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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.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

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April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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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
Symmetric inequality with two directions
Tintarn   21
N 9 minutes ago by Maximilian113
Source: Germany 2017, Problem 5
Prove that for all non-negative numbers $x,y,z$ satisfying $x+y+z=1$, one has
\[1 \le \frac{x}{1-yz}+\frac{y}{1-zx}+\frac{z}{1-xy} \le \frac{9}{8}.\]
21 replies
Tintarn
May 4, 2017
Maximilian113
9 minutes ago
Do not try to case bash lol
ItzsleepyXD   4
N 13 minutes ago by straight
Source: Own , Mock Thailand Mathematic Olympiad P3
Let $n,d\geqslant 6$ be a positive integer such that $d\mid 6^{n!}+1$ .
Prove that $d>2n+6$ .
4 replies
ItzsleepyXD
Yesterday at 9:08 AM
straight
13 minutes ago
Show that CK is parallel to AB
Math-lover123   45
N 34 minutes ago by reni_wee
Source: Sharygin First Round 2013, Problem 16
The incircle of triangle $ABC$ touches $BC$, $CA$, $AB$ at points $A_1$, $B_1$, $C_1$, respectively. The perpendicular from the incenter $I$ to the median from vertex $C$ meets the line $A_1B_1$ in point $K$. Prove that $CK$ is parallel to $AB$.
45 replies
1 viewing
Math-lover123
Apr 8, 2013
reni_wee
34 minutes ago
Olympiad with ADHD
nunoarala   0
42 minutes ago
Disclaimer: I'm not sure about whether this is the right forum for this topic; feel free to move this elsewhere if appropriate.

I would like to get some insight into the experience of doing Math Olympiads (say, at the IMO level or close) while dealing with ADHD or similar conditions. If there are any people out there who have experienced that and are comfortable talking about it publicly, I would be interested in answers to the following.

(1) Were you formally diagnosed with ADHD?

(2) If so, was this before or after you started engaging seriously with Math Olympiads?

(3) Do you feel that ADHD hurt your olympiad performance at some point? If so, how? (E.g. by leading to more careless mistakes in exams? By making you unable to sit straight for 4h30?)

(4) Did you get some treatment that helped reduce the impact of ADHD on your ability to perform at a contest? How much do you think that helped? What suggestions would you give to a fellow olympiad contestant struggling with ADHD?

Thanks in advance!
0 replies
nunoarala
42 minutes ago
0 replies
Infinite sum of the angles made by the points A_i,B_i
WakeUp   2
N an hour ago by Rohit-2006
Source: Baltic Way 1997
On two parallel lines, the distinct points $A_1,A_2,A_3,\ldots $ respectively $B_1,B_2,B_3,\ldots $ are marked in such a way that $|A_iA_{i+1}|=1$ and $|B_iB_{i+1}|=2$ for $i=1,2,\ldots $. Provided that $A_1A_2B_1=\alpha$, find the infinite sum $\angle A_1B_1A_2+\angle A_2B_2A_3+\angle A_3B_3A_4+\ldots $
2 replies
WakeUp
Jan 28, 2011
Rohit-2006
an hour ago
Distributing coins in a circle
quacksaysduck   1
N an hour ago by BR1F1SZ
Source: JOM 2025 Mock 1 P4
There are $n$ people arranged in a circle, and $n^{n^n}$ coins are distributed among them, where each person has at least $n^n$ coins. Each person is then assigned a random index number in $\{1,2,...n\}$ such that no two people have the same number. Then every minute, if $i$ is the number of minutes passed, the person with index number congruent to $i$ mod $n$ will give a coin to the person on his left or right. After some time, everyone has the same number of coins.

For what $n$ is this always possible, regardless of the original distribution of coins and index numbers?

(Proposed by Ho Janson)
1 reply
quacksaysduck
Jan 26, 2025
BR1F1SZ
an hour ago
The product of two p-pods is a p-pod
MellowMelon   10
N 2 hours ago by Mathandski
Source: USA TST 2011 P3
Let $p$ be a prime. We say that a sequence of integers $\{z_n\}_{n=0}^\infty$ is a $p$-pod if for each $e \geq 0$, there is an $N \geq 0$ such that whenever $m \geq N$, $p^e$ divides the sum
\[\sum_{k=0}^m (-1)^k {m \choose k} z_k.\]
Prove that if both sequences $\{x_n\}_{n=0}^\infty$ and $\{y_n\}_{n=0}^\infty$ are $p$-pods, then the sequence $\{x_ny_n\}_{n=0}^\infty$ is a $p$-pod.
10 replies
MellowMelon
Jul 26, 2011
Mathandski
2 hours ago
Nordic squares!
mathisreaI   36
N 2 hours ago by awesomehuman
Source: IMO 2022 Problem 6
Let $n$ be a positive integer. A Nordic square is an $n \times n$ board containing all the integers from $1$ to $n^2$ so that each cell contains exactly one number. Two different cells are considered adjacent if they share a common side. Every cell that is adjacent only to cells containing larger numbers is called a valley. An uphill path is a sequence of one or more cells such that:

(i) the first cell in the sequence is a valley,

(ii) each subsequent cell in the sequence is adjacent to the previous cell, and

(iii) the numbers written in the cells in the sequence are in increasing order.

Find, as a function of $n$, the smallest possible total number of uphill paths in a Nordic square.

Author: Nikola Petrović
36 replies
mathisreaI
Jul 13, 2022
awesomehuman
2 hours ago
Monochromatic bipartite subgraphs
L567   4
N 2 hours ago by ihategeo_1969
Source: STEMS Mathematics 2023 Category B P6
For a positive integer $n$, let $f(n)$ denote the largest integer such that for any coloring of a $K_{n,n}$ with two colors, there exists a monochromatic subgraph of $K_{n,n}$ isomorphic to $K_{f(n), f(n)}$. Is it true that for each positive integer $m$ we can find a natural $N$ such that for any integer $n \geqslant N$, $f(n) \geqslant m$?

Proposed by Suchir
4 replies
L567
Jan 8, 2023
ihategeo_1969
2 hours ago
Tilted Students Thoroughly Splash Tiger part 2
DottedCaculator   18
N 2 hours 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
2 hours ago
Find area!
ComplexPhi   4
N 2 hours 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
2 hours ago
Easy integer functional equation
MarkBcc168   93
N 3 hours 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
3 hours ago
-2 belongs to S
WakeUp   3
N 3 hours 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
3 hours ago
Short combi omg
Davdav1232   5
N 3 hours 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
3 hours ago
For triangles with equal inradii
Ankoganit   3
N Jan 13, 2020 by Kitkat2207
Source: India Postal Set 1 P3 2016
Four points lie on a plane such that no three of them are collinear. Consider the four triangles formed by taking any three points at a time. If the inradii of these four triangles are all equal, prove that the four triangles are congruent.
3 replies
Ankoganit
Jan 18, 2017
Kitkat2207
Jan 13, 2020
For triangles with equal inradii
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G H BBookmark kLocked kLocked NReply
Source: India Postal Set 1 P3 2016
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Ankoganit
3070 posts
#1 • 3 Y
Y by PRO2000, jonlin1000, Adventure10
Four points lie on a plane such that no three of them are collinear. Consider the four triangles formed by taking any three points at a time. If the inradii of these four triangles are all equal, prove that the four triangles are congruent.
This post has been edited 1 time. Last edited by Ankoganit, Jan 18, 2017, 3:08 AM
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WizardMath
2487 posts
#2 • 2 Y
Y by Adventure10, Mango247
The solution I submitted:
$\textbf{Solution}$:
Let the quadrilateral be $ABCD$ with the points lying in this order and that any side doesn't cross any other.
Assume that the quadrilateral $ABCD$ is concave, without loss of generality let it be concave at $D$. Then since the incircle is the largest circle that is completely inside a triangle, and since $D$ lies strictly inside $\triangle ABC$, the inradius of $ABC$ is strictly greater than that of $BDC$, which is contrary to our assumption. Thus the quadrilateral is convex.
By $[F]$ we denote the area of a figure $F$, by $s_{MNP}$ we denote the semiperimeter of the triangle $MNP$, by $r_{MNP}$ we denote the inradius of the triangle $MNP$.
Note that by the condition, $[ABC]/s_{ABC}=[ABD]/s_{ABD}=[DBC]/s_{DBC}=[ADC]/s_{ADC}=[ABCD]/(s_{ABC}+s_{ACD})=[ABCD]/(s_{ABD}+s_{BCD})$, which implies that the sum of the semiperimeters of triangles $ABC$ and $ACD$ is the same as that of $ABD$ and $BCD$.
$\Longrightarrow AB + BC + CA + AC + CD + DA = AB + BD + DA + BC + CD + DB \Longrightarrow AC = BD$.
Now we show that $\triangle ADC \cong \triangle BCD$, from which the others follow similarly.
Reflect $A$ about the perpendicular bisector of $CD$ to $B'$. Draw the semicircle $(D)$ with center as $D$ and radius $DB'$, with $CD$ lying on one of the diameters. Clearly $B$ should be on $(D)$ because $DB= AC= DB'$. Suppose $B \neq B'$.
Select an arbitrary point $X$ on the $(D)$. The inradius of $XDC$ is given by
$$r_{XDC}=\frac{m \sin x}{k+\sqrt{k^2-2m(1+ \cos x)}}$$where $m = XD.DC = BD.DC$, $k = XD + DC = BD + DC$, both of which are constant, and $x=\angle XDC$.
Let $AC$ intersect $(D)$ at $J$. We claim that as a function of $x$ there is atmost one X such that $f'(x)=0$ where the function is $r_{XDC}$ as $X$ varies from ray $DC \cap (D) = Y$ to $J$, excluding $Y$, (noting that the function is continuous and differentiable in the interval), for then, either the inradius increases from $B'$ to the maximum and then decreases till $J$ or the inradius increases till $J$. In either case, the inradius of $B'DC$ is not the inradius of any $\triangle XBC$, because the inradius strictly decreases in the second case, and in the first case because the radius can't decrease as much to equal the inradius of $B'DC$ because the inradius of $JDC$ is more than that of $B'DC$ as $B'DC$ is inside $JDC$, and due to the condition we prove, because otherwise it would increase after decreasing, which would violate the condition, for the slope of the tangent of the graph of the function will be 0 again after becoming 0 once. This analysis would imply that $B$ would not lie in the interval $(Y,J)$, and thus that $AC$ and $BD$ intersect outside the quadrilateral, which would be a contradiction, because it would violate the ordering we gave to the quadrilateral, and thus leading to a contradiction to our assumption that $B \neq B'$
Now we prove that there is exactly one $x$ in the given interval such that $f'(x)=0$. $f'(x)=0$ is equivalent to
$$mk \cos x +m \cos x \sqrt{k^2-2m(1+ \cos x)} = \frac {m^2 \sin^2 x}{\sqrt{k^2-2m(1+ \cos x)}}$$Note that $x$ can't be in the interval $[\pi/2,\pi]$ because the LHS would otherwise be negative or 0, while the RHS is not at the same time. So the only such possibility is when $x \in (0,\pi/2)$.
On substituting $\cos x=t$, the equation becomes $mt^2+t(2m-k^2)-tk\sqrt{k^2-2m(1+t)}+m=0$. Define $P(t)=mt^2+t(2m-k^2)-tk\sqrt{k^2-2m(1+t)}+m$. This is clearly continuous. When $x=0$, $t=1$, and $P(1)=-(BD-DC)^2-(BD+DC)|BD-DC| \leq 0$ but when $BD=DC$, $x\ne 0$ because then the triangle degenerates. When $x=\pi/2$, $t=0$, and $P(0) = m>0.$ So there exists an odd number of solutions for acute $x$, because in the graph of $f'(t)$, the places where it cuts the $x$ axis are precisely the points of change of sign. Since the equation on separating the expression with the radical on the other side and squaring yields a quartic expression which can have at most 4 real roots, there can be either 1 or 3 roots for which $x$ is acute. In particular, the expression is $m^2t^4 + 4m^2 t^3 + \_ t^2 + \_ t + m^2=0$ where the $\_$ stand for some coefficients we don't have to work with. From here, we can see that 3 roots don't exist for which $x$ is acute, because when we expand the quartic, we observe that the sum of roots is -4, and the product of the roots is 1, from which if there were 3 real roots, then there would have been 4 real roots, as complex roots occur in conjugates, and at least 2 roots must be negative since product is 1 and sum is negative, which gives us $\leq$ 2 roots for which $x$ is in the interval $(0,\pi/2)$, which is a contradiction, which leads to $B=B'$ and thus to the fact that $\triangle ABD \cong \triangle ADC$ due to the reflection. Similarly we show that the other pairs of triangles having the same base (as one of the sides of the quadrilateral) are congruent, and due to the transitivity of congruences, we conclude that all triangles are congruent. \ \ \ $ \blacksquare$
$\textbf{Note}:$ We can also prove the fact that $ABCD$ is a rectangle using the conclusion of the solution.
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Ankoganit
3070 posts
#3 • 2 Y
Y by Adventure10, Mango247
This is also Japan MO 2011 P5.
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Kitkat2207
98 posts
#4 • 1 Y
Y by Adventure10
An alternative approach:
We know that [ABC]+[ACD]=[BAD]+[BCD]....using [area]=$s*r$ here, we get that the diagonals are equal,lets call them x.
Using s*r=( (s)(s-a)(s-b)(s-x) )^1\2 for any one of the triangles where s=(a+b+x)/2 and a and b are two corresponding adjacent sides and simplifying gives us a cubic in x: $x^3-x^2(b+c)-x(a^2+b^2-2ab+(a+b)r^2)+(a+b)( (a-b)^2-r^2 )$
We know that x is same for any two sides=> cubic has same coefficients wrt any 2 adjacent sides. So,the cubic reveals that sum of adjacent sides is equal. Which implies that opposite sides of ABCD are equal. :jump:

As two adjacent sides and third side(diagonal) are equal, the triangles are congruent by SSS test.
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