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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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
Need help on this simple looking problem
TheGreatEuler   0
2 minutes ago
Show that 1+2+3+4....n divides 1^k+2^k+3^k....n^k when k is odd. Is this possible to prove without using congruence modulo or binomial coefficients?
0 replies
TheGreatEuler
2 minutes ago
0 replies
Geometry
Lukariman   5
N 30 minutes ago by Lukariman
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that $\angle HDM$ = 2∠AMP.
5 replies
Lukariman
Yesterday at 12:43 PM
Lukariman
30 minutes ago
inq , not two of them =0
win14   0
36 minutes ago
Let a,b,c be non negative real numbers such that no two of them are simultaneously equal to 0
$$\frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a} \ge \frac{5}{2\sqrt{ab + bc + ca}}.$$
0 replies
1 viewing
win14
36 minutes ago
0 replies
IMO Genre Predictions
ohiorizzler1434   62
N 41 minutes ago by ehuseyinyigit
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
62 replies
ohiorizzler1434
May 3, 2025
ehuseyinyigit
41 minutes ago
Iranians playing with cards module a prime number.
Ryan-asadi   1
N an hour ago by ItzsleepyXD
Source: Iranian Team Selection Test - P2
Let $p$ be an arbitrary prime number.we have a deck of cards which a number is written on the back of each of them. such that for every $i \in \{1,…,p-1\}$ we have at most one card with number $i$ written on its back and we also have exactly one card with number zero on its back.we want to design a game in which we need to decide for every two cards $X,Y$ witch one wins from the other one using the following rules.

$(I)$: If $x$ wins from $y$ and also $y$ wins from $z$ , then $x$ wins from $z$.
$(II)$: If $x$ doesn’t fail $y$ and $z$ doesn’t fail $t$ , then in case of existing of both cards $y+t$ and $x+z$ module $p$, card $x+z$ also doesn’t fail $y+t$.

What is the maximum number of cards which designing such game is possible?

1 reply
Ryan-asadi
an hour ago
ItzsleepyXD
an hour ago
Number theory
MathsII-enjoy   5
N an hour ago by MathsII-enjoy
Prove that when $x^p+y^p$ | $(p^2-1)^n$ with $x,y$ are positive integers and $p$ is prime ($p>3$), we get: $x=y$
5 replies
MathsII-enjoy
Monday at 3:22 PM
MathsII-enjoy
an hour ago
Coloring plane in black
Ryan-asadi   0
an hour ago
Source: Iran Team Selection Test - P3
We are given $n>10$ lines in the plane such that no two of them are parallel and no three of them are concurrent. We color at least $\frac{n^2}{8}+1$ regions in black from all limited regions which have constructed in the plane. We call a triangle constructed by three lines “black-less” if there exist exactly one black region inside that. Prove that there exist at least $\frac{n}{2}$ “black-lass” triangles in plane.
0 replies
Ryan-asadi
an hour ago
0 replies
Number theory
Foxellar   0
an hour ago
It is known that for all positive integers $k$,
\[
1^2 + 2^2 + 3^2 + \ldots + k^2 = \frac{k(k + 1)(2k + 1)}{6}
\]Find the smallest positive integer $k$ such that $1^2 + 2^2 + 3^2 + \ldots + k^2$ is divisible by 200.
0 replies
Foxellar
an hour ago
0 replies
Combinatorics
P162008   4
N 2 hours ago by cazanova19921
Let $m,n \in \mathbb{N}.$ Let $[n]$ denote the set of natural numbers less than or equal to $n.$

Let $f(m,n) = \sum_{(x_1,x_2,x_3, \cdots, x_m) \in [n]^{m}} \frac{x_1}{x_1 + x_2 + x_3 + \cdots + x_m} \binom{n}{x_1} \binom{n}{x_2} \binom{n}{x_3} \cdots \binom{n}{x_m} 2^{\left(\sum_{i=1}^{m} x_i\right)}$

Compute the sum of the digits of $f(4,4).$
4 replies
P162008
Today at 5:38 AM
cazanova19921
2 hours ago
An analytic sequence
Ryan-asadi   0
2 hours ago
Source: Iran Team Selection Test - P1
Let $a_n$ be a sequence of real positive numbers such that for all $n>2025$ :
$$a_n = \max_{1 \le i \le 2025}{a_{n-i}}-\min_{1 \le i \le 2025}{a_{n-i}} $$Prove that for all large enough natural $n$ we have that $a_n < \frac{1}{1404}$.
0 replies
Ryan-asadi
2 hours ago
0 replies
Aime type Geo
ehuseyinyigit   4
N 2 hours ago by ehuseyinyigit
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
4 replies
ehuseyinyigit
Monday at 9:04 PM
ehuseyinyigit
2 hours ago
n variables with n-gon sides
mihaig   1
N 2 hours ago by mihaig
Source: Own
Let $n\geq3$ and let $a_1,a_2,\ldots, a_n\geq0$ be reals such that $\sum_{i=1}^{n}{\frac{1}{2a_i+n-2}}=1.$
Prove
$$\frac{24}{(n-1)(n-2)}\cdot\sum_{1\leq i<j<k\leq n}{a_ia_ja_k}\geq3\sum_{i=1}^{n}{a_i}+n.$$
1 reply
mihaig
Apr 25, 2025
mihaig
2 hours ago
Geometry
gggzul   5
N 2 hours ago by nabodorbuco2
In trapezoid $ABCD$ segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle A$ and $\angle C$ meet at $P$. Angle bisectors of $\angle B$ and $\angle D$ meet at $Q$. Prove that $ABPQ$ is cyclic
5 replies
gggzul
Yesterday at 8:22 AM
nabodorbuco2
2 hours ago
Inspired by lgx57
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b>0, a^4+ab+b^4=60 $. Prove that
$$  2\sqrt{15} \leq a^2+ab+b^2 \leq \frac{3(\sqrt{481}-1)}{4}$$$$\frac{\sqrt{481}-1}{4}\leq a^2-ab+b^2  \leq 2\sqrt{15} $$Let $ a,b>0, a^4-ab+b^4=60 $. Prove that
$$ 2\sqrt{15} \leq a^2+ab+b^2 \leq \frac{3(\sqrt{481}+1)}{4}$$$$ 5<a^2-ab+b^2 \leq 2\sqrt{15} $$
1 reply
sqing
3 hours ago
sqing
2 hours ago
Similar triangles formed by angular condition
Mahdi_Mashayekhi   5
N Apr 21, 2025 by sami1618
Source: Iran 2025 second round P3
Point $P$ lies inside of scalene triangle $ABC$ with incenter $I$ such that $:$
$$ 2\angle ABP = \angle BCA , 2\angle ACP = \angle CBA $$Lines $PB$ and $PC$ intersect line $AI$ respectively at $B'$ and $C'$. Line through $B'$ parallel to $AB$ intersects $BI$ at $X$ and line through $C'$ parallel to $AC$ intersects $CI$ at $Y$. Prove that triangles $PXY$ and $ABC$ are similar.
5 replies
Mahdi_Mashayekhi
Apr 19, 2025
sami1618
Apr 21, 2025
Similar triangles formed by angular condition
G H J
G H BBookmark kLocked kLocked NReply
Source: Iran 2025 second round P3
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Mahdi_Mashayekhi
695 posts
#1 • 2 Y
Y by Rounak_iitr, Parsia--
Point $P$ lies inside of scalene triangle $ABC$ with incenter $I$ such that $:$
$$ 2\angle ABP = \angle BCA , 2\angle ACP = \angle CBA $$Lines $PB$ and $PC$ intersect line $AI$ respectively at $B'$ and $C'$. Line through $B'$ parallel to $AB$ intersects $BI$ at $X$ and line through $C'$ parallel to $AC$ intersects $CI$ at $Y$. Prove that triangles $PXY$ and $ABC$ are similar.
Z K Y
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gghx
1072 posts
#2
Y by
Let $S=BP\cap CI$ and $T=BI\cap CP$. Note that $$\angle BPC=\angle ABP+\angle ACP=\angle ACI+\angle ABI=\angle BIC,$$hence $BPIC$ is cyclic.

We now prove that $S$ is the circumcenter of $\triangle B'IX$. This is true because $$\angle SB'I=\angle BAI + \angle ABP=\frac{1}{2}(\angle A + \angle C)=\angle IAC+\angle ACI = \angle SIB',$$hence $SI=SB'$. Furthermore, $$\angle ISB'=180^\circ-\angle A - \angle C=\angle B=2\angle ABI=2\angle B'XI,$$hence $S$ is the circumcenter of $\triangle B'IX$ as desired.

Now, $\angle SXI=\angle SIX=\angle SIT=\angle SPT$, hence $SXTP$ is cyclic. Similarly, $SYTP$ is cyclic, so $SXTPY$ is cyclic.

We are now ready to finish the question. We have $$\angle YPX=\angle XSI=180^\circ-2\angle SIX=\angle A$$and $$\angle YXP=180^\circ-\angle PSI=\angle ABP+\angle ACI=\angle C,$$so triangles $PXY$ and $ABC$ are similar.
Z K Y
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ItzsleepyXD
132 posts
#3
Y by
quite similar (or same IDK) to @above

since $\triangle ABB' \sim \triangle ACI$ and $\triangle ACC' \sim \triangle ABI$
implies that $\triangle BB'I \sim \triangle CIC'$

Let $E = BB' \cap IY , F= CC' \cap IX$ .
So $EB'=EI , FC'=FI$
Known that $\angle B'EI = \angle ABC = 2 \angle B'XI$ implies that $E =$ center of $(B'IX)$.
so $ \angle EXI = \angle EIX = \angle EPF$ so $E,X,F,P$ concyclic.
same as $E,Y,F,P$ concyclic .
so $E,Y,X,F,P$ concyclic.
thus $\angle PXY = \angle PEY = \angle ABC$ and $ \angle PYX = 180^{\circ} - \angle PFX = \angle ACB$
Conclude that $\triangle PXY \sim \triangle ABC$ . done $\square$
Z K Y
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mathuz
1524 posts
#4
Y by
Consider the intersection $C'Y\cap B'X = O(.)$. Then $O$ is the circumcenter of $PB'C'$, and it suffices to show that $O$ lies on the circumcircle of $PXY$.
Z K Y
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bin_sherlo
719 posts
#5
Y by
Let $BP\cap CI=U,PC\cap BI=V,AI\cap BC=D$. Note that $\measuredangle CPB=90+\frac{\measuredangle A}{2}=\measuredangle CIB$ hence $B,I,C,P$ are concyclic.
By Menelaus at $B'UBDCI$ we get $\frac{B'B}{IC}=\frac{\sin \measuredangle C}{\sin \measuredangle B}$. Hence $BX=BB'.\frac{\sin \frac{\measuredangle C}{2}}{\sin \frac{\measuredangle B}{2}}=\frac{BB'.IB}{IC}=\frac{\sin \measuredangle C}{\sin \measuredangle B}.IB$ Also $\frac{BV}{BP}=\frac{\cos \frac{\measuredangle A}{2}}{\sin \measuredangle C}$ thus,
\[\frac{BX.BV}{BP}=\frac{BI}{\sin \measuredangle B}.\cos \frac{\measuredangle A}{2}=BU\]which implies $X\in (PUV)$. Similarily $Y\in (PUV)$. Hence $\measuredangle XYP=\measuredangle XVC=\measuredangle C$ and $\measuredangle PXY=\measuredangle PUY=\measuredangle B$ as desired.$\blacksquare$
Z K Y
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sami1618
907 posts
#6
Y by
Here is a different approach rephrasing the problem in terms of reference triangle $PB'C'$. When I first drew the diagram the points $P$, $X$, and $Y$ were all very close together so this solution was motivated by drawing a diagram consisting of the points other than $A$, $B$, and $C$. :)
It is not hard to show that $\angle B'PC'=90^{\circ}-\tfrac{1}{2}\angle A$, $\angle PB'C'=90^{\circ}-\tfrac{1}{2}\angle B$, and $\angle PC'B'=90^{\circ}-\tfrac{1}{2}\angle C$. Since $\angle BPC=\angle BIC$, it must be that $I$ lies on the interior of segment $B'C'$. Because of this, it is also not hard to see that $X$, $Y$, and $P$ all lie on the same side of segment $AI$.
[asy]

import geometry;

size(10cm);
pair A = dir(110);
pair B = dir(200);
pair C = dir(340);
pair I = incenter(A, B, C);
pair D = foot(I, B, C);
pair Ep=B+C-D;
pair J=I+Ep-D;
pair P=isogonalconjugate(triangle(A,B,C),J);
pair Bp=intersectionpoint(line(B,P),line(A,I));
pair Cp=intersectionpoint(line(C,P),line(A,I));
pair X=intersectionpoint(line(Bp,B+Bp-A), line(B,I));
pair Y=intersectionpoint(line(Cp,C+Cp-A), line(C,I));
pair O=intersectionpoint(line(Cp,Y), line(Bp,X));
point[] Ap=intersectionpoints(circle(Bp,O,Cp),line(O,I));
pair Ap=Ap[1];
pair Op=circumcenter(Ap,Bp,Cp);
pair M_a=2*Op-O;
point[] N_c=intersectionpoints(circle(Bp,O,Cp),line(P,Cp));
pair N_c=N_c[0];
point[] N_b=intersectionpoints(circle(Bp,O,Cp),line(P,Bp));
pair N_b=N_b[0];


draw(A--B--C--cycle, black);
draw(B--Bp,black); draw(C--P,black); draw(A--Cp,black); draw(B--I,black); 
draw(C--Y,black); draw(Bp--X); draw(Cp--Y);  draw(P--Bp--Cp--cycle, black);


dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(50));

dot("$P$", P, dir(140));
dot("$B'$", Bp, dir(40));
dot("$C'$", Cp, dir(290));
dot("$X$", X, dir(310));
dot("$Y$", Y, dir(110));
[/asy]
Now notice that $\triangle B'IX\sim\triangle AIB$ and $\triangle C'IY\sim \triangle AIC$. We will now focus on the acute reference triangle $PB'C'$. Let $O$ be the circumcenter of $PB'C'$. Let $X'$ be the second intersection of line $B'O$ with the circumcircle of triangle $POC'$ and let $Y'$ be the second intersection of line $C'O$ with the circumcircle of triangle $POB'$. It is easy to show that $\triangle PX'C'\sim$ $\triangle PB'Y'\sim$ $\triangle ABC$. Now notice that $\triangle B'C'X'\sim$ $\triangle AIB\sim$ $\triangle B'IX$. Thus $X$ is the point on segment $B'X'$ with $IX\parallel C'X'$. Similarly, $Y$ is the point along segment $C'Y'$ with $IY\parallel B'Y'$.
[asy]
import geometry;

size(10cm);

pair P=dir(100);
pair Bp=dir(270+55);
pair Cp=dir(270-55);
pair O=(0,0);
pair I=intersectionpoint(line(O,P),line(Bp,Cp));
pair Xp[]=intersectionpoints(line(Bp,O),circle(P,O,Cp));
pair Xp=Xp[1];
pair Yp[]=intersectionpoints(line(Cp,O),circle(P,O,Bp));
pair Yp=Yp[1];
pair X=intersectionpoint(line(I,I+Xp-Cp), line(Bp,Xp));
pair Y=intersectionpoint(line(I,I+Yp-Bp), line(Cp,Yp));

fill(P--Xp--Cp--cycle,palered+white);
fill(P--Yp--Bp--cycle,palered+white);
fill(P--X--Y--cycle, palered);

draw(P--X--Y--cycle);
draw(P--Bp--Cp--cycle);
draw(circle(P,O,Cp));
draw(circle(P,O,Bp));
draw(Bp--Xp); draw(Cp--Yp);
draw(X--I--Y);
draw(Cp--Xp--P--Yp--Bp);

dot("P",P,2*dir(P));
dot("B'",Bp,dir(270));
dot("C'",Cp,dir(270));

dot("O",O,dir(270));
dot("I",I,dir(270));
dot("X'",Xp,dir(150));
dot("Y'",Yp,dir(30));
dot("X",X,dir(220));
dot("Y",Y,dir(-40));
[/asy]
Then we have that $$\frac{X'X}{XB'}=\frac{C'I}{IB'}=\frac{C'Y}{YY'}.$$Thus triangle $PXY$ is a linear combination of triangle $PX'C'$ and $PB'Y'$. Since these triangles are both similar to $ABC$, it is a well-known result that $PXY$ must be similar to $ABC$ as well.
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