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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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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 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
Plz give me the solution
Madunglecha   1
N 5 minutes ago by top1vien
For given M
h(n) is defined as the number of which is relatively prime with M, and 1 or more and n or less.
As B is h(M)/M, prove that there are at least M/3 or more N such that satisfying the below inequality
|h(N)-BN| is under 1+sqrt(B×2^((the number of prime factor of M)-3))
1 reply
Madunglecha
3 hours ago
top1vien
5 minutes ago
King's Constrained Walk
Hellowings   2
N 18 minutes ago by Hellowings
Source: Own
Given an n x n chessboard, with a king starting at any square, the king's task is to visit each square in the board exactly once (essentially an open path); this king moves how a king in chess would.
However, we are allowed to place k numbers on the board of any value such that for each number A we placed on the board, the king must be in the position of that number A on its Ath square in its journey, with the starting square as its 1st square.
Suppose after we placed k numbers, there is one and only one way to complete the king's task (this includes placing the king in a starting square), find the minimum value of k set by n.

Didn't know I could post it here xd; I'm unsure how hard this question could be.
2 replies
Hellowings
3 hours ago
Hellowings
18 minutes ago
Inspired by qrxz17
sqing   9
N an hour ago by sqing
Source: Own
Let $a, b,c>0 ,(a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 27 $. Prove that $$a+b+c\geq 3\sqrt {3}$$
9 replies
sqing
Yesterday at 8:50 AM
sqing
an hour ago
Interesting inequality
sqing   0
an hour ago
Source: Own
Let $  a, b,c>0,b+c\geq 3a$. Prove that
$$ \sqrt{\frac{a}{b+c-a}}-\frac{ 2a^2-b^2-c^2}{(a+b)(a+c)}\geq \frac{2}{5}+\frac{1}{\sqrt 2}$$$$ \frac{3}{2}\sqrt{\frac{a}{b+c-a}}-\frac{ 2a^2-b^2-c^2}{(a+b)(a+c)}\geq \frac{2}{5}+\frac{3}{2\sqrt 2}$$
0 replies
sqing
an hour ago
0 replies
9 Isogonal and isotomic conjugates
V0305   13
N 2 hours ago by ohiorizzler1434
1. Do you think isogonal conjugates should be renamed to angular conjugates?
2. Do you think isotomic conjugates should be renamed to cevian conjugates?

Please answer truthfully :)

Credit to Stead for this renaming idea
13 replies
V0305
May 26, 2025
ohiorizzler1434
2 hours ago
Prove atleast one from a,b,c is 2
Darealzolt   2
N 3 hours ago by sqing
Let \(a,b,c\) be real numbers, such that
\[
a^2+b^2+c^2+abc=5
\]\[
a+b+c=3
\]Prove that atleast one of the numbers \(a,b,c\) is equal to \( 2\).
2 replies
Darealzolt
Yesterday at 11:31 AM
sqing
3 hours ago
Interesting Geometry
captainmath99   4
N Yesterday at 8:01 PM by captainmath99
Let ABC be a right triangle such that $\angle{C}=90^\circ, CA=6, CB=4$. A circle O with center C has a radius of 2. Let P be a point on the circle O.

a)What is the minimum value of $(AP+\dfrac{1}{2}BP)$?
Answer Check

b) What is the minimum value of $(\dfrac{1}{3}AP+BP)$?
Answer Check
4 replies
captainmath99
May 25, 2025
captainmath99
Yesterday at 8:01 PM
Looking for even one person to study math.
abduqahhor_math   2
N Yesterday at 6:25 PM by EaZ_Shadow
Hi guys,I am looking for a person to study math topics related to olympiad.I have just finished 10th grade
2 replies
abduqahhor_math
Yesterday at 5:22 PM
EaZ_Shadow
Yesterday at 6:25 PM
Great Geometry with Squares on sides of triangles
SomeonecoolLovesMaths   3
N Yesterday at 6:14 PM by sunken rock
Three squares are drawn on the sides of triangle \(ABC\) (i.e., the square on \(AB\) has \(AB\) as one of its sides and lies outside \(ABC\)). Show that the lines drawn from the vertices \(A\), \(B\), and \(C\) to the centers of the opposite squares are concurrent.

IMAGE
3 replies
SomeonecoolLovesMaths
May 22, 2025
sunken rock
Yesterday at 6:14 PM
rare creative geo problem spotted in the wild
abbominable_sn0wman   0
Yesterday at 6:04 PM
The following is the construction of the twindragon fractal.

Let $I_0$ be the solid square region with vertices at
\[
(0, 0), \left(\frac{1}{2}, \frac{1}{2}\right), (1, 0), \left(\frac{1}{2}, -\frac{1}{2}\right).
\]
Recursively, the region $I_{n+1}$ consists of two copies of $I_n$: one copy which is rotated $45^\circ$ counterclockwise around the origin and scaled by a factor of $\frac{1}{\sqrt{2}}$, and another copy which is also rotated $45^\circ$ counterclockwise around the origin and scaled by a factor of $\frac{1}{\sqrt{2}}$, and then translated by $\left(\frac{1}{2}, -\frac{1}{2}\right)$.

We have displayed $I_0$ and $I_1$ below.

Let $I_\infty$ be the limiting region of the sequence $I_0, I_1, \dots$.

The area of the smallest convex polygon which encloses $I_\infty$ can be written as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$. Find $a + b$.
0 replies
abbominable_sn0wman
Yesterday at 6:04 PM
0 replies
Inequalities
lgx57   0
Yesterday at 3:55 PM
Let $a,b,c,d,e \ge 0$,$\sum \dfrac{1}{a+4}=1$.Prove that:
$$\sum \dfrac{a}{a^2+4} \le 1$$
Let $x,y,z>0$.Prove that:
$$\sum (y+z)\sqrt{\dfrac{yz}{(z+x)(y+x)}} \ge x+y+z$$
0 replies
lgx57
Yesterday at 3:55 PM
0 replies
Find the number of ordered triples (p,q,r)
Darealzolt   2
N Yesterday at 2:25 PM by elizhang101412
Let \(p,q,r\) be prime numbers such that
\[
\frac{1}{pq}+\frac{1}{qr}+\frac{1}{pr}=\frac{1}{839}
\]Hence find the numbers of ordered triples \(\{p,q,r\}\)
2 replies
Darealzolt
Yesterday at 1:52 PM
elizhang101412
Yesterday at 2:25 PM
Find the sum of all the products a_i a_j
Darealzolt   1
N Yesterday at 2:13 PM by alexheinis
Among the 100 constants \( \{ a_1,a_2,a_3,\dots,a_{100} \} \),there are \(39\) equal to \( -1\), and \(61\) equal to \(1\). Find the sum of all the products \(a_i a_j\) , where \(a \leq i < j \leq 100\).
1 reply
Darealzolt
Yesterday at 11:24 AM
alexheinis
Yesterday at 2:13 PM
IOQM P26 2024
SomeonecoolLovesMaths   5
N Yesterday at 1:11 PM by SomeonecoolLovesMaths
The sum of $\lfloor x \rfloor$ for all real numbers $x$ satisfying the equation $16 + 15x + 15x^2 = \lfloor x \rfloor ^3$ is:
5 replies
SomeonecoolLovesMaths
Sep 8, 2024
SomeonecoolLovesMaths
Yesterday at 1:11 PM
FE solution too simple?
Yiyj1   9
N Apr 23, 2025 by jasperE3
Source: 101 Algebra Problems from the AMSP
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the equality $$f(f(x)+y) = f(x^2-y)+4f(x)y$$holds for all pairs of real numbers $(x,y)$.

My solution

I feel like my solution is too simple. Is there something I did wrong or something I missed?
9 replies
Yiyj1
Apr 9, 2025
jasperE3
Apr 23, 2025
FE solution too simple?
G H J
G H BBookmark kLocked kLocked NReply
Source: 101 Algebra Problems from the AMSP
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Yiyj1
1267 posts
#1
Y by
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the equality $$f(f(x)+y) = f(x^2-y)+4f(x)y$$holds for all pairs of real numbers $(x,y)$.

My solution

I feel like my solution is too simple. Is there something I did wrong or something I missed?
Z K Y
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InterLoop
279 posts
#2 • 1 Y
Y by Yiyj1
You cannot immediately "cancel" the $f$ without further conclusions.

For example $f(3) = f(2) = 1$ is possible for a function - this does not mean that $3 = 2$.
The property that leads you to $f(a) = f(b) \implies a = b$ or the "cancellation" of $f$ is called injectivity. You have to prove the function is injective first before cancellation.

Another example is simply the fact that you have not "excluded" the solution $f(x) \equiv 0$ from the equation $f(f(x)) =f(x^2)$ in any way - so $f(x) = x^2$ is wrong for that function as well. (thus $f(x) \equiv 0$ is not injective)
This post has been edited 2 times. Last edited by InterLoop, Apr 9, 2025, 3:39 AM
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Yiyj1
1267 posts
#3
Y by
InterLoop wrote:
You cannot immediately "cancel" the $f$ without further conclusions.

For example $f(3) = f(2) = 1$ is possible for a function - this does not mean that $3 = 2$.
The property that leads you to $f(a) = f(b) \implies a = b$ or the "cancellation" of $f$ is called injectivity. You have to prove the function is injective first before cancellation.

ahh ic. I'll try to prove the injectivity. ty!
Z K Y
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AshAuktober
1012 posts
#4
Y by
This is in fact from Iran TST.
Z K Y
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davichu
8 posts
#5
Y by
Clearly, $f(x)\equiv0$ is a trivial solution, from now on, we assume it is not the case
Let $P(x,y)$ denote the assertion $f(f(x)+y) = f(x^2-y)+4f(x)y$
$$P(x,-f(x))\rightarrow f(0)=f(x^2+f(x))-4f(x)^2$$$$P(x,x^2)\rightarrow f(x^2+f(x))=f(0)+4f(x)x^2$$Adding these two together we get:
$4f(x)^2=4f(x)x^2\rightarrow f(x)^2=f(x)x^2$
Since $f(x)\neq0$,we can divide by $f(x)$ on both sides to get $f(x)=x^2$
so the only solutions are $f(x)\equiv0$ and $f(x)=x^2\forall x \in \mathbb{R}$
Z K Y
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Primeniyazidayi
117 posts
#6
Y by
davichu wrote:
Since $f(x)\neq0$,we can divide by $f(x)$ on both sides to get $f(x)=x^2$

You must at first prove that $f(x) =0 \text{ iff } x=0$(or simply avoid pointwise trap).
This post has been edited 1 time. Last edited by Primeniyazidayi, Apr 22, 2025, 11:12 AM
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Primeniyazidayi
117 posts
#7
Y by
The finish for @2above(hopefully correct):We will avoid pointwise trap.We of course have $f(0) =0$.Let $f(t) =0$ for $t \neq 0$.$P(t,y)$ gives $f(y) =f(t^2-y)$.Take some $u$ such that $f(u) =u^2 \neq 0$.Then we have $u^2=t^2(t^2-2u) +u^2$ or $u=\frac{t^2}{2}$.But $P(0, x) $ gives that $f$ is even which means $\frac{t^2}{2}=-\frac{t^2}{2}$ or $t=0$, contradiction. Thus we are done.
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ariopro1387
26 posts
#8
Y by
Let $P(x,y)$ be the assertion of the problem.
$P(x,\frac{x^2-f(x)}{2});$ $\frac{x^2-f(x)}{2}.f(x) = 0$
$\forall x \in \mathbb{R}$
1. $f(x)\equiv0$
2. $f(x)=x^2$
we have to just check that both won't happen:
if $f(x_{1}) = 0:$
$P(x_{1},y);$ $f(y) = f(x_{1}^2-y)$
then by changing $y$ value we get that $x_{1} = 0$ or $f(x)\equiv C$ (Just $C=0$ works).
This post has been edited 1 time. Last edited by ariopro1387, Apr 22, 2025, 4:06 PM
Reason: edit
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lksb
182 posts
#9 • 1 Y
Y by Yiyj1
one-liner
This post has been edited 1 time. Last edited by lksb, Apr 22, 2025, 7:15 PM
Reason: typo
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jasperE3
11388 posts
#10
Y by
lksb wrote:
one-liner

pointwise trap
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