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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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0 replies
jlacosta
Apr 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
Tilted Students Thoroughly Splash Tiger part 2
DottedCaculator   18
N 24 minutes ago by MathLuis
Source: ELMO 2024/5
In triangle $ABC$ with $AB<AC$ and $AB+AC=2BC$, let $M$ be the midpoint of $\overline{BC}$. Choose point $P$ on the extension of $\overline{BA}$ past $A$ and point $Q$ on segment $\overline{AC}$ such that $M$ lies on $\overline{PQ}$. Let $X$ be on the opposite side of $\overline{AB}$ from $C$ such that $\overline{AX} \parallel \overline{BC}$ and $AX=AP=AQ$. Let $\overline{BX}$ intersect the circumcircle of $BMQ$ again at $Y \neq B$, and let $\overline{CX}$ intersect the circumcircle of $CMP$ again at $Z \neq C$. Prove that $A$, $Y$, and $Z$ are collinear.

Tiger Zhang
18 replies
DottedCaculator
Jun 21, 2024
MathLuis
24 minutes ago
Find area!
ComplexPhi   4
N 27 minutes ago by TigerOnion
Let $O_1$ be a point in the exterior of the circle $\omega$ of center $O$ and radius $R$ , and let $O_1N$ , $O_1D$ be the tangent segments from $O_1$ to the circle. On the segment $O_1N$ consider the point $B$ such that $BN=R$ .Let the line from $B$ parallel to $ON$ intersect the segment $O_1D$ at $C$ . If $A$ is a point on the segment $O_1D$ other than $C$ so that $BC=BA=a$ , and if the incircle of the triangle $ABC$ has radius $r$ , then find the area of $\triangle ABC$ in terms of $a ,R ,r$.
4 replies
ComplexPhi
Feb 4, 2015
TigerOnion
27 minutes ago
Easy integer functional equation
MarkBcc168   93
N 41 minutes ago by ray66
Source: APMO 2019 P1
Let $\mathbb{Z}^+$ be the set of positive integers. Determine all functions $f : \mathbb{Z}^+\to\mathbb{Z}^+$ such that $a^2+f(a)f(b)$ is divisible by $f(a)+b$ for all positive integers $a,b$.
93 replies
MarkBcc168
Jun 11, 2019
ray66
41 minutes ago
-2 belongs to S
WakeUp   3
N an hour ago by Burmf
Source: Baltic Way 1996 Q12
Let $S$ be a set of integers containing the numbers $0$ and $1996$. Suppose further that any integer root of any non-zero polynomial with coefficients in $S$ also belongs to $S$. Prove that $-2$ belongs to $S$.
3 replies
WakeUp
Mar 19, 2011
Burmf
an hour ago
Short combi omg
Davdav1232   5
N an hour ago by fagot
Source: Israel TST 2025 test 4 p3
Let \( n \) be a positive integer. A graph on \( 2n - 1 \) vertices is given such that the size of the largest clique in the graph is \( n \). Prove that there exists a vertex that is present in every clique of size \( n\)
5 replies
Davdav1232
Feb 3, 2025
fagot
an hour ago
Isi 2016 geometry
zizou10   22
N an hour ago by kamatadu
Source: ISI BSTAT 2016 #5
Prove that there exists a right angle triangle with rational sides and area $d$ if and only if $x^2,y^2$ and $z^2$ are squares of rational numbers and are in Arithmetic Progression

Here $d$ is an integer.
22 replies
zizou10
May 8, 2016
kamatadu
an hour ago
If ab+1 is divisible by A then so is a+b
ravengsd   3
N an hour ago by trigadd123
Source: Romania EGMO TST 2025 Day 2, Problem 4
Find the greatest positive integer $A$ such that, for all positive integers $a$ and $b$, if $A$ divides $ab+1$, then $A$ divides $a+b$.
3 replies
ravengsd
5 hours ago
trigadd123
an hour ago
IMO Shortlist 2012, Geometry 2
lyukhson   88
N an hour ago by zuat.e
Source: IMO Shortlist 2012, Geometry 2
Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ meet at $E$. The extensions of the sides $AD$ and $BC$ beyond $A$ and $B$ meet at $F$. Let $G$ be the point such that $ECGD$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $AD$. Prove that $D,H,F,G$ are concyclic.
88 replies
lyukhson
Jul 29, 2013
zuat.e
an hour ago
Trivial fun Equilateral
ItzsleepyXD   5
N 2 hours ago by reni_wee
Source: Own , Mock Thailand Mathematic Olympiad P1
Let $ABC$ be a scalene triangle with point $P$ and $Q$ on the plane such that $\triangle BPC , \triangle CQB$ is an equilateral . Let $AB$ intersect $CP$ and $CQ$ at $X$ and $Z$ respectively and $AC$ intersect $BP$ and $BQ$ at $Y$ and $W$ respectively .
Prove that $XY\parallel ZW$
5 replies
ItzsleepyXD
Yesterday at 9:05 AM
reni_wee
2 hours ago
Geometry..Pls
Jackson0423   2
N 2 hours ago by Royal_mhyasd
In equilateral triangle \( ABC \), let \( AB = 10 \). Point \( D \) lies on segment \( BC \) such that \( BC = 4 \cdot DC \). Let \( O \) and \( I \) be the circumcenter and incenter of triangle \( ABD \), respectively. Let \( O' \) and \( I' \) be the circumcenter and incenter of triangle \( ACD \), respectively. Suppose that lines \( OI \) and \( O'I' \) intersect at point \( X \). Find the length of \( XD \).
2 replies
+1 w
Jackson0423
5 hours ago
Royal_mhyasd
2 hours ago
4 variables with quadrilateral sides 2
mihaig   4
N 2 hours ago by arqady
Source: Own
Let $a,b,c,d\geq0$ satisfying
$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}=2.$$Prove
$$\left(a+b+c+d-2\right)^2+8\geq3\left(abc+abd+acd+bcd\right).$$
4 replies
mihaig
Tuesday at 8:47 PM
arqady
2 hours ago
BMO 2015 #1: Inequality on a,b,c.
MathKnight16   25
N 2 hours ago by Rayvhs
Source: BMO 2015 problem 1
If ${a, b}$ and $c$ are positive real numbers, prove that

\begin{align*}
 a ^ 3b ^ 6 + b ^ 3c ^ 6 + c ^ 3a ^ 6 + 3a ^ 3b ^ 3c ^ 3 &\ge{ abc \left (a ^ 3b ^ 3 + b ^ 3c ^ 3 + c ^ 3a ^ 3 \right) + a ^ 2b ^ 2c ^ 2 \left (a ^ 3 + b ^ 3 + c ^ 3 \right)}.
\end{align*}

(Montenegro).
25 replies
MathKnight16
May 5, 2015
Rayvhs
2 hours ago
4 lines concurrent
Zavyk09   5
N 2 hours ago by tomsuhapbia
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 $OH$.
5 replies
Zavyk09
Apr 9, 2025
tomsuhapbia
2 hours ago
SL 2015 G1: Prove that IJ=AH
Problem_Penetrator   136
N 3 hours ago by Mathgloggers
Source: IMO 2015 Shortlist, G1
Let $ABC$ be an acute triangle with orthocenter $H$. Let $G$ be the point such that the quadrilateral $ABGH$ is a parallelogram. Let $I$ be the point on the line $GH$ such that $AC$ bisects $HI$. Suppose that the line $AC$ intersects the circumcircle of the triangle $GCI$ at $C$ and $J$. Prove that $IJ = AH$.
136 replies
Problem_Penetrator
Jul 7, 2016
Mathgloggers
3 hours ago
Concurrent
Omid Hatami   10
N Aug 17, 2023 by Tafi_ak
Source: Iran 2005
Suppose $H$ and $O$ are orthocenter and circumcenter of triangle $ABC$. $\omega$ is circumcircle of $ABC$. $AO$ intersects with $\omega$ at $A_1$. $A_1H$ intersects with $\omega$ at $A'$ and $A''$ is the intersection point of $\omega$ and $AH$. We define points $B',\ B'',\ C'$ and $C''$ similiarly. Prove that $A'A'',B'B''$ and $C'C''$ are concurrent in a point on the Euler line of triangle $ABC$.
10 replies
Omid Hatami
Aug 27, 2005
Tafi_ak
Aug 17, 2023
Concurrent
G H J
Source: Iran 2005
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Omid Hatami
1275 posts
#1 • 2 Y
Y by Adventure10, Mango247
Suppose $H$ and $O$ are orthocenter and circumcenter of triangle $ABC$. $\omega$ is circumcircle of $ABC$. $AO$ intersects with $\omega$ at $A_1$. $A_1H$ intersects with $\omega$ at $A'$ and $A''$ is the intersection point of $\omega$ and $AH$. We define points $B',\ B'',\ C'$ and $C''$ similiarly. Prove that $A'A'',B'B''$ and $C'C''$ are concurrent in a point on the Euler line of triangle $ABC$.
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grobber
7849 posts
#2 • 2 Y
Y by Adventure10, Mango247
I think the statement obscures what's really going on: if you replace the circumcircle with a circumconic, and $O,H$ with any two points $U,V$, you still have a valid problem :). So let's restate it (it won't be the exact analog, but something equivalent):

Problem: Let $BB_1B''B',CC_1C''C'$ be two quadrilaterals inscirbed in a conic $\mathcal T$ having the same intersection $V$ of the diagonals (i.e. $V=B_1B'\cap BB''=C_1C'\cap CC''$). Put $U=CC_1\cap BB_1,T=C'C''\cap B'B''$. Then $U,V,T$ are collinear.

If we define $A_1,A',A''$ in a similar manner and prove the analogous statements for the pairs of configurations corresponding to $(A,B),(A,C)$, we will have shown that each two among the lines $A'A'',B'B'',C'C''$ intersect on $UV$, which, except maybe for some uninteresting limit cases, is different from all of them, meaning that all three of them actually concur on $UV$, as desired.

Now for a proof of the problem stated above:

Fix the quadrilateral $BB_1B''B'$ and move $C$ on $\mathcal T$. This gives rise to a self-map $C'\mapsto C''$ of $\mathcal T$ which is clearly projective. When $C$ reaches the position that $C_1$ initially had, $C'$ and $C''$ are interchanged, so this map is an involution, meaning that $C'C''$ passes through a fixed point. When $C=B_1$, the line $C'C''$ coincides with $B''B'$, so this fixed point is actually the intersection $T$ between $C'C''$ and $B'B''$. Finally, when $C=B''$, the collinearity of $U,V,T$ follows from a direct application of Pascal's Theorem applied to the hexagon $CC_1C'C''B_1B'$ (in this configuration $C=B'',C''=B$).

There's probably a slicker and more direct proof of the problem based on a couple of applications of Pascal's Theorem as well, but I didn't want to go point-chasing :).
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shobber
3498 posts
#3 • 2 Y
Y by Adventure10, Mango247
what is point chasing?
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mecrazywong
606 posts
#4 • 1 Y
Y by Adventure10
shobber wrote:
what is point chasing?
I believe it is not mathematically defined. Anyway, grobber's meaning of point chasing is understandable here: Applying Pascal Theorem to one orientation of six of $B,B_1,B'',B',C,C_1,C'',C'$. I've tried it last night(GMT+8), and applying two times is enough.
Anyway, I prefer using trigo rather than Pascal to solve it-they are just the same ;)
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darij grinberg
6555 posts
#5 • 2 Y
Y by Adventure10 and 1 other user
Just wanted to note:

The points H, O, H lie on the Euler line of triangle ABC.
The triangle ABC is the circumcevian triangle of the point H with respect to triangle A''B''C''.
The triangle $A_1B_1C_1$ is the circumcevian triangle of the point O with respect to triangle ABC.
The triangle A'B'C' is the circumcevian triangle of the point H with respect to triangle $A_1B_1C_1$.

Thus, the statement of the problem, namely that the lines A''A', B''B', C''C' concur at one point on the Euler line of triangle ABC, follows from the "circumcevian pingpong theorem".

Darij
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jensen
572 posts
#6 • 1 Y
Y by Adventure10
here is my solution:
if we use Pascal theorem about inscribed hexagons we will have:
in $A_1AA'C_1CC'$ $\rightarrow$
$[(A_1A , C_1C)=O]$& $[(AA',CC')=H]$ & $[(A_1C' , C_1A')=X]$ are on one line.(2)
and in $C_1A'A''A_1C'C''$ $\rightarrow$
$[(C_1A' , C'A_1)=X]$ &
$[(A'A'' , C'C'')=Q]$ & $[(C_1C'' , A_1A'')=H]$ are on one line.(1)
and we have $[(C_1A' , C'A_1)=(A_1C' , C_1A')]$ AND
$[(C_1C'' , A_1A'')=(AA' , CC')=H]$ (3)
and from (1),(2) and (3) we get that $H,O,Q,X$ are on one line.
and it means that $A'A'',C'C'',OH$ are concurrent.
Similary we get that $B'B'',C'C'',OH$ are concurrent and we R done
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H.HAFEZI2000
328 posts
#7 • 2 Y
Y by Adventure10, Mango247
PRETTY EASILY USING COMPLEX
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drmzjoseph
445 posts
#8 • 2 Y
Y by Adventure10, Mango247
Trivial cause using a composition of three involutions in $A'$ to $A$ to $A_1$ to $A''$ with focus $H,O,H$ resp, makes $A',A"$ an involution with focus at $OH$
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Yaghi
412 posts
#9 • 2 Y
Y by Adventure10, Mango247
Another way to do this is poles and polar:
Let $X_a$ be the pole of $A'A"$.Define $X_b,X_c$ similarly.Since $(A',A";B,C)=-1 \implies X_a \in BC$.Note that
$$X_aO^2 -X_aH^2=X_aA'^2+R^2-X_aH^2=R^2$$So by Carnot's theorem,$X_a,X_b,X_c$ are collinear and the line passing through them is perpendicular to $OH$,so their polar are concurrent on $OH$ and we are done.
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enhanced
515 posts
#10 • 2 Y
Y by Adventure10, Mango247
It can also be proven very easily using inversion .
This post has been edited 2 times. Last edited by enhanced, Jul 7, 2019, 2:21 AM
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Tafi_ak
309 posts
#11
Y by
Perform an inversion centered at $H$ with radius $\sqrt{-HA\cdot HA''}$. So this inversion fixes the circumcircle.

After this inversion, it is sufficient to show circumferences $(A''H)$, $(B''H)$, $(C'C''H)$ are coaxial for the concurrency of $A'A''$, $B'B''$, $C'C''$. Notice that \[ \angle A'A''H=\angle A'A_1O=\angle A_1A'O \]So $OA'$ is tangent to $(A'A''H)$, similarity for the other two. Hence $O$ has equal power wrt these three circles, so $OH$ is the common radical axis of these three circles. So these circles are coaxial. Done for the first part.

Now suppose $A'A''$, $B'B''$, $C'C''$ concur at $X$. By power of point notice that $X$ lies on their common radical axis which is $OH$. Done.
This post has been edited 1 time. Last edited by Tafi_ak, Aug 17, 2023, 5:13 PM
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