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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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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
\sqrt{a^2+b^2+2}+\sqrt{b^2+c^2+2 }+\sqrt{c^2+a^2+2}\ge 6
parmenides51   19
N 6 minutes ago by NicoN9
Source: JBMO Shortlist 2017 A1
Let $a, b, c$ be positive real numbers such that $a + b + c + ab + bc + ca + abc = 7$. Prove
that $\sqrt{a^2 + b^2 + 2 }+\sqrt{b^2 + c^2 + 2 }+\sqrt{c^2 + a^2 + 2 } \ge 6$ .
19 replies
parmenides51
Jul 25, 2018
NicoN9
6 minutes ago
Inspired by Austria 2025
sqing   1
N 14 minutes ago by sqing
Source: Own
Let $ a,b\geq 0 ,a,b\neq 1$ and $  a^2+b^2=1. $ Prove that$$   (a + b ) \left( \frac{a}{(b -1)^2} + \frac{b}{(a - 1)^2} \right) \geq 12+8\sqrt 2$$
1 reply
sqing
19 minutes ago
sqing
14 minutes ago
IMO Genre Predictions
ohiorizzler1434   51
N 23 minutes ago by ethan2011
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
51 replies
+1 w
ohiorizzler1434
May 3, 2025
ethan2011
23 minutes ago
Concurrency from isogonal Mittenpunkt configuration
MarkBcc168   17
N 2 hours ago by Ilikeminecraft
Source: Fake USAMO 2020 P3
Let $\triangle ABC$ be a scalene triangle with circumcenter $O$, incenter $I$, and incircle $\omega$. Let $\omega$ touch the sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ at points $D$, $E$, and $F$ respectively. Let $T$ be the projection of $D$ to $\overline{EF}$. The line $AT$ intersects the circumcircle of $\triangle ABC$ again at point $X\ne A$. The circumcircles of $\triangle AEX$ and $\triangle AFX$ intersect $\omega$ again at points $P\ne E$ and $Q\ne F$ respectively. Prove that the lines $EQ$, $FP$, and $OI$ are concurrent.

Proposed by MarkBcc168.
17 replies
MarkBcc168
Apr 28, 2020
Ilikeminecraft
2 hours ago
Inequality with a,b,c
GeoMorocco   7
N 2 hours ago by lele0305
Source: Morocco Training
Let $   a,b,c   $ be positive real numbers such that : $   ab+bc+ca=3   $ . Prove that : $$\frac{\sqrt{1+a^2}}{1+ab}+\frac{\sqrt{1+b^2}}{1+bc}+\frac{\sqrt{1+c^2}}{1+ca}\ge \sqrt{\frac{3(a+b+c)}{2}}$$
7 replies
GeoMorocco
Apr 11, 2025
lele0305
2 hours ago
Property of the divisors of k^3 - 2
Scilyse   2
N 3 hours ago by Assassino9931
Source: KoMaL A. 892
Given two integers, $k$ and $d$ such that $d$ divides $k^3 - 2$. Show that there exists integers $a$, $b$, $c$ satisfying $d = a^3 + 2b^3 + 4c^3 - 6abc$.

Proposed by Csongor Beke and László Bence Simon, Cambridge
2 replies
Scilyse
Jan 13, 2025
Assassino9931
3 hours ago
Centroid, altitudes and medians, and concyclic points
BR1F1SZ   1
N 3 hours ago by sami1618
Source: Austria National MO Part 1 Problem 2
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)
1 reply
BR1F1SZ
5 hours ago
sami1618
3 hours ago
Something nice
KhuongTrang   31
N 3 hours ago by NguyenVanHoa29
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
31 replies
KhuongTrang
Nov 1, 2023
NguyenVanHoa29
3 hours ago
Nordic 2025 P3
anirbanbz   8
N 4 hours ago by lksb
Source: Nordic 2025
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. Let $E$ and $F$ be points on the line segments $AC$ and $AB$ respectively such that $AEHF$ is a parallelogram. Prove that $\vert OE \vert = \vert OF \vert$.
8 replies
anirbanbz
Mar 25, 2025
lksb
4 hours ago
another functional inequality?
Scilyse   32
N 4 hours ago by ihategeo_1969
Source: 2023 ISL A4
Let $\mathbb R_{>0}$ be the set of positive real numbers. Determine all functions $f \colon \mathbb R_{>0} \to \mathbb R_{>0}$ such that \[x \big(f(x) + f(y)\big) \geqslant \big(f(f(x)) + y\big) f(y)\]for every $x, y \in \mathbb R_{>0}$.
32 replies
Scilyse
Jul 17, 2024
ihategeo_1969
4 hours ago
Mount Inequality erupts in all directions!
BR1F1SZ   1
N 4 hours ago by sami1618
Source: Austria National MO Part 1 Problem 1
Let $a$, $b$ and $c$ be pairwise distinct nonnegative real numbers. Prove that
\[
(a + b + c) \left( \frac{a}{(b - c)^2} + \frac{b}{(c - a)^2} + \frac{c}{(a - b)^2} \right) > 4.
\](Karl Czakler)
1 reply
1 viewing
BR1F1SZ
5 hours ago
sami1618
4 hours ago
Division involving difference of squares
BR1F1SZ   1
N 4 hours ago by grupyorum
Source: Austria National MO Part 1 Problem 4
Determine all integers $n$ that can be written in the form
\[
n = \frac{a^2 - b^2}{b},
\]where $a$ and $b$ are positive integers.

(Walther Janous)
1 reply
BR1F1SZ
5 hours ago
grupyorum
4 hours ago
Erasing the difference of two numbers
BR1F1SZ   0
5 hours ago
Source: Austria National MO Part 1 Problem 3
Consider the following game for a positive integer $n$. Initially, the numbers $1, 2, \ldots, n$ are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers $3,6,11$ and $17$ are on the board, then $3$ can be erased as $6 - 3=3$, or $6$ as $17 - 11=6$, or $11$ as $17 - 6=11$.)

For which values of $n$ is it possible to end with only one number remaining on the board?

(Michael Reitmeir)
0 replies
BR1F1SZ
5 hours ago
0 replies
4 wise men and 100 hats. 3 must guess their numbers
NO_SQUARES   2
N 5 hours ago by NO_SQUARES
Source: 239 MO 2025 10-11 p5
There are four wise men in a row, each sees only those following him in the row, i.e. the $1$st sees the other three, the $2$nd sees the $3$rd and $4$th, and the $3$rd sees only the $4$th. The devil has $100$ hats, numbered from $1$ to $100$, he puts one hat on each wise man, and hides the extra $96$ hats. After that, each wise man (in turn: first the first, then the second, etc.) loudly calls a number, trying to guess the number of his hat. The numbers mentioned should not be repeated. When all the wise men have spoken, they take off their hats and check which one of them has guessed. Can the sages to act in such a way that at least three of them knowingly guessed?
2 replies
NO_SQUARES
Yesterday at 5:44 PM
NO_SQUARES
5 hours ago
Cevian Triangle with Perpendiculars
v_Enhance   4
N Jul 7, 2020 by amar_04
Source: ELMO 2014 Shortlist G10, by Sammy Luo
We are given triangles $ABC$ and $DEF$ such that $D\in BC, E\in CA, F\in AB$, $AD\perp EF, BE\perp FD, CF\perp DE$. Let the circumcenter of $DEF$ be $O$, and let the circumcircle of $DEF$ intersect $BC,CA,AB$ again at $R,S,T$ respectively. Prove that the perpendiculars to $BC,CA,AB$ through $D,E,F$ respectively intersect at a point $X$, and the lines $AR,BS,CT$ intersect at a point $Y$, such that $O,X,Y$ are collinear.

Proposed by Sammy Luo
4 replies
v_Enhance
Jul 24, 2014
amar_04
Jul 7, 2020
Cevian Triangle with Perpendiculars
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G H BBookmark kLocked kLocked NReply
Source: ELMO 2014 Shortlist G10, by Sammy Luo
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v_Enhance
6877 posts
#1 • 5 Y
Y by narutomath96, swamih, Gaussian_cyber, Adventure10, swynca
We are given triangles $ABC$ and $DEF$ such that $D\in BC, E\in CA, F\in AB$, $AD\perp EF, BE\perp FD, CF\perp DE$. Let the circumcenter of $DEF$ be $O$, and let the circumcircle of $DEF$ intersect $BC,CA,AB$ again at $R,S,T$ respectively. Prove that the perpendiculars to $BC,CA,AB$ through $D,E,F$ respectively intersect at a point $X$, and the lines $AR,BS,CT$ intersect at a point $Y$, such that $O,X,Y$ are collinear.

Proposed by Sammy Luo
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XmL
552 posts
#2 • 4 Y
Y by narutomath96, Adventure10, Mango247, and 1 other user
We know that $AD,BE,CF$ concur at the orthocenter of $DEF$, denoted by $H$. Let $X'$ denote the isogonal conjugate of $H$ wrt $ABC$, If $DEF$ isn't the pedal triangle of $X'$, then its pedal triangle is homothetic with $DEF$ which is impossible since the lines that connect the corresponding points of the two homothetic triangles don't concur. Hence $X'\equiv X$. It's a well known property of isogonal conjugates that $O,X,H$ are collinear and $RST$ is the pedal triangle of $H$.

We will prove that $BS\cap CT=Y'$ lies on line $OXH$, in which it's clear that the concurrence at $Y$ follows. Apply Pappus' theorem on $CES,BFT$ and we have $H,Y',ET\cap FS$ are collinear. Let $HS,HT\cap (O)=S',T'$ again and since $SS'\cap TT'=O$, by Pascal theorem we have $O,H,ET\cap FS$ are collinear and hence $O,X,H,ET\cap FS,Y'$ are collinear and we are done.
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TelvCohl
2312 posts
#3 • 5 Y
Y by Gaussian_cyber, Ru83n05, Adventure10, Mango247, and 1 other user
My solution:

Since the lines through $ A, B, C $ and perpendicular to $ EF, FD, DE $ are concurrent at the orthocenter of $\triangle DEF $ ,
so $ \triangle DEF $ and $ \triangle ABC $ are orthologic $ \Longrightarrow $ the perpendiculars through $ D, E, F $ to $ BC, CA, AB $ are concurrent at $ X $ .

Since $ AD, BE, CF $ are concurrent ,
so by Terquem theorem we get $ AR, BS, CT $ are concurrent at $ Y $ .

Since $ \triangle RST $ is the pedal triangle of $ X' $ ($ X' $ is the isogonal conjugate of $ X $ ) ,
so the line passing $ A, B, C $ and perpendicular to $ ST, TR, RS $ are concurrent at $ X $ ,
hence by Sondat theorem (for $\triangle ABC $ and $\triangle RST $ ) we get $ Y, X', X $ are collinear.
ie. $ X, O, Y $ are collinear

Q.E.D

Remark:

(1) Easy to see $ X' = AD \cap BE \cap BF $ which is the orthocenter of $\triangle DEF $

(2) Another interesting property in this configuration :
Denote $ O' $ as the circumcenter of $ \triangle ABC $ , then the isogonal conjugate of $ Y $ lie on $ O'H $ .
(see http://www.artofproblemsolving.com/community/c6h366335 )
This post has been edited 1 time. Last edited by TelvCohl, Apr 13, 2015, 12:53 AM
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JuanOrtiz
366 posts
#4 • 2 Y
Y by Adventure10, Mango247
Here is my solution.

Let $H$ be the orthocenter of $DEF$, which clearly lies on $AD,BE,CF$. By Desargues on $EDF,CH_DB$, where $H_D$ is the foot of the perpendicular from $D$ to $EF$ (define $H_E,H_F$ similarly), we get that $H_EH_F,EF,CB$ are concurrent and therefore $D(T,E,H,F)$ is a harmonic pencil, and so $TEH_DF$ is a harmonic quadrilateral. Thus if $H_D'$ is the reflection of $H_D$ across the perpendicular bisector of $EF$ (which is also clearly the point diametrically opposite to $D$), since $TH_D$ is symmedian of $TEF$ we get that $TH_D'$ is a median of $TEF$. Thus $T$ lies on $MH_D'$, where $M$ is the midpoint of $EF$, but $H$ also lies here, and so these 4 points are collinear. Therefore, $\boxed{\angle HTF=90}$ (since $DH_D'$ is a diameter).

Reflecting $\ell$, the line perpendicular to $CB$ that passes through $D$, through $O$, it is easily seen that it transforms into line $TH$. The analogous happens for $E$ and $F$. Thus the three original lines are concurrent at the reflection of $H$ through $O$. Therefore, $X$ is the reflection of $H$ through $O$. We must now prove $Y$ exists and lies on the Euler line of $\triangle DEF$.

To this end, we invert through $H$ with any radius. Let $Z'$ denote the inverse of $Z$ for any point $Z \neq H$. Lines $HD,HE,HF$ invert to themselves. Notice that the circle $HEH_DH_FS$ has a diameter $HE$, and hence its inverse is a line perpendicular to $HE$ (the analogous applies to $HD,HF$). From this we gather that $D',E',F'$ are the feet of the altitudes in a triangle, and that $S,R,T$ lie on the sides. Since $DEFSRT$ are concyclic, so are their inverses and so $S', R', T'$ are the midpoints of said triangle, and their circumcenter is the 9-point circle. Finally, since $\angle HSA=\angle HTA=90$, we gather that $STAH$ are cyclic, so $S',T',A'$ are collinear. From this, $A'=T'S' \cap HD'$. The analogous occurs for $B',C'$. From now on, for convenience, let $Z$ denote what we previously denoted by $Z'$, for any letter $Z$. Let the big triangle be $XYZ$.

The question is now to prove that the circumcenters of $HTC,HRA,HSB$ have the Euler line of $XYZ$ as a common radical axis (since $O$ inverted to a point on this line, because it also contains the circumcenter of $DEF$, and $H$ and $O$ are collinear with this circumcenter by symmetry). Let $G$ be the barycenter of $XYZ$, let's prove that $G$ has the same power of a point to all 3 circles.

Let $Z_0$ be the intersection of $ZT$ and the 9-point circle. Then $(ZZ_0)(ZT)=(ZS)(ZE)=(ZC)(ZH)$ since $HCSE$ is cyclic because $\angle HCS= \angle HES=90$. Then $THCZ_0$ are concyclic. But $ZT$ is the median of $XYZ$, and so $G$ lies on it. Therefore $G$ has the same power of a point to the circumcircle of $THCZ_0$ than to the 9-point circle, and by analogy we are done. $\blacksquare \text{ Q.E.D}$.
This post has been edited 2 times. Last edited by JuanOrtiz, Apr 13, 2015, 12:09 AM
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amar_04
1915 posts
#5 • 6 Y
Y by GeoMetrix, Bumblebee60, Gaussian_cyber, Ru83n05, Mango247, Mango247
ELMOSL 2014 G10 wrote:
We are given triangles $ABC$ and $DEF$ such that $D\in BC, E\in CA, F\in AB$, $AD\perp EF, BE\perp FD, CF\perp DE$. Let the circumcenter of $DEF$ be $O$, and let the circumcircle of $DEF$ intersect $BC,CA,AB$ again at $R,S,T$ respectively. Prove that the perpendiculars to $BC,CA,AB$ through $D,E,F$ respectively intersect at a point $X$, and the lines $AR,BS,CT$ intersect at a point $Y$, such that $O,X,Y$ are collinear.

Proposed by Sammy Luo

Clearly $AD,BE,CF$ are concurrent at the Orthocenter $U$ of $\Delta DEF$. So, $AR,BS,CT$ are concurrent at the Cyclocevian Conjugate $(Y)$ of $U$ WRT $\Delta ABC$ $(\bigstar)$. Also Clearly $\{\Delta ABC,\Delta DEF\}$ are Orthologic, so the perpendiculars to $BC,CA,AB$ through $D,E,F$ concurs at $X$. Now notice that the Perpendiculars from $R,S,T$ to $BC,CA,AB$ respectively concurs at a point $X^*$ which is the Isogonal Conjugate of $X$. Also the perpendiculars from $A,B,C$ to $ST,RT,SR$ are concurrent at $X$. Hence, $\{\Delta ABC,\Delta RST\}$ are Orthologic with centers of Orthology $\{X,X^*\}$. Also from $(\bigstar)$ we get that $\{\Delta ABC,\Delta RST\}$ are perspective. So, by Sondat's Theorem $Y\in\overline{XX^*}$ but $O\in\overline{XX^*}$. Hence. $\overline{O-X-Y}$. $\blacksquare$
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