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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
m^4+3^m is a perfect square number
Havu   4
N 5 minutes ago by ReticulatedPython
Find a positive integer m such that $m^4+3^m$ is a perfect square number.
4 replies
Havu
30 minutes ago
ReticulatedPython
5 minutes ago
Functional equation
Nima Ahmadi Pour   100
N 8 minutes ago by jasperE3
Source: ISl 2005, A2, Iran prepration exam
We denote by $\mathbb{R}^+$ the set of all positive real numbers.

Find all functions $f: \mathbb R^ + \rightarrow\mathbb R^ +$ which have the property:
\[f(x)f(y)=2f(x+yf(x))\]
for all positive real numbers $x$ and $y$.

Proposed by Nikolai Nikolov, Bulgaria
100 replies
Nima Ahmadi Pour
Apr 24, 2006
jasperE3
8 minutes ago
how can I solve this FE
Jackson0423   6
N 9 minutes ago by jasperE3

Let \( f : \mathbb{R} \to \mathbb{R} \) be a function that satisfies the following equation for all real numbers \( x \):
\[
f(x^2 + x + 3) + 2f(x^2 - 3x + 5) = 6x^2 - 10x + 17.
\]Find the value of \( f(100) \).
6 replies
Jackson0423
2 hours ago
jasperE3
9 minutes ago
a does not divide b^n - n for all positive integers n?
N.T.TUAN   28
N 10 minutes ago by alexanderhamilton124
Source: USA Team Selection Test 2007
Determine whether or not there exist positive integers $ a$ and $ b$ such that $ a$ does not divide $ b^n - n$ for all positive integers $ n$.
28 replies
N.T.TUAN
Dec 8, 2007
alexanderhamilton124
10 minutes ago
Urgent. Need them quick
sealight2107   1
N 11 minutes ago by arqady
With $a,b,c>1$ and $a+b+c=2abc$. Prove that:
$\sqrt[3]{ab-1}+\sqrt[3]{bc-1}+\sqrt[3]{ca-1} \le \sqrt[3]{(a+b+c)^2}$
1 reply
sealight2107
21 minutes ago
arqady
11 minutes ago
Divisibility..
Sadigly   3
N 36 minutes ago by Jackson0423
Source: Azerbaijan Junior MO 2025 P2
Find all $4$ consecutive even numbers, such that the square of their product is divisible by the sum of their squares.
3 replies
Sadigly
Today at 7:37 AM
Jackson0423
36 minutes ago
Divisibilty...
Sadigly   9
N an hour ago by Jackson0423
Source: My (fake) translation error
Find all $4$ consecutive even numbers, such that the square of their product divides the sum of their squares.
9 replies
Sadigly
2 hours ago
Jackson0423
an hour ago
Inspired by Kosovo 2010
sqing   1
N an hour ago by ytChen
Source: Own
Let $ a,b>0  , a+b\leq k $. Prove that
$$\left(1+\frac{1}{a(b+1)}\right)\left(1+\frac{1}{b(a+1)}\right)\geq\left(1+\frac{4}{k(k+2)}\right)^2$$$$\left(1+\frac {a}{b(a+1)}\right)\left(1+\frac {b}{a(b+1)}\right) \geq\left(1+\frac{2}{k+2}\right)^2$$Let $ a,b>0  , a+b\leq 2 $. Prove that
$$\left(1+\frac{1}{a(b+1)}\right)\left(1+\frac{1}{b(a+1)}\right)\geq \frac{9}{4} $$$$\left(1+\frac {a}{b(a+1)}\right)\left(1+\frac {b}{a(b+1)}\right) \geq \frac{9}{4} $$
1 reply
sqing
Today at 3:56 AM
ytChen
an hour ago
Number Theory
VicKmath7   5
N an hour ago by Adywastaken
Source: Archimedes Junior 2014
Let $p$ prime and $m$ a positive integer. Determine all pairs $( p,m)$ satisfying the equation: $ p(p+m)+p=(m+1)^3$
5 replies
VicKmath7
Mar 17, 2020
Adywastaken
an hour ago
x+y in B iff x,y in A
fattypiggy123   5
N an hour ago by Math2030
Source: China Mathematical Olympiad 2015 Q3
Let $n \geq 5$ be a positive integer and let $A$ and $B$ be sets of integers satisfying the following conditions:

i) $|A| = n$, $|B| = m$ and $A$ is a subset of $B$
ii) For any distinct $x,y \in B$, $x+y \in B$ iff $x,y \in A$

Determine the minimum value of $m$.
5 replies
fattypiggy123
Dec 20, 2014
Math2030
an hour ago
IMO Genre Predictions
ohiorizzler1434   65
N an hour ago by Oksutok
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
65 replies
ohiorizzler1434
May 3, 2025
Oksutok
an hour ago
k^2/p for k =1 to (p-1)/2
truongphatt2668   1
N an hour ago by Double07
Let $p$ be a prime such that: $p = 4k+1$. Simplify:
$$\sum_{k=1}^{\frac{p-1}{2}}\begin{Bmatrix}\dfrac{k^2}{p}\end{Bmatrix}$$
1 reply
truongphatt2668
3 hours ago
Double07
an hour ago
Interesting inequality
imnotgoodatmathsorry   1
N an hour ago by Bergo1305
Let $x,y,z > \frac{1}{2}$ and $x+y+z=3$.Prove that:
$\sqrt{x^3+y^3+3xy-1}+\sqrt{y^3+z^3+3yz-1}+\sqrt{z^3+x^3+3zx-1}+\frac{1}{4}(x+5)(y+5)(z+5) \le 60$
1 reply
imnotgoodatmathsorry
2 hours ago
Bergo1305
an hour ago
every lucky set of values {a_1,a_2,..,a_n} satisfies a_1+a_2+...+a_n >n2^{n-1}
parmenides51   6
N an hour ago by jonh_malkovich
Source: 2020 International Olympiad of Metropolises P3
Let $n>1$ be a given integer. The Mint issues coins of $n$ different values $a_1, a_2, ..., a_n$, where each $a_i$ is a positive integer (the number of coins of each value is unlimited). A set of values $\{a_1, a_2,..., a_n\}$ is called lucky, if the sum $a_1+ a_2+...+ a_n$ can be collected in a unique way (namely, by taking one coin of each value).
(a) Prove that there exists a lucky set of values $\{a_1, a_2, ..., a_n\}$ with $$a_1+ a_2+...+ a_n < n \cdot 2^n.$$(b) Prove that every lucky set of values $\{a_1, a_2,..., a_n\}$ satisfies $$a_1+ a_2+...+ a_n >n \cdot 2^{n-1}.$$
Proposed by Ilya Bogdanov
6 replies
parmenides51
Dec 19, 2020
jonh_malkovich
an hour ago
4 lines concurrent
Zavyk09   7
N May 2, 2025 by bin_sherlo
Source: Homework
Let $ABC$ be triangle with circumcenter $(O)$ and orthocenter $H$. $BH, CH$ intersect $(O)$ again at $K, L$ respectively. Lines through $H$ parallel to $AB, AC$ intersects $AC, AB$ at $E, F$ respectively. Point $D$ such that $HKDL$ is a parallelogram. Prove that lines $KE, LF$ and $AD$ are concurrent at a point on $OH$.
7 replies
Zavyk09
Apr 9, 2025
bin_sherlo
May 2, 2025
4 lines concurrent
G H J
Source: Homework
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Zavyk09
13 posts
#1 • 1 Y
Y by PikaPika999
Let $ABC$ be triangle with circumcenter $(O)$ and orthocenter $H$. $BH, CH$ intersect $(O)$ again at $K, L$ respectively. Lines through $H$ parallel to $AB, AC$ intersects $AC, AB$ at $E, F$ respectively. Point $D$ such that $HKDL$ is a parallelogram. Prove that lines $KE, LF$ and $AD$ are concurrent at a point on $OH$.
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aidenkim119
33 posts
#2 • 1 Y
Y by PikaPika999
............
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aidenkim119
33 posts
#3 • 1 Y
Y by PikaPika999
First three are trivial by pascal, but AD looks a bit hard / '
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ItzsleepyXD
140 posts
#4 • 1 Y
Y by PikaPika999
Redefine
Let $ABC$ be triangle with circumcenter $(O)$ and orthocenter $H$. $A'$ is antipode of $A$ . $O'$ is circumcenter of $(BOC)$ . Point $E,F$ satisfied $OE // A'F // AB , OF // A'E // AC$ then prove $OH,O'A',BE,CF$ concurrent .

Let $B',C'$ be antipode of $B,C$ respectively.
MMP: Fix $(O),B,C$ . Move $A$ on $(O)$ deg 2.
Since $A',E,C'$ collinear and $A',F,B'$ collinear
By $\angle C'EO = \angle BAC = \angle BC'C = \angle C'BO$ so $C',B,O,E$ concyclic.
implies that $E$ deg 2. Also $F$ deg 2.
So line $BE,CF$ deg 1.
$H=$ reflection of $A \infty_{\perp BC} \cap (O)$ across $BC$
Since $H,A'$ deg 2. implies that line $O'A'.OH$ deg 2.
We want to prove $BE,CF,OH$ concurrent first and $BE,CF,O'A'$ concurrent.
but both have deg 1+1+2+1 = 5 .

choose $A= B,C,B',C'$ and midpoint of arc $BC$
the rest of problem is easy. $\square$
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pingupignu
49 posts
#5 • 1 Y
Y by PikaPika999
My solution may not be elegant but here's some DDIT spam for you guys to enjoy :blush:.
Let $X = KE \cap LF$. I will show that $X, A, D$ and $X, O, H$ are collinear.

Part 1:
I first claim that $\angle \infty_{CH}XL = \angle \infty_{BH}XK$. This follows from
$$\angle \infty_{CH}XL = \angle FLH = \angle FHL = 90^\circ - \angle BHL = 90^\circ - \angle CHK = \angle EHK = \angle XKH = \angle \infty_{BH}XK.$$Then, applying DDIT on $X \cup LDKH$ we see that $$(XK, XL), (XH, XD), (X\infty_{BH}, X\infty_{CH})$$are reciprocal pairs under some involution on $\mathcal{P}_X$. This involution must be a reflection in the angle bisector of $\angle KXL$. Hence $XH, XD$ are isogonal in $\angle KXL$.

Next, since $AK=AL$ (well-known), $LF=FH=AE$, $AF=EH=EK$, we yield $\triangle LAF \cong \triangle AKE$.
I claim that $X\infty_{AB}, X\infty_{AC}$ are isogonal in $\angle XKL$. This is because $$\angle \infty_{AB}XF = 180^\circ - \angle AFL = 180^\circ - \angle AEK = \angle AEX = \angle \infty_{AC}XE.$$Applying DDIT on $X \cup AEHF$ would then give $XA, XH$ are isogonal in $\angle EXF$. Since $XH, XD$, $XH, XA$ are isogonal in $\angle KXL = \angle EXF$ we conclude that $XA \equiv XD$, or $X \in AD$.

Part 2:
I first prove that $X\infty_{CH}, OL, AK$ concur at a point $S$ on $(XLK)$. For this, let $S = X\infty_{CH} \cap OL$, where from a short angle chase we get $$\angle XSL = \angle OLH = B-A = (90^\circ - A) - (90^\circ - B) = \angle AKL - \angle LCB$$$$= \angle AKL - \angle LAF = \angle AKL - \angle  AKE = \angle EKL = \angle XKL$$Hence $XSKL$ are cyclic, and from
$$\angle SKX = \angle SLX = \angle FLO = \angle ALO - \angle ALF = A - \angle EAK = A - (90^\circ - C)$$which equals
$$= A+C-90^\circ = 90^\circ - B = \angle LAB = \angle AKE = \angle AKX$$$\implies S \in AK$. The claim is proven.

From DDIT in $X \cup AKOL$ we have the reciprocal pairs $$(XA, XO), (XK, XL), (XS, XT)$$where $T = AL \cap KO \cap X \infty_{BH}$ (similarly).
since $(XS, XT) = (X\infty_{CH}, X\infty_{BH})$ and we have established $(XA, XH), (X\infty_{CH}, X\infty_{BH})$ are isogonal in $\angle KXL$ we get $XH \equiv XO \implies X \in OH$. The problem is solved. $\blacksquare$
Attachments:
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tomsuhapbia
6 posts
#6 • 1 Y
Y by Amkan2022
We need two well-known lemmas about the isogonal line:
1. Given a $\triangle ABC$ and a point $P$ satisfied $\angle ABP=\angle ACP$. Let $Q$ be the reflection of $P$ in the midpoint of $BC$. Then $AP$ and $AQ$ are isogonals wrt $\angle BAC$.
2. Given a trapezoid $ABCD$ $(AB\parallel CD)$ is inscribed $(O)$. Let $E,F$ be the intersections of $BC$ and $AD$; $AC$ and $BD$. Let $S$ be a abitary point on $(O)$. Then $SE,SG$ are isogonals wrt $\angle ASB$.

Back to the problem: Let $X,Y$ be the intersections of $OL$ and $AK$; $AL$ and $OK$. By symmetric, we have $AK=AH=AL$ so $AO$ is the perpendicular bisector of $KL$ then $XKLY$ is a trapezoid. Let $LK$ intersects $KE$ at $Z$.

We have
$$\angle FZK=\angle ZEH-\angle ZFC=180^\circ-\angle KEH-\angle HFC=180^\circ-3\angle BAC$$and
$$\angle LYK=\angle AOK-\angle OAL=2(90^\circ-\angle OAL)-\angle OAL=180^\circ-3\angle OAL=180^\circ-3\angle BAC=\angle FZK\,(2)$$since
$$\angle OAL=\dfrac{1}{2}\angle KAL=\dfrac{1}{2}(\angle KAB+\angle BAH+\angle HAC+\angle CAL)=\angle BAC$$$(2)$ leads to $Z$ lies on the circumcircle of $XKLY$. From the lemma 2, we obtain that $ZA,ZO$ are isogonals wrt $\angle LZK$ $(1)$. We also have $\angle ZFE=\angle LFE=\angle KEH=180^\circ-\angle ZEH$ and $\angle HEZ=\angle HKE=\angle FLH=180^\circ-\angle HLZ$, using lemma 1 and get $(XD,XH)$ and $(XH,XA)$ are two isogonal pairs wrt $\angle LZK\equiv\angle FZE$, so $A,D,Z$ are collinear. Combine with $(1)$ and we conclude $Z,O,H$ are collinear or $AD,LF,KE,OH$ are concurrent at $Z$.

https://i.postimg.cc/hJQdTNdW/image.png
This post has been edited 2 times. Last edited by tomsuhapbia, May 1, 2025, 5:25 PM
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hectorleo123
344 posts
#7
Y by
I apologize for the complex bash, but I couldn't find another way.

Let \( B' \) and \( C' \) be the antipodes of \( B \) and \( C \).
Since \( L \) is the reflection of \( H \) over \( AB \), we have \( \angle FLB = \angle FHB = 90^\circ \) (since \( FH \parallel AC \perp BH \)).
Analogously, \( \angle EKC = 90^\circ \).
\(\Rightarrow B', F, L \) are collinear and \( C', E, K \) are collinear.
By Pascal's Theorem on
\[ \binom{B, L, C'}{C, K, B'} \]we get that \( KE, LF \), and \( OH \) are concurrent.
Now it suffices to prove that \( AD, KE \), and \( LF \) are concurrent.

We use complex numbers, where \( (ABC) \) is the unit circle and \( a = 1 \).
Let
\[
k = -\frac{c}{b}, \quad l = -\frac{b}{c}, \quad c' = -c, \quad b' = -b, \quad h = b + c + 1, \quad o = 0
\]\[
d + b + c + 1 = d + h = l + k = -\frac{b}{c} - \frac{c}{b}
\]\[
\Rightarrow d = -\frac{b^2 + c^2 + b^2c + bc^2 + bc}{bc}
\]
Let \( X = KC' \cap LB' \).
We have:
\[
\frac{x + c}{\overline{x + c}} = \frac{c - \frac{c}{b}}{\overline{c - \frac{c}{b}}} = -\frac{c^2}{b}
\Rightarrow \overline{x} = -\frac{xb + bc + c}{c^2}
\]Analogously,
\[
\overline{x} = -\frac{xc + bc + b}{b^2}
\]Equating both expressions:
\[
(b - c)(b^2c + bc^2 + bc - x(b^2 + bc + c^2)) = 0
\Rightarrow x = -\frac{bc(b + c + 1)}{b^2 + bc + c^2}
\]
Points \( A, D, X \) are collinear if and only if
\[
\frac{d - 1}{x - 1} \in \mathbb{R}
\]Substituting:
\[
\frac{(b^2 + c^2 + b^2c + bc^2 + 2bc)/bc}{(b^2c + bc^2 + 2bc + b^2 + c^2)/(b^2 + bc + c^2)} = \frac{b^2 + bc + c^2}{bc}\in \mathbb{R}_\blacksquare
\]
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bin_sherlo
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Let $B',C'$ be the antipodes of $B,C$ on $(ABC)$. Also let $LC'\cap KB'=W,KC'\cap LB'=P$.
Claim: $B',F,L$ and $C',E,K$ are collinear.
Proof: Pascal at $KB'LCAB$ yields $AC_{\infty},B'L\cap AB,H$ are collinear thus, $B'L\cap AB=F$. Similarily $C',E,K$ are collinear.
Claim: $P$ lies on $OH$.
Proof: Pascal at $BKC'CLB'$ gives $H,P,O$ are collinear.
Claim: $A,D,P$ are collinear.
Proof: Notice that $W,L,H,K$ lie on the circle with diameter $WH$ and since $AH=AK=AL$, $A$ must be the circumcenter of $(KLHW)$. Hence $W,A,H$ are collinear.
DDIT at $DLHK$ implies $(\overline{AD},\overline{AH}),(\overline{AK},\overline{AL}),(\overline{AC'},\overline{AB'})$ is an involution. DDIT at $B'KC'L$ gives $(\overline{AP},\overline{AH}),(\overline{AK},\overline{AL}),(\overline{AC'},\overline{AB'})$ is an involution. Combining these implies $AD\equiv AP$ hence $A,D,P$ are collinear as desired.$\blacksquare$
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