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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
Two lines meeting on circumcircle
Zhero   54
N 13 minutes ago by Ilikeminecraft
Source: ELMO Shortlist 2010, G4; also ELMO #6
Let $ABC$ be a triangle with circumcircle $\omega$, incenter $I$, and $A$-excenter $I_A$. Let the incircle and the $A$-excircle hit $BC$ at $D$ and $E$, respectively, and let $M$ be the midpoint of arc $BC$ without $A$. Consider the circle tangent to $BC$ at $D$ and arc $BAC$ at $T$. If $TI$ intersects $\omega$ again at $S$, prove that $SI_A$ and $ME$ meet on $\omega$.

Amol Aggarwal.
54 replies
Zhero
Jul 5, 2012
Ilikeminecraft
13 minutes ago
Help me this problem. Thank you
illybest   3
N 38 minutes ago by jasperE3
Find f: R->R such that
f( xy + f(z) ) = (( xf(y) + yf(x) )/2) + z
3 replies
illybest
Today at 11:05 AM
jasperE3
38 minutes ago
line JK of intersection points of 2 lines passes through the midpoint of BC
parmenides51   4
N an hour ago by reni_wee
Source: Rioplatense Olympiad 2018 level 3 p4
Let $ABC$ be an acute triangle with $AC> AB$. be $\Gamma$ the circumcircle circumscribed to the triangle $ABC$ and $D$ the midpoint of the smallest arc $BC$ of this circle. Let $E$ and $F$ points of the segments $AB$ and $AC$ respectively such that $AE = AF$. Let $P \neq A$ be the second intersection point of the circumcircle circumscribed to $AEF$ with $\Gamma$. Let $G$ and $H$ be the intersections of lines $PE$ and $PF$ with $\Gamma$ other than $P$, respectively. Let $J$ and $K$ be the intersection points of lines $DG$ and $DH$ with lines $AB$ and $AC$ respectively. Show that the $JK$ line passes through the midpoint of $BC$
4 replies
parmenides51
Dec 11, 2018
reni_wee
an hour ago
AGM Problem(Turkey JBMO TST 2025)
HeshTarg   3
N an hour ago by ehuseyinyigit
Source: Turkey JBMO TST Problem 6
Given that $x, y, z > 1$ are real numbers, find the smallest possible value of the expression:
$\frac{x^3 + 1}{(y-1)(z+1)} + \frac{y^3 + 1}{(z-1)(x+1)} + \frac{z^3 + 1}{(x-1)(y+1)}$
3 replies
HeshTarg
2 hours ago
ehuseyinyigit
an hour ago
book/resource recommendations
walterboro   1
N 2 hours ago by Konigsberg
hi guys, does anyone have book recs (or other resources) for like aime+ level alg, nt, geo, comb? i want to learn a lot of theory in depth
also does anyone know how otis or woot is like from experience?
1 reply
walterboro
Yesterday at 8:57 PM
Konigsberg
2 hours ago
Compilation of functions problems
Saucepan_man02   5
N 2 hours ago by Konigsberg
Could anyone post some handout/compilation of problems related to functions (difficulty similar to AIME/ARML/HMMT etc)?

Thanks..
5 replies
Saucepan_man02
May 7, 2025
Konigsberg
2 hours ago
polynomials book recs
sunshine_12   2
N 2 hours ago by Konigsberg
hi all! I know pretty much all of the basic high school algebra upto 11th grade- quadratics, solving equations, matrices nd determinants, etc. I was looking for book recs or handouts on polynomials, but pls know that I have no previous experience whatsoever in olympiad algebra. I did try from an excursion in mathematics but couldn't really approach the problems. any help would be rlly appreciated.
xx
2 replies
sunshine_12
5 hours ago
Konigsberg
2 hours ago
Continued fraction
ReticulatedPython   4
N 3 hours ago by jasperE3
Find the exact value of the continued fraction $$1^2+\frac{1}{2^2+\frac{1}{3^2+\frac{1}{4^2+\frac{1}{5^2+\cdots}}}}
$$
I know that it is approximately $1.2432$ but I am looking for the exact value. Does anyone know how to solve this problem?
4 replies
ReticulatedPython
5 hours ago
jasperE3
3 hours ago
Weird locus problem
Sedro   3
N 4 hours ago by Sedro
Points $A$ and $B$ are in the coordinate plane such that $AB=2$. Let $\mathcal{H}$ denote the locus of all points $P$ in the coordinate plane satisfying $PA\cdot PB=2$, and let $M$ be the midpoint of $AB$. Points $X$ and $Y$ are on $\mathcal{H}$ such that $\angle XMY = 45^\circ$ and $MX\cdot MY=\sqrt{2}$. The value of $MX^4 + MY^4$ can be expressed in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
3 replies
Sedro
Yesterday at 3:12 AM
Sedro
4 hours ago
Can someone catch my mistake?
jL56L06B9   3
N 5 hours ago by RegalSparrow
$4$ knights and $4$ jokers are to be seated around a round table. They randomly pick the particular seats they want to sit at. What's the probability they'd be seated alternately?
sol
what I did
3 replies
jL56L06B9
Aug 2, 2020
RegalSparrow
5 hours ago
进制问题1234
yyhloveu1314   0
Today at 2:28 PM
I want to represent a decimal number using an unconventional base (-4) numeral system, and I wish to prove the uniqueness and existence of this representation.
0 replies
yyhloveu1314
Today at 2:28 PM
0 replies
dirichlet
spiralman   0
Today at 9:36 AM
Let n be a positive integer. Consider 2n+1 distinct positive integers whose total sum is less than (n+1)(3n+1). Prove that among these 2n+1 numbers, there exist two numbers whose sum is 2n+1.
0 replies
spiralman
Today at 9:36 AM
0 replies
Inequalities
sqing   0
Today at 9:08 AM
Let $ a, b, c >0, a^2 + \frac{b}{a}  = 8 $ and $ 3a + b + c \geq  9\sqrt{3} .$ Prove that $$   ab + c^2\geq 18$$
0 replies
sqing
Today at 9:08 AM
0 replies
Plz help
Bet667   2
N Today at 7:42 AM by jasperE3
f:R-->R for any integer x,y
f(yf(x)+f(xy))=(x+f(x))f(y)
find all function f
(im not good at english)
2 replies
Bet667
Jan 28, 2024
jasperE3
Today at 7:42 AM
easy inequality with a^2+b^2+c^2=3abc
Sayan   9
N Nov 13, 2019 by Math-wiz
Let $a,b,c$ be positive reals such that $a^2+b^2+c^2=3abc$. Prove that
\[\frac{a}{b^2c^2}+\frac{b}{c^2a^2}+\frac{c}{a^2b^2} \geq \frac{9}{a+b+c}\]
9 replies
Sayan
Jun 29, 2012
Math-wiz
Nov 13, 2019
easy inequality with a^2+b^2+c^2=3abc
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Sayan
2130 posts
#1 • 3 Y
Y by integrated_JRC, Adventure10, Mango247
Let $a,b,c$ be positive reals such that $a^2+b^2+c^2=3abc$. Prove that
\[\frac{a}{b^2c^2}+\frac{b}{c^2a^2}+\frac{c}{a^2b^2} \geq \frac{9}{a+b+c}\]
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oldbeginner
3428 posts
#2 • 4 Y
Y by ayan_mathematics_king, Adventure10, Mango247, ehuseyinyigit
Sayan wrote:
Let $a,b,c$ be positive reals such that $a^2+b^2+c^2=3abc$. Prove that
\[\frac{a}{b^2c^2}+\frac{b}{c^2a^2}+\frac{c}{a^2b^2} \geq \frac{9}{a+b+c}\]
$LHS=\frac{a^4}{a^3b^2c^2}+\frac{b^4}{b^3c^2a^2}+\frac{c^4}{c^3a^2b^2}\ge\frac{(a^2+b^2+c^2)^2}{a^2b^2c^2(a+b+c)}=\frac{9}{a+b+c}$
I used C-S and $a^2+b^2+c^2=3abc$
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hatchguy
555 posts
#3 • 4 Y
Y by RISH2909, Adventure10, Mango247, and 1 other user
Probably the same but it just looks nicer...

Divide by $abc$ the condition and we get $\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}=3$

By C-S $(\frac{a}{b^2c^2}+\frac{b}{a^2c^2}+\frac{c}{a^2b^2}) (a+b+c) \ge (\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab})^2 = 9$ and we are done.
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sahadian
112 posts
#4 • 2 Y
Y by Adventure10, Mango247
we can change it to
$(9.a^3/(a^2+b^2+c^2)^2)+(9.b^3/(a^2+b^2+c^2)^2)+(9.(c^3/a^2+b^2+c^2)^2)>=(9/a+b+c)$so we have
$a^3+b^3+c^3/(a^2+b^2+c^2)^2>=1/a+b+c$
$(a^3+b^3+c^3).(a+b+c)>=(a^2+b^2+c^2)^2$
and its a famous inequality
This post has been edited 1 time. Last edited by sahadian, Jan 22, 2013, 11:06 AM
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sahadian
112 posts
#5 • 2 Y
Y by Adventure10, Mango247
i think these all answer are like each other
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nicusorz
1158 posts
#6 • 2 Y
Y by Adventure10, Mango247
we have
$ \frac{a^{3}+b^{3}+c^{3}}{a^{2}b^{2}c^{2}}\geq \frac{9}{a+b+c}\Leftrightarrow (\sum a)(\sum a^{3})\geq (\sum a^{2})^{2}  $
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mathbuzz
803 posts
#7 • 2 Y
Y by Adventure10, Mango247
$ \sum \frac{a}{b^2.c^2}$=$\sum\frac{a^4}{a^3.b^2.c^2} \ge \frac{[\sum a^2]^2}{a^2.b^2.c^2(a+b+c)}$=$\frac{9}{a+b+c}$
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shivangjindal
676 posts
#8 • 2 Y
Y by Adventure10, Mango247
Sayan wrote:
Let $a,b,c$ be positive reals such that $a^2+b^2+c^2=3abc$. Prove that
\[\frac{a}{b^2c^2}+\frac{b}{c^2a^2}+\frac{c}{a^2b^2} \geq \frac{9}{a+b+c}\]


$ \frac{a}{b^2c^2}+\frac{b}{c^2a^2}+\frac{c}{a^2b^2} = \frac{a^4}{a^3b^2c^2}+\frac{b^4}{b^3c^2a^2}+\frac{c^3}{c^3a^2b^2} \geq_ \frac{(a^2+b^2+c^2)^2}{a^3b^2c^2 + b^3c^2a^2 + c^3a^2b^2} =  \frac{(a^2+b^2+c^2)^2}{a^2b^2c^2(a+b+c)} $

since $a^2+b^2+c^2 = 3abc$

$  \frac{(a^2+b^2+c^2)^2}{a^2b^2c^2(a+b+c)} =  \frac{(3abc)^2}{a^2b^2c^2(a+b+c)} = \frac{9}{a+b+c} $
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sayantanchakraborty
505 posts
#9 • 2 Y
Y by Adventure10, Mango247
Easy but nice problem. The inequality transforms to $(a^3+b^3+c^3)(a+b+c)>=9a^2b^2c^2$. Use cauchy scwarz inequality.
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Math-wiz
6107 posts
#10 • 2 Y
Y by Adventure10, Mango247
Clearing the denominators, we need to prove
$$(a+b+c)(a^3+b^3+c^3)\geq 9a^2b^2c^2$$which is true by C-S
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