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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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0 replies
jlacosta
Apr 2, 2025
0 replies
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Weird Inequality Problem
Omerking   1
N 6 minutes ago by Nuran2010
Following inequality is given:
$$3\geq ab+bc+ca\geq \dfrac{1}{3}$$Find the range of values that can be taken by :
$1)a+b+c$
$2)abc$

Where $a,b,c$ are positive reals.
1 reply
Omerking
26 minutes ago
Nuran2010
6 minutes ago
J
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Weakly-periodic functions compared with lines
Miquel-point   2
N 7 minutes ago by dgrozev
Source: KoMaL A. 895
Let's call a function $f:\mathbb R\to\mathbb R$ weakly periodic if it is continuous and $f(x+1)=f(f(x))+1$ for all $x\in\mathbb R$.
a) Does there exist a weakly periodic function such that $f(x)>x$ for all $x\in\mathbb R$?
b) Does there exist a weakly periodic function such that $f(x)<x$ for all $x\in\mathbb R$?

Proposed by: András Imolay, Budapest
2 replies
Miquel-point
Jan 13, 2025
dgrozev
7 minutes ago
J
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Beatty sequences of continued fractions
Miquel-point   1
N 31 minutes ago by starchan
Source: KoMaL A. 903
Let the irrational number
\[\alpha =1-\cfrac{1}{2a_1-\cfrac{1}{2a_2-\cfrac{1}{2a_3-\cdots}}}\]where coefficients $a_1, a_2, \ldots$ are positive integers, infinitely many of which are greater than $1$. Prove that for every positive integer $N$ at least half of the numbers $\lfloor \alpha\rfloor, \lfloor 2\alpha\rfloor, \ldots, \lfloor N\alpha\rfloor$ are even.

Proposed by Géza Kós, Budapest
1 reply
Miquel-point
Yesterday at 5:41 PM
starchan
31 minutes ago
J
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Easy Inceter related geometry
Pluto1708   10
N 36 minutes ago by ihategeo_1969
Source: STEMS 2021/CAT B/P3
Let $ABC$ be a triangle with $I$ as incenter.The incircle touches $BC$ at $D$.Let $D'$ be the antipode of $D$ on the incircle.Make a tangent at $D'$ to incircle.Let it meet $(ABC)$ at $X,Y$ respectively.Let the other tangent from $X$ meet the other tangent from $Y$ at $Z$.Prove that $(ZBD)$ meets $IB$ at the midpoint of $IB$
10 replies
Pluto1708
Jan 24, 2021
ihategeo_1969
36 minutes ago
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No more topics!
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Abcd is a square
Fermat -Euler   3
N Dec 24, 2022 by dheerstar12
Source: India National Olympiad 2003, Problem 1
$ABCD$ is a convex quadrilateral. $K$, $L$, $M$, $N$ are the midpoints of the sides $AB$, $BC$, $CD$, $DA$. $BD$ bisects $KM$ at $Q$. $QA = QB = QC = QD$ , and$\frac{LK}{LM} = \frac{CD}{CB}$. Prove that $ABCD$ is a square
3 replies
Fermat -Euler
Oct 10, 2005
dheerstar12
Dec 24, 2022
J
Abcd is a square
G H J
Reveal topic tags
geometry
parallelogram
geometry proposed
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G H BBookmark kLocked kLocked NReply
Source: India National Olympiad 2003, Problem 1
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Fermat -Euler
444 posts
Fermat -Euler
#1 h  Oct 10, 2005, 5:19 PM • 2 Y
Y by Adventure10, Mango247
$ABCD$ is a convex quadrilateral. $K$, $L$, $M$, $N$ are the midpoints of the sides $AB$, $BC$, $CD$, $DA$. $BD$ bisects $KM$ at $Q$. $QA = QB = QC = QD$ , and$\frac{LK}{LM} = \frac{CD}{CB}$. Prove that $ABCD$ is a square
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ashwath.rabindranath
437 posts
ashwath.rabindranath
#2 h  Oct 11, 2005, 2:55 AM • 4 Y
Y by Adventure10, Adventure10, Mango247, Mathlover112358
Fermat -Euler wrote:
$ABCD$ is a convex quadrilateral. $K$, $L$, $M$, $N$ are the midpoints of the sides $AB$, $BC$, $CD$, $DA$. $BD$ bisects $KM$ at $Q$. $QA = QB = QC = QD$ , and$\frac{LK}{LM} = \frac{CD}{CB}$. Prove that $ABCD$ is a square

Erm this was the easiest INMO problem ever, I think.
We prove that the triangles QKB and QMD are congruent.
which implies that MD = KB and so AB=CD
also AD is parallel to CD as a pair of alternate interior angles are equal.From the above congruency.
So,ABCD will become a parallelogram.
With this we can show that the angles QMD,QMC,QMA,QMB are all equal to $\pi/2$(this is a direct result of congruency) and so
all the angles of the parallelogram will be $\pi/2$ and now the quadrileteral becomes a rectangle. As by congruency LK/LM=1,BC/CD=1, i.e BC=CD .
Now ABCD becomes a square,and we are done.
Ashwath :D
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Sahil
661 posts
Sahil
#3 h  Feb 2, 2012, 4:26 PM • 2 Y
Y by Adventure10, Mango247
Agreed , easiest ever :) .
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dheerstar12
18 posts
dheerstar12
#4 h  Dec 24, 2022, 11:53 AM
Y by
how did this make it to inmo :huh:
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