
Source: 2012 Yokohama National University entrance exam/Economic, #1
Given $n$ points $P_k(x_k,\ y_k)\ (k=1,\ 2,\ 3,\ \cdots,\ n)$ on the $xy$-plane.
Let $a=\sum_{k=1}^n x_k^2,\ b=\sum_{k=1}^n y_k^2,\ c=\sum_{k=1}^{n} x_ky_k$. Denote by $d_k$ the distance between $P_k$ and the line $l : x\cos \theta +y\sin \theta =0$. Let $L=\sum_{k=1}^n d_k^2$.

Answer the following questions:

(1) Express $L$ in terms of $a,\ b,\ c,\ \theta$.

(2) When $\theta$ moves in the range of $0\leq \theta <\pi$, express the maximum and minimum value of $L$ in terms of $a,\ b,\ c$.
1 reply
Kunihiko_Chikaya
Feb 27, 2012
Mathzeus1024
Apr 24, 2025
Max and min of Sum of d_k^2
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Source: 2012 Yokohama National University entrance exam/Economic, #1
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Kunihiko_Chikaya
14514 posts
#1 • 2 Y
Y by Adventure10, Mango247
Given $n$ points $P_k(x_k,\ y_k)\ (k=1,\ 2,\ 3,\ \cdots,\ n)$ on the $xy$-plane.
Let $a=\sum_{k=1}^n x_k^2,\ b=\sum_{k=1}^n y_k^2,\ c=\sum_{k=1}^{n} x_ky_k$. Denote by $d_k$ the distance between $P_k$ and the line $l : x\cos \theta +y\sin \theta =0$. Let $L=\sum_{k=1}^n d_k^2$.

Answer the following questions:

(1) Express $L$ in terms of $a,\ b,\ c,\ \theta$.

(2) When $\theta$ moves in the range of $0\leq \theta <\pi$, express the maximum and minimum value of $L$ in terms of $a,\ b,\ c$.
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Mathzeus1024
894 posts
#2
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Let $l: y=-\cot(\theta)x$ and define $d_{k}$ as the perpendicular distance from point $P(x_{k},y_{k})$ to $l$. These intersection points $P'$ on $l$ are found to be:

$y=-\cot(\theta)x$ and $y-y_{k} = \tan(\theta)(x-x_{k}) \Rightarrow P'\left(\frac{\tan(\theta)x_{k}-y_{k}}{\tan(\theta)+\cot(\theta)},\frac{-x_{k}+\cot(\theta)y_{k}}{\tan(\theta)+\cot(\theta)}\right)$;

and $d_{k}^{2}= \left(x_{k} - \frac{\tan(\theta)x_{k}-y_{k}}{\tan(\theta)+\cot(\theta)}\right)^2 + \left(y_{k} - \frac{-x_{k}+\cot(\theta)y_{k}}{\tan(\theta)+\cot(\theta)}\right)^2 = \left(\frac{\cot(\theta)x_{k} +y_{k}}{\tan(\theta)+\cot(\theta)}\right)^2 + \left(\frac{x_{k}+\tan(\theta)y_{k}}{\tan(\theta)+\cot(\theta)}\right)^2$;

or $d_{k}^{2} = \frac{[1+\cot^{2}(\theta)]x_{k}^{2} + 2[\tan(\theta)+\cot(\theta)]x_{k}y_{k} + [1+\tan^{2}(\theta)]y_{k}^{2}}{[\tan(\theta)+\cot(\theta)]^2}$;

or $d_{k}^{2} = \frac{\csc^{2}(\theta)x_{k}^{2} + 2[\tan(\theta)+\cot(\theta)]x_{k}y_{k} + \sec^{2}(\theta)y_{k}^{2}}{[\tan(\theta)+\cot(\theta)]^2} = \cos^{2}(\theta)x_{k}^{2} + 2\sin(\theta)\cos(\theta)x_{k}y_{k} +\sin^{2}(\theta)y_{k}^{2}$;

and $L = \sum_{k=1}^{n} d_{k}^{2} = \textcolor{red}{a\cos^{2}(\theta) +b\sin^{2}(\theta) + c\sin(2\theta)}$.

Taking $L'(\theta)= 0$ yields the critical values:

$L'(\theta) = -a\sin(2\theta)+b\sin(2\theta)+2c\cos(2\theta)=0 \Rightarrow \frac{2c}{a-b} = \tan(2\theta)$ or $\frac{a-b}{\sqrt{4c^2+(a-b)^2}} = \cos(2\theta)$ or $\frac{2c}{\sqrt{4c^2+(a-b)^2}} = \sin(2\theta)$;

and a second derivative check of $L$ against these critical values yields:

$L''(\theta) = -2a\cos(2\theta)+2b\cos(2\theta)-4c\sin(2\theta) \Rightarrow \frac{-2a^2+2b^2-8c^2}{\sqrt{4c^2+(a-b)^2}}$;

which $L'' < 0$ for $2b^2 < 2a^2+8c^2$ (hence, a maximum) and $L'' > 0$ for $2b^2 > 2a^2+8c^2$ (hence, a minimum). Thus, we ultimately obtain:

$L\left(\frac{1}{2}\arctan\left(\frac{2c}{a-b}\right)\right) = a\cos^{2}\left(\frac{1}{2}\arctan\left(\frac{2c}{a-b}\right)\right) +b\sin^{2}\left(\frac{1}{2}\arctan\left(\frac{2c}{a-b}\right)\right) +2\sin\left(\frac{1}{2}\arctan\left(\frac{2c}{a-b}\right)\right) \cos\left(\frac{1}{2}\arctan\left(\frac{2c}{a-b}\right)\right)$;

or $L\left(\frac{1}{2}\arctan\left(\frac{2c}{a-b}\right)\right) = \textcolor{red}{\frac{(a+b)\sqrt{4c^2+(a-b)^2}+(a-b)^2+4c^2}{2\sqrt{4c^2+(a-b)^2}}}$.
This post has been edited 12 times. Last edited by Mathzeus1024, Apr 24, 2025, 11:16 AM
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