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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
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[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
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0 replies
jlacosta
Mar 2, 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
Geometry Problem #42
vankhea   2
N 18 minutes ago by kaede_Arcadia
Source: Van Khea
Let $P$ be any point. Let $D, E, F$ be projection point from $P$ to $BC, CA, AB$. Circumcircle $(ABC)$ cuts circumcircle $(AEF), (BFD), (CDE)$ at $A_1, B_1, C_1$. Let $A_2, B_2, C_2$ be antipode of $A_1, B_1, C_1$ wrt $(AEF), (BFD), (CDE)$. Prove that $A_2, B_2, C_2, P$ are cyclic.
2 replies
1 viewing
vankhea
Sep 6, 2023
kaede_Arcadia
18 minutes ago
divisibility
srnjbr   3
N 21 minutes ago by srnjbr
Find all natural numbers n such that there exists a natural number l such that for every m members of the natural numbers the number m+m^2+...m^l is divisible by n.
3 replies
srnjbr
3 hours ago
srnjbr
21 minutes ago
Very easy inequality
pggp   5
N 30 minutes ago by ionbursuc
Source: Polish Junior MO Second Round 2019
Let $x$, $y$ be real numbers, such that $x^2 + x \leq y$. Prove that $y^2 + y \geq x$.
5 replies
pggp
Oct 26, 2020
ionbursuc
30 minutes ago
Solve in gaussian integers
CHESSR1DER   0
39 minutes ago
Solve in gaussian integers.
$
\sin\left(\ln\left(x^{x^{x^2}}\right)\right) = x^4
$
0 replies
CHESSR1DER
39 minutes ago
0 replies
Inequality and function
srnjbr   4
N an hour ago by srnjbr
Find all f:R--R such that for all x,y, yf(x)+f(y)>=f(xy)
4 replies
srnjbr
3 hours ago
srnjbr
an hour ago
Problem 4
blug   3
N an hour ago by sunken rock
Source: Polish Junior Math Olympiad Finals 2025
In a rhombus $ABCD$, angle $\angle ABC=100^{\circ}$. Point $P$ lies on $CD$ such that $\angle PBC=20^{\circ}$. Line parallel to $AD$ passing trough $P$ intersects $AC$ at $Q$. Prove that $BP=AQ$.
3 replies
blug
Mar 15, 2025
sunken rock
an hour ago
Simple vector geometry existence
AndreiVila   2
N 2 hours ago by sunken rock
Source: Romanian District Olympiad 2025 9.1
Let $ABCD$ be a parallelogram of center $O$. Prove that for any point $M\in (AB)$, there exist unique points $N\in (OC)$ and $P\in (OD)$ such that $O$ is the center of mass of $\triangle MNP$.
2 replies
AndreiVila
Mar 8, 2025
sunken rock
2 hours ago
CMI Entrance 19#6
bubu_2001   5
N 2 hours ago by quasar_lord
$(a)$ Compute -
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} \bigg[ \int_{0}^{e^x} \log ( t ) \cos^4 ( t ) \mathrm{d}t \bigg]
\end{align*}$(b)$ For $x > 0 $ define $F ( x ) = \int_{1}^{x} t \log ( t ) \mathrm{d}t . $

$1.$ Determine the open interval(s) (if any) where $F ( x )$ is decreasing and all the open interval(s) (if any) where $F ( x )$ is increasing.

$2.$ Determine all the local minima of $F ( x )$ (if any) and all the local maxima of $F ( x )$ (if any) $.$
5 replies
bubu_2001
Nov 1, 2019
quasar_lord
2 hours ago
a! + b! = 2^{c!}
parmenides51   6
N 2 hours ago by ali123456
Source: 2023 Austrian Mathematical Olympiad, Junior Regional Competition , Problem 4
Determine all triples $(a, b, c)$ of positive integers such that
$$a! + b! = 2^{c!}.$$
(Walther Janous)
6 replies
parmenides51
Mar 26, 2024
ali123456
2 hours ago
Inequality
srnjbr   0
3 hours ago
a^2+b^2+c^2+x^2+y^2=1. Find the maximum value of the expression (ax+by)^2+(bx+cy)^2
0 replies
srnjbr
3 hours ago
0 replies
Graph Theory
JetFire008   1
N 3 hours ago by JetFire008
Prove that for any Hamiltonian cycle, if it contain edge $e$, then it must not contain edge $e'$.
1 reply
1 viewing
JetFire008
3 hours ago
JetFire008
3 hours ago
Inspired by hunghd8
sqing   1
N 3 hours ago by sqing
Source: Own
Let $ a,b,c\geq 0 $ and $ a+b+c\geq 2+abc . $ Prove that
$$a^2+b^2+c^2- abc\geq \frac{7}{4}$$$$a^2+b^2+c^2-2abc \geq 1$$$$a^2+b^2+c^2- \frac{1}{2}abc\geq \frac{31}{16}$$$$a^2+b^2+c^2- \frac{8}{5}abc\geq \frac{34}{25}$$
1 reply
sqing
3 hours ago
sqing
3 hours ago
Assisted perpendicular chasing
sarjinius   2
N 3 hours ago by chisa36
Source: Philippine Mathematical Olympiad 2025 P7
In acute triangle $ABC$ with circumcenter $O$ and orthocenter $H$, let $D$ be an arbitrary point on the circumcircle of triangle $ABC$ such that $D$ does not lie on line $OB$ and that line $OD$ is not parallel to line $BC$. Let $E$ be the point on the circumcircle of triangle $ABC$ such that $DE$ is perpendicular to $BC$, and let $F$ be the point on line $AC$ such that $FA = FE$. Let $P$ and $R$ be the points on the circumcircle of triangle $ABC$ such that $PE$ is a diameter, and $BH$ and $DR$ are parallel. Let $M$ be the midpoint of $DH$.
(a) Show that $AP$ and $BR$ are perpendicular.
(b) Show that $FM$ and $BM$ are perpendicular.
2 replies
sarjinius
Mar 9, 2025
chisa36
3 hours ago
Find min
hunghd8   4
N 3 hours ago by imnotgoodatmathsorry
Let $a,b,c$ be nonnegative real numbers such that $ a+b+c\geq 2+abc $. Find min
$$P=a^2+b^2+c^2.$$
4 replies
hunghd8
Today at 12:10 PM
imnotgoodatmathsorry
3 hours ago
Show that ab is a perfect square
mruczek   12
N May 1, 2023 by Amiralizakeri2007
Source: XIII Polish Junior MO 2018 Finals - Problem 1
Positive odd integers $a, b$ are such that $a^bb^a$ is a perfect square. Show that $ab$ is a perfect square.
12 replies
mruczek
Apr 24, 2018
Amiralizakeri2007
May 1, 2023
Show that ab is a perfect square
G H J
G H BBookmark kLocked kLocked NReply
Source: XIII Polish Junior MO 2018 Finals - Problem 1
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mruczek
67 posts
#1 • 2 Y
Y by Adventure10, Mango247
Positive odd integers $a, b$ are such that $a^bb^a$ is a perfect square. Show that $ab$ is a perfect square.
This post has been edited 1 time. Last edited by mruczek, Apr 24, 2018, 1:02 AM
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skyerzym27
54 posts
#2 • 8 Y
Y by programjames1, Wizard_32, hansu, mrdriller, NO_SQUARES, howcanicreateacccount, Adventure10, Mango247
WLOG \(a \ge b\),
We know that \(a^{b}b^{a}\) is perfect square.
Then notice that \(a^{b}b^{a}=(ab)^{b}b^{a-b}\). Because \(a\) and \(b\) are odds, then \(a-b\) is even. Therefore, \(b^{a-b}\) is perfect square.
Because \(a^{b}b^{a}\) is perfect square, then we get \((ab)^{b}\) is perfect square. Because \(b\) is odd, then \(ab\) is perfect square.
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khan.academy
633 posts
#3 • 2 Y
Y by Adventure10, Mango247
Redacted...
This post has been edited 2 times. Last edited by khan.academy, Aug 8, 2018, 4:32 AM
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Vrangr
1600 posts
#4 • 1 Y
Y by Adventure10
@above, by definition of squares...
\[b^{a-b} = (b^{\frac{a-b}{2}})^2\]$\frac{a-b}2$ natural number since, $a - b$ is even and $a\ge b$.
This post has been edited 1 time. Last edited by Vrangr, Aug 8, 2018, 11:53 AM
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skyerzym27
54 posts
#5 • 2 Y
Y by Adventure10, Mango247
khan.academy wrote:
skyerzym27 wrote:
WLOG \(a \ge b\),
We know that \(a^{b}b^{a}\) is perfect square.
Then notice that \(a^{b}b^{a}=(ab)^{b}b^{a-b}\). Because \(a\) and \(b\) are odds, then \(a-b\) is even. Therefore, \(b^{a-b}\) is perfect square.
Because \(a^{b}b^{a}\) is perfect square, then we get \((ab)^{b}\) is perfect square. Because \(b\) is odd, then \(ab\) is perfect square.

How $a-b=\text{even}\implies b^{a-b}$ is a perfect square?

The number is said perfect square if and only if that number can be expressed as \(x^{2k}\) where \(x\) and \(k\) are nonnegative integers.
This post has been edited 1 time. Last edited by skyerzym27, Mar 18, 2019, 7:27 AM
Z K Y
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skyerzym27
54 posts
#6 • 2 Y
Y by Adventure10, Mango247
Vrangr wrote:
@above, by definition of squares...
\[b^{a-b} = (b^{\frac{a-b}{2}})^2\]$\frac{a-b}2$ natural number since, $a - b$ is odd and $a\ge b$.

do you mean \(\frac{a-b}{2}\) is even
Z K Y
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Devastator
348 posts
#7 • 2 Y
Y by Adventure10, Mango247
But then $(b^{\frac{a-b}{2}})^2$ is equal to $b^{2\cdot \frac{a-b}{2}}$, and we know that (a-b)/2 is an integer so it is indeed a square
This post has been edited 1 time. Last edited by Devastator, Apr 24, 2018, 11:30 AM
Reason: Wrong latex
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Vrangr
1600 posts
#8 • 2 Y
Y by Adventure10, Mango247
Vrangr wrote:
@above, by definition of squares...
\[b^{a-b} = (b^{\frac{a-b}{2}})^2\]$\frac{a-b}2$ natural number since, $a - b$ is even and $a\ge b$.
skyerzym27 wrote:
do you mean \(\frac{a-b}{2}\) is even
Mistyped it earlier
This post has been edited 1 time. Last edited by Vrangr, Apr 24, 2018, 11:41 AM
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razor406
6 posts
#9 • 2 Y
Y by Adventure10, Mango247
\(a^{b}b^{a}=a^{b-1}b^{a-1}ab\)
both b-1 and a-1 is even, so ab is a perfect square
This post has been edited 5 times. Last edited by razor406, Aug 8, 2018, 4:38 AM
Reason: format change
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ATGY
2502 posts
#10
Y by
Notice that:
$$\implies a^b b^a = a^b b^b \cdot b^{(a - b)} = (ab)^b\cdot b^{(a - b)}.$$We know that $a$ and $b$ are odd, hence:
$$\implies 2|(a - b).$$This clearly implies that $b^{(a - b)}$ is a perfect square as $b$ is raised to an even power.

For $a^b b^a$ to be a perfect square, we need $(ab)^b$ to be a perfect square as we just showed that $b^{(a - b)}$ is always a perfect square.

It is given in the question that $b$ is odd, which implies that $(ab)^b$ can never be a perfect square, unless $(ab)$ itself is a perfect square.

Thus $(ab)$ must be a perfect square.

Hence proved.

credits to my blog
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Keith50
464 posts
#11 • 4 Y
Y by llplp, mrdriller, PNT, trying_to_solve_br
For every prime $p|a^bb^a$, we have \[2| v_p(a^bb^a)=bv_p(a)+av_p(b)\]and since $b,a$ are odd, we must have $v_p(a),v_p(b)$ to be the same parity in order for $a^bb^a$ to a perfect square. Thus, $2|v_p(a)+v_p(b)=v_p(ab). \square$
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lian_the_noob12
173 posts
#13
Y by
$\color{blue} \boxed{\textbf{SOLUTION}}$

Let, $a=2k+1, b=2k'+1$
So,
$a^{b} b^{a} = (2k+1)^{2k'+1} (2k'+1)^{2k+1}$
$\implies x^2=[(2k+1)^{k'}(2k'+1)^{k}]^{2} (2k+1)(2k'+1)$
$\implies x^2=m^{2} ab$
$\implies ab$ is a perfect square $\blacksquare$
Z K Y
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Amiralizakeri2007
45 posts
#14
Y by
Note that for all $p$, $bv_p(a)+av_p(b) \equiv 0 (mod2)$, thus $v_p(a)+v_p(b) \equiv 0 (mod 2)$ $\forall p$. thus we are done.
Z K Y
N Quick Reply
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