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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.

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[*]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
Batman chases the Joker on a square board
Lukaluce   1
N an hour ago by navier3072
Source: 2025 Junior Macedonian Mathematical Olympiad P1
Batman, Robin, and The Joker are in three of the vertex cells in a square $2025 \times 2025$ board, such that Batman and Robin are on the same diagonal (picture). In each round, first The Joker moves to an adjacent cell (having a common side), without exiting the board. Then in the same round Batman and Robin move to an adjacent cell. The Joker wins if he reaches the fourth "target" vertex cell (marked T). Batman and Robin win if they catch The Joker i.e. at least one of them is on the same cell as The Joker.

If in each move all three can see where the others moved, who has a winning strategy, The Joker, or Batman and Robin? Explain the answer.

Comment. Batman and Robin decide their common strategy at the beginning.

IMAGE
1 reply
Lukaluce
Yesterday at 3:23 PM
navier3072
an hour ago
Interesting inequalities
sqing   12
N an hour ago by ytChen
Source: Own
Let $ a,b,c\geq 0 , (a+k )(b+c)=k+1.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2k-3+2\sqrt{k+1}}{3k-1}$$Where $ k\geq \frac{2}{3}.$
Let $ a,b,c\geq 0 , (a+1)(b+c)=2.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 2\sqrt{2}-1$$Let $ a,b,c\geq 0 , (a+3)(b+c)=4.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{7}{4}$$Let $ a,b,c\geq 0 , (3a+2)(b+c)= 5.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2(2\sqrt{15}-5)}{3}$$
12 replies
sqing
May 10, 2025
ytChen
an hour ago
Foot from vertex to Euler line
cjquines0   31
N 2 hours ago by awesomeming327.
Source: 2016 IMO Shortlist G5
Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $ABC$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $AB$ and $AC$ at $X$ and $Y$ respectively. Let $P$ be the foot of altitude from $A$ to $BC$, and let $M$ be the midpoint of $BC$. Prove that the circumcentre of triangle $XSY$ is equidistant from $P$ and $M$.
31 replies
cjquines0
Jul 19, 2017
awesomeming327.
2 hours ago
Integer polynomial commutes with sum of digits
cjquines0   43
N 2 hours ago by Ilikeminecraft
Source: 2016 IMO Shortlist N1
For any positive integer $k$, denote the sum of digits of $k$ in its decimal representation by $S(k)$. Find all polynomials $P(x)$ with integer coefficients such that for any positive integer $n \geq 2016$, the integer $P(n)$ is positive and $$S(P(n)) = P(S(n)).$$
Proposed by Warut Suksompong, Thailand
43 replies
cjquines0
Jul 19, 2017
Ilikeminecraft
2 hours ago
3^x+4xy=5^y diophantine
parmenides51   8
N 3 hours ago by shendrew7
Source: 2020 ℕumber Theory Contest (USAJMO level) #1 https://artofproblemsolving.com/community/c594864h2339943p18855098
Find all ordered pairs of natural numbers $(x,y)$ such that$$3^x+4xy=5^y.$$
Proposed by i3435
8 replies
parmenides51
Dec 3, 2023
shendrew7
3 hours ago
Grand finale of 2021 Iberoamerican MO
jbaca   5
N 3 hours ago by MathLuis
Source: 2021 Iberoamerican Mathematical Olympiad, P6
Consider a $n$-sided regular polygon, $n \geq 4$, and let $V$ be a subset of $r$ vertices of the polygon. Show that if $r(r-3) \geq n$, then there exist at least two congruent triangles whose vertices belong to $V$.
5 replies
jbaca
Oct 20, 2021
MathLuis
3 hours ago
IMO Shortlist 2010 - Problem N1
Amir Hossein   50
N 3 hours ago by shendrew7
Find the least positive integer $n$ for which there exists a set $\{s_1, s_2, \ldots , s_n\}$ consisting of $n$ distinct positive integers such that
\[ \left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \cdots \left( 1 - \frac{1}{s_n} \right) = \frac{51}{2010}.\]

Proposed by Daniel Brown, Canada
50 replies
Amir Hossein
Jul 17, 2011
shendrew7
3 hours ago
Computing functions
BBNoDollar   3
N 4 hours ago by BBNoDollar
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )
3 replies
BBNoDollar
Yesterday at 5:25 PM
BBNoDollar
4 hours ago
Oh no! Inequality again?
mathisreaI   109
N 4 hours ago by da-rong_wae
Source: IMO 2022 Problem 2
Let $\mathbb{R}^+$ denote the set of positive real numbers. Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for each $x \in \mathbb{R}^+$, there is exactly one $y \in \mathbb{R}^+$ satisfying $$xf(y)+yf(x) \leq 2$$
109 replies
mathisreaI
Jul 13, 2022
da-rong_wae
4 hours ago
Mmmmmm...Tasty!
whatshisbucket   35
N 4 hours ago by shendrew7
Source: 2017 ELMO #4
An integer $n>2$ is called tasty if for every ordered pair of positive integers $(a,b)$ with $a+b=n,$ at least one of $\frac{a}{b}$ and $\frac{b}{a}$ is a terminating decimal. Do there exist infinitely many tasty integers?

Proposed by Vincent Huang
35 replies
whatshisbucket
Jun 26, 2017
shendrew7
4 hours ago
Floor double summation
CyclicISLscelesTrapezoid   53
N 4 hours ago by ezpotd
Source: ISL 2021 A2
Which positive integers $n$ make the equation \[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\]true?
53 replies
CyclicISLscelesTrapezoid
Jul 12, 2022
ezpotd
4 hours ago
Sets with Polynomials
insertionsort   27
N 5 hours ago by ezpotd
Source: ISL 2020 A2
Let $\mathcal{A}$ denote the set of all polynomials in three variables $x, y, z$ with integer coefficients. Let $\mathcal{B}$ denote the subset of $\mathcal{A}$ formed by all polynomials which can be expressed as
\begin{align*}
(x + y + z)P(x, y, z) + (xy + yz + zx)Q(x, y, z) + xyzR(x, y, z)
\end{align*}with $P, Q, R \in \mathcal{A}$. Find the smallest non-negative integer $n$ such that $x^i y^j z^k \in \mathcal{B}$ for all non-negative integers $i, j, k$ satisfying $i + j + k \geq n$.
27 replies
insertionsort
Jul 20, 2021
ezpotd
5 hours ago
Constructing two sets from conditions on their intersection, union and product
jbaca   17
N 5 hours ago by MathLuis
Source: 2021 Iberoamerican Mathematical Olympiad, P5
For a finite set $C$ of integer numbers, we define $S(C)$ as the sum of the elements of $C$. Find two non-empty sets $A$ and $B$ whose intersection is empty, whose union is the set $\{1,2,\ldots, 2021\}$ and such that the product $S(A)S(B)$ is a perfect square.
17 replies
jbaca
Oct 20, 2021
MathLuis
5 hours ago
Functional Inequality Implies Uniform Sign
peace09   34
N 5 hours ago by MathIQ.
Source: 2023 ISL A2
Let $\mathbb{R}$ be the set of real numbers. Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function such that \[f(x+y)f(x-y)\geqslant f(x)^2-f(y)^2\]for every $x,y\in\mathbb{R}$. Assume that the inequality is strict for some $x_0,y_0\in\mathbb{R}$.

Prove that either $f(x)\geqslant 0$ for every $x\in\mathbb{R}$ or $f(x)\leqslant 0$ for every $x\in\mathbb{R}$.
34 replies
peace09
Jul 17, 2024
MathIQ.
5 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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