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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
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
Prove XBY equal to angle C
nataliaonline75   2
N 13 minutes ago by starchan
Let $M$ be the midpoint of $BC$ on triangle $ABC$. Point $X$ lies on segment $AC$ such that $AX=BX$ and $Y$ on line $AM$ such that $XY//AB$. Prove that $\angle XBY = \angle ACB$.
2 replies
nataliaonline75
Yesterday at 2:47 PM
starchan
13 minutes ago
Unique number to make a square of a rational
Zavyk09   2
N 26 minutes ago by Zavyk09
Source: Homework
Find all positive integers $n$ there exists a unique positive integers $m$ such that $\frac{n+m}{m}$ is a square of a rational number.
2 replies
Zavyk09
an hour ago
Zavyk09
26 minutes ago
Classical NT using modular arithmetic
electrovector   7
N 28 minutes ago by Blackbeam999
Source: 2022 Turkey TST P1 Day 1 + 2023 Dutch BxMO TST, Problem 5
Find all pairs of prime numbers $(p,q)$ for which
\[2^p = 2^{q-2} + q!.\]
7 replies
electrovector
Mar 13, 2022
Blackbeam999
28 minutes ago
Inequality
lgx57   9
N 31 minutes ago by lgx57
Source: Own
$a,b,c>0,ab+bc+ca=1$. Prove that

$$\sum \sqrt{8ab+1} \ge 5$$
(I don't know whether the equality holds)
9 replies
lgx57
Saturday at 3:14 PM
lgx57
31 minutes ago
old one but good one
Sunjee   2
N 32 minutes ago by ehuseyinyigit
If $x_1,x_2,...,x_n $ are positive numbers, then prove that
$$\frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+\cdots+ \frac{x_n}{1+x_1^2+\cdots+x_n^2}\geq \sqrt{n}$$
2 replies
Sunjee
2 hours ago
ehuseyinyigit
32 minutes ago
Inspired by giangtruong13
sqing   0
36 minutes ago
Source: Own
Let $ a,b>0  .$ Prove that$$ \frac{a}{b}+\frac{b}{a}+\frac{a^3}{2b^3+kab^2}+\frac{2b^3}{a^3+b^3+kab^2} \geq \frac{2k+7}{k+2}$$Where $ k\geq 0. $
0 replies
sqing
36 minutes ago
0 replies
Functional equation
Math-wiz   24
N 38 minutes ago by nmoon_nya
Source: IMOC SL A1
Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that for all $x,y\in\mathbb{R}$,
$$f(xy+f(x))=f(xf(y))+x$$
24 replies
Math-wiz
Dec 15, 2019
nmoon_nya
38 minutes ago
we can find one pair of a boy and a girl
orl   18
N an hour ago by bin_sherlo
Source: Vietnam TST 2001 for the 42th IMO, problem 3
Some club has 42 members. It’s known that among 31 arbitrary club members, we can find one pair of a boy and a girl that they know each other. Show that from club members we can choose 12 pairs of knowing each other boys and girls.
18 replies
orl
Jun 26, 2005
bin_sherlo
an hour ago
geometry
blug   0
an hour ago
In trapezius $ABCD$, segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle B$ and $\angle D$ intersect at $P$. Circumcircles of $ABP$ and $CDP$ meet again at $Q$. Angle bisector of $\angle D$ cuts $AB$ at $S$. Prove that
$$1. QD=QS,$$$$2. \angle DCQ=\angle BCQ,$$$$3. \angle BAQ=\angle QAD.$$
0 replies
blug
an hour ago
0 replies
Eulerline problem
Retemoeg   0
an hour ago
Source: Extension from a problem I read in a book
Show that the isogonal conjugate of the isotomic conjugate of the orthocenter lies on the Euler line.
0 replies
Retemoeg
an hour ago
0 replies
Nice numer theory
GeoArt   4
N an hour ago by Blackbeam999
$p$ is a prime number, $m, x, y$ are natural numbers ($m, x, y > 1$). It is known that $\frac{x^p + y^p}{2}$ $=$ $(\frac{x+y}{2} )^m$. Prove that $p = m$.
4 replies
GeoArt
Jan 7, 2021
Blackbeam999
an hour ago
Inequality
MathsII-enjoy   6
N 2 hours ago by MathsII-enjoy
A interesting problem generalized :-D
6 replies
MathsII-enjoy
Saturday at 1:59 PM
MathsII-enjoy
2 hours ago
Number Theory
fasttrust_12-mn   8
N 2 hours ago by ErTeeEs06
Source: Pan African Mathematics Olympiad P1
Find all positive intgers $a,b$ and $c$ such that $\frac{a+b}{a+c}=\frac{b+c}{b+a}$ and $ab+bc+ca$ is a prime number
8 replies
fasttrust_12-mn
Aug 15, 2024
ErTeeEs06
2 hours ago
My Unsolved FE on R+
ZeltaQN2008   4
N 2 hours ago by mashumaro
Source: IDK
Give $a>0$. Find all funcitions $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that for all any $x,y\in (0,\infty):$
$$f(xf(y)+a)=yf(x+y+a)$$
4 replies
ZeltaQN2008
4 hours ago
mashumaro
2 hours ago
Equal angles
stergiu   11
N Jan 21, 2019 by Ditzymathstar
Source: Israel 1995
I found this problem in the site of Hong Kong. There is a collection with proposed problems which can be solved with inversion. I post one of them. Any other solution is also welcome.
11 replies
stergiu
May 1, 2007
Ditzymathstar
Jan 21, 2019
Source: Israel 1995
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stergiu
1648 posts
#1 • 4 Y
Y by Adventure10, Mango247, and 2 other users
I found this problem in the site of Hong Kong. There is a collection with proposed problems which can be solved with inversion. I post one of them. Any other solution is also welcome.
Attachments:
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pohoatza
1145 posts
#2 • 5 Y
Y by bestwillcui1, Adventure10, Mango247, and 2 other users
Very nice problem, Stergiu!

Denote $K$ the point where the semicircle cuts the circle $O$, and $T$ the tangenty point where $AB$ cuts the circle $O$.

Let's extend the semircircle and denote $D$ the intersection of $AB$ with the circle $H$.

Thus, Because $AD$ is tangent to the circle $O$, which is also tangent to the circle $H$, then $KT$ is bisects the angle $AKD$. Therefore $Q$ is the midpoint of the arc $\widehat{AD}$.

Now because $\angle{PBT}=\angle{PKT}=90$, we have that the quadrilateral $PBTK$ is cyclic, therefore $QB \cdot QP=QT \cdot QK=QC^{2}$ (the power of point $A$, w.r.t to circle $O$)

But in the right triangle $APQ$, $AB$ is the altitude, therefore $AQ^{2}=QB \cdot QP$, therefore from the above equality we have $QC=QA$.

Now the problem is dead, because let's note $\angle{BAC}=x$, therefore $\angle{ACB}=90-x$, but from $QC=QA$, we have also $\angle{CAQ}=90-x$, therefore $\angle{BAQ}=90-2x$, so $\angle{BQA}=2x$, but $\angle{PAB}=\angle{BQA}$, therefore $\angle{PAC}=2x-x=x$, so $AC$ bisects $\angle{PAB}$, and thus the problem in solved!


Another way for solving the problem from where we found out that $QC=QA$, is just considering the circle $O'$ tangent to circle $H$, $BD$, and $BP$, so the construction is basicly simetric, therefore $CQ=AQ=DQ$, and $PB$ is the bisector of $\angle{APD}$ in triangle $APD$, therefore $C$ is the incenter of triangle $\triangle{APD}$, so $AC$ bisects $\angle{PAB}$.

Generalization:
Let $ABC$ be a triangle, and $D \in (BC)$ such that $\angle{BAD}=\angle{ACB}$, and consider the circle $H$ tangent to the circumcircle of $ABC$ at $K$, to $BD$ at $P$, and to $AD$ at $P'$. Show that $AP$ is the bisector of $\angle{BAD}$.

Another enuntiation of the generalization: (***) - own
In a triangle $ABC$, denote it's circumcircle $w$, and consider the Thebault circles $k_{1}, k_{2}$ corresponding to the bisector $AD$ of $\angle{BAC}$. Prove that $k_{1}$ and $k_{2}$ are tangent at $I$, where $I$ is the incircle of $ABC$.
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M4RI0
639 posts
#3 • 2 Y
Y by Adventure10, Mango247
Here's a link: http://www.mathlinks.ro/Forum/viewtopic.php?t=123139
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pohoatza
1145 posts
#4 • 2 Y
Y by Adventure10, Mango247
I guess that now you got your geometric solution, Mario! :)
This post has been edited 1 time. Last edited by pohoatza, Apr 24, 2008, 4:41 PM
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M4RI0
639 posts
#5 • 2 Y
Y by Adventure10, Mango247
Yeah :D Finally after a long time, but now I wonder how can it be solved with inversion :?:
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Virgil Nicula
7054 posts
#6 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Very nice the Pohoatza's generalization of this cool problem !

Proof. Suppose w.l.o.g. that $B\in (OP)$ and $C\in (PB)$. Denote : the circles $\mathcal H=C(O,R)$ and $\mathcal O=C(I,r)$ ;

the reflection $A'$ of the point $A$ w.r.t. the diameter $[PQ]$ and the incenter $S$ of the $P$- isosceles triangle $APA'$.

Prove easily that $QC=QA=R+\sqrt{R(R-2r)}$ $\implies$ $C\equiv S$, i.e. the ray $[AC$ bisects the angle $\widehat{PAB}$.

Indeed, $OC^{2}=OI^{2}-IC^{2}=$ $(R-r)^{2}-r^{2}\implies$ $\left\{\begin{array}{c}OC=\sqrt{R(R-2r)}\\\\ QC=QO+OC\end{array}\right\|$ $\implies$ $\left\{\begin{array}{c}\boxed{QC=R+\sqrt{R(R-2r)}}\\\\ QB=QC-BC\end{array}\right\|$ $\implies$

$\left\{\begin{array}{c}QB=R-r+\sqrt{R(R-2r)}\\\\ PQ=2R\ ,\ QA^{2}=PQ\cdot QB\end{array}\right\|$ $\implies$ $QA^{2}=$ $R^{2}+2R\sqrt{R(R-2r)}+\left(\sqrt{R(R-2r)}\right)^{2}$ $\implies$ $\boxed{QA=R+\sqrt{R(R-2r)}}$.
This post has been edited 2 times. Last edited by Virgil Nicula, May 2, 2007, 9:00 PM
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SpongeBob
188 posts
#7 • 4 Y
Y by Adventure10, Mango247, and 2 other users
This is wanted solution with inversion.

We use inversion with center $C$. Then circle $O$ maps onto a line
$O^{*}$ parallel with $PQ$, and semicircle $H$ maps into a semi circle $H^{*}$ with center
on line $PQ$, and diameter $P^{*}Q^{*}$ which touches $O^{*}$. $A^{*}$ is intersection of the
line $CA$ and $H^{*}$. Line $AB$ maps into a circle that goes through
$C$ and $A^{*}$, with center on $CB$ (because $CB\bot AB$) and
touches line $O^{*}$. $B^{*}$ is the second intersection point of that
circle and $PQ$. Now, we have two equal circles $(AB)^{*}$ and $H^{*}$
(because they have their centers on the same line, and they touch
the line parallel with the line on which their centers are), and
their intersection is $A^{*}$, so $A^{*}P^{*}=A^{*}B^{*}$, in other words $\angle A^{*}P^{*}C=\angle CB^{*}A^{*}$. But $\angle CP^{*}A^{*}=\angle CAP$ and $\angle CB^{*}A^{*}=\angle CAB$ so this is the end of a proof.

Please ask if something isn't quite clear, I'm not so experienced
in writing solutions with inversion so I don't know in how many
details I should go.

Bye
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stergiu
1648 posts
#8 • 2 Y
Y by Adventure10, Mango247
I had forgotten this problem :( . In my archiev I have a synthetiacal solution. I think I have sent it to this forum in the past. It is different from those you sent. I had not a solution with inversion.Now I feel happy because we have many solutions to the problem.
Thank you all for your nice solutions. I'll try to send the next problem from this collection.

Babis
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e.lopes
349 posts
#9 • 1 Y
Y by Adventure10
This can solved with a lot of computations!

One friend say to try with inversion. I see one nice and easy solution, basically the same of SpongeBob (i think that also the same of the book 'Mathematical Olympiads Around The World, 1995/06')
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April
1270 posts
#10 • 3 Y
Y by drEdrE, Adventure10, Mango247
Quote:
$AB$ is a half chord perpendicular to the diameter $PQ$ of the semicircle $C(O,\,R).$ A circle $C(O',\,R')$ is inscribed, toucher $PQ$ at $C.$ Prove that: $AC$ is angle bisector of $\angle PAB$
Proof. We have:
$OO'=OC^{2}+O'C^{2}\Longrightarrow (R-R')^{2}=(QB+R'-R)^{2}+R'^{2}\\ \Longrightarrow 2\cdot QB\cdot R=(QB+R')^{2}\Longrightarrow QB\cdot QQ=(QB+BC)^{2}\Longrightarrow QA^{2}=QC^{2}\\ \Longrightarrow QA=QC\Longrightarrow \angle QCA=\angle QAC\Longrightarrow \angle CAB=\angle PAC$
Attachments:
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stergiu
1648 posts
#11 • 2 Y
Y by Adventure10, Mango247
April wrote:
Quote:
$AB$ is a half chord perpendicular to the diameter $PQ$ of the semicircle $C(O,\,R).$ A circle $C(O',\,R')$ is inscribed, toucher $PQ$ at $C.$ Prove that: $AC$ is angle bisector of $\angle PAB$
Proof. We have:
$OO'=OC^{2}+O'C^{2}\Longrightarrow (R-R')^{2}=(QB+R'-R)^{2}+R'^{2}\\ \Longrightarrow 2\cdot QB\cdot R=(QB+R')^{2}\Longrightarrow QB\cdot QQ=(QB+BC)^{2}\Longrightarrow QA^{2}=QC^{2}\\ \Longrightarrow QA=QC\Longrightarrow \angle QCA=\angle QAC\Longrightarrow \angle CAB=\angle PAC$

April ,

thank you . I did not know this simple and nice solution. Just Pythagorian theorem and ... done.Very nice!

Babis
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Ditzymathstar
14 posts
#12 • 1 Y
Y by Adventure10
https://artofproblemsolving.com/community/c6h1770754p11619140
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