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k a June Highlights and 2025 AoPS Online Class Information
jlacosta   0
3 hours ago
Congratulations to all the mathletes who competed at National MATHCOUNTS! If you missed the exciting Countdown Round, you can watch the video at this link. Are you interested in training for MATHCOUNTS or AMC 10 contests? How would you like to train for these math competitions in half the time? We have accelerated sections which meet twice per week instead of once starting on July 8th (7:30pm ET). These sections fill quickly so enroll today!

[list][*]MATHCOUNTS/AMC 8 Basics
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0 replies
+1 w
jlacosta
3 hours ago
0 replies
smallest a so that S(n)-S(n+a) = 2018, where S(n)=sum of digits
parmenides51   3
N 12 minutes ago by TheBaiano
Source: Lusophon 2018 CPLP P3
For each positive integer $n$, let $S(n)$ be the sum of the digits of $n$. Determines the smallest positive integer $a$ such that there are infinite positive integers $n$ for which you have $S (n) -S (n + a) = 2018$.
3 replies
parmenides51
Sep 13, 2018
TheBaiano
12 minutes ago
Ducks can play games now apparently
MortemEtInteritum   35
N an hour ago by pi271828
Source: USA TST(ST) 2020 #1
Let $a$, $b$, $c$ be fixed positive integers. There are $a+b+c$ ducks sitting in a
circle, one behind the other. Each duck picks either rock, paper, or scissors, with $a$ ducks
picking rock, $b$ ducks picking paper, and $c$ ducks picking scissors.
A move consists of an operation of one of the following three forms:

[list]
[*] If a duck picking rock sits behind a duck picking scissors, they switch places.
[*] If a duck picking paper sits behind a duck picking rock, they switch places.
[*] If a duck picking scissors sits behind a duck picking paper, they switch places.
[/list]
Determine, in terms of $a$, $b$, and $c$, the maximum number of moves which could take
place, over all possible initial configurations.
35 replies
1 viewing
MortemEtInteritum
Nov 16, 2020
pi271828
an hour ago
2017 IGO Advanced P3
bgn   18
N an hour ago by Circumcircle
Source: 4th Iranian Geometry Olympiad (Advanced) P3
Let $O$ be the circumcenter of triangle $ABC$. Line $CO$ intersects the altitude from $A$ at point $K$. Let $P,M$ be the midpoints of $AK$, $AC$ respectively. If $PO$ intersects $BC$ at $Y$, and the circumcircle of triangle $BCM$ meets $AB$ at $X$, prove that $BXOY$ is cyclic.

Proposed by Ali Daeinabi - Hamid Pardazi
18 replies
bgn
Sep 15, 2017
Circumcircle
an hour ago
Own made functional equation
JARP091   1
N 2 hours ago by JARP091
Source: Own (Maybe?)
\[
\text{Find all functions } f : \mathbb{R} \to \mathbb{R} \text{ such that:} \\
f(a^4 + a^2b^2 + b^4) = f\left((a^2 - f(ab) + b^2)(a^2 + f(ab) + b^2)\right)
\]
1 reply
JARP091
May 31, 2025
JARP091
2 hours ago
Floor of Cube Root
Magdalo   1
N 3 hours ago by Magdalo
Find the amount of natural numbers $n<1000$ such that $\lfloor \sqrt[3]{n}\rfloor\mid n$.
1 reply
Magdalo
3 hours ago
Magdalo
3 hours ago
Approximating nonlinear recurrence
KSH31415   0
3 hours ago
I was working on finding the general solution to this problem with $n$ red and $n$ blue balls. Its pretty simple to get the recurrence
$$E_n=1+\frac{n}{2n-1}E_{n-1}$$where $E_n=1$. I tried to find an explicit form for $E_n$ but none of my attempts worked and the fractions looked increasingly complex. I don't think finding an explicit form is possible (I would love to be disproven), so I decided to look at the relation asymptotically.

First $\lim_{n\rightarrow \infty}E_n=2$. This is pretty clear since $\frac{n}{2n-1}$ tends to $\frac 12$, in which case $E_n=\frac 12$ is a fixed point. This is also obvious when looking at the original problem, since the probability of drawing a match gets closer to $\frac 12$ for large $n$, so the expected number of draws gets closer to $\frac{1}{\frac 12}=2$. Graphing the sequence and subtracting $2$, I found that $E_n-2\approx \frac 1n$. I haven't, however, been able to match $E_n-2-\frac 1n$ to any function.

Here is a link to the Desmos graph if you want to play around.

This exploration leaves me with a some questions about the recurrence. Does and explicit form exist? How far can we keep approximating $E_n$? Could we write $E_n$ as some infinite sum?
0 replies
KSH31415
3 hours ago
0 replies
[PMO21 Areas] I.19
Magdalo   1
N 3 hours ago by Magdalo
How many distinct numbers are there from $\left\lfloor\dfrac{1^2}{2018}\right\rfloor,\left\lfloor\dfrac{2^2}{2018}\right\rfloor,\dots,\left\lfloor\dfrac{2018^2}{2018}\right\rfloor$?
1 reply
Magdalo
3 hours ago
Magdalo
3 hours ago
[PMO22 Qualifying] II.19
Magdalo   1
N 3 hours ago by Magdalo
For a real number $t$, $\lfloor t\rfloor$ is the greatest integer less than or equal to $t$. How many natural numbers $n$ are there such that $\left\lfloor\dfrac{n^3}{9}\right\rfloor$ is prime?
1 reply
Magdalo
3 hours ago
Magdalo
3 hours ago
Inequalities
sqing   8
N 4 hours ago by cj13609517288
Let $ a,b> 0 ,\frac{a}{2b+1}+\frac{b}{3}+\frac{1}{2a+1} \leq 1.$ Prove that
$$  a^2+b^2 -ab\leq 1$$$$ a^2+b^2 +ab \leq3$$Let $ a,b,c> 0 , \frac{a}{2b+1}+\frac{b}{2c+1}+\frac{c}{2a+1} \leq 1.$ Prove that
$$    a +b +c +abc \leq 4$$
8 replies
sqing
May 24, 2025
cj13609517288
4 hours ago
Sipnayan 2025 SHS Orals Final Round VD-FriedChicken
qrxz17   0
4 hours ago
Problem: Let \( a, b, c \) be complex numbers. For a complex number \( z = p + qi \) where \( i = \sqrt{-1} \), define the norm \( |z| \) to be the distance of \( z \) from the origin, or \( |z| = \sqrt{p^2 + q^2} \). Let \( m \) be the minimum value and \( M \) be the maximum value of
\[
\frac{ |a + b| + |b + c| + |c + a| }{ |a| + |b| + |c| }
\]for all complex numbers \( a, b, c \) where \( |a| + |b| + |c| \ne 0 \). Find \( M + m \).

Answer: Click to reveal hidden text

Solution: Let us first get the maximum value \(M\). When two complex numbers \(a\) and \(b\) are added, their sum \(a + b\) is also a vector. Geometrically, this is represented by placing the tail of \(b\) at the head of \(a\) or vice versa, following the parallelogram rule. The vector \(a + b\) then extends from the origin to the head of the translated \(b\), forming a triangle with the origin and the head of \(a\).

Then, we can apply the triangle inequality. We have
\begin{align*}
        |a+b| & \leq |a|+|b| \\
        |b+c| & \leq |b|+|c| \\
        |c+a| & \leq |c| + |a| 
    \end{align*}
The given equation can then be rewritten to
\begin{align*}
        \frac{ |a + b| + |b + c| + |c + a| }{ |a| + |b| + |c| } \leq \frac{ |a| + |b| + |b| + |c| + |c| + |a| }{ |a| + |b| + |c| } \leq \frac{ 2(|a| + |b| + |c |) }{ |a| + |b| + |c| } \leq 2
    \end{align*}
Thus, the maximum value M is equal to 2.

Now, let us determine the minimum value \(m\). To find the minimum value, we can look for configurations where ``cancellation" in the vector sums is maximized. This typically happens when the complex numbers are collinear but point in opposite directions. To do this, let \(a\) and \(b\) be at the same point, and let \(c\) be the same distance from the origin as \(a\) and \(b\), but in the opposite direction.

We have
\begin{align*}
        a =\text{ } &b \text{ } =-c \\
        |a| =|&b|=|c|.
    \end{align*}Substituting this in the given equation, we get

\begin{align*}
        \frac{ |a + b| + |b + c| + |c + a| }{ |a| + |b| + |c| } = \frac{ |a+a| + |a + -a| + |-a + a| }{ |a| + |a| + |a| } =\frac{ |2a| }{ 3|a| }=\frac{2}{3}.
    \end{align*}
Thus, the minimum value \(m\) is equal to \(\frac{2}{3}\).

Therefore, \(M+m=2+\frac{2}{3}=\boxed{\frac{8}{3}}\).
0 replies
qrxz17
4 hours ago
0 replies
MATHirang MATHibay 2012 Eliminations Average A1
qrxz17   0
4 hours ago
Problem. Determine the sum of all real and complex solutions to the equation
\[
x^2 + 2|x| - 6x + 15 = 0.
\](Note: the modulus of a complex number \( x = a + bi \) is \( |x| = \sqrt{a^2 + b^2} \).)
Answer: Click to reveal hidden text
Solution: Substituting
\begin{align*}
        x = a + bi \text{ and } |x| = \sqrt{a^2 + b^2}
    \end{align*}
into the equation, we get
\begin{align*}
        (a^2 - b^2 + 2\sqrt{a^2 + b^2} - 6a + 15) + i(2ab - 6b) = 0
    \end{align*}
For this equation to hold,
\begin{align*}
    a^2 - b^2 + 2\sqrt{a^2 + b^2} - 6a + 15 &= 0 \text{  } \text{ and}\\
    2ab - 6b &=0.
    \end{align*}
Solving this system of equations, we get \(a= 3\) and \(b=\pm 4\).

Thus, we have the solutions \(x = 3+4i\) and \(x = 3-4i\).

Summing these solutions, we get \(\boxed{6}\).
0 replies
qrxz17
4 hours ago
0 replies
Inequalities
toanrathay   0
4 hours ago
Prove that this inequality holds for all positive reals $a,b,c$ \[
\frac{ab + bc + ca}{a^2 + b^2 + c^2} + \frac{1}{6} \left( \frac{(a - b)^2}{a^2 + b^2} + \frac{(b - c)^2}{b^2 + c^2} + \frac{(c - a)^2}{c^2 + a^2} \right) \leq 1.
\]
0 replies
toanrathay
4 hours ago
0 replies
[Sipnayan 2021 SHS SF-E2] Sum of 256th powers
aops-g5-gethsemanea2   1
N 6 hours ago by aops-g5-gethsemanea2
If $r_1,r_2,\dots,r_{256}$ are the 256 roots (not necessarily distinct) of the equation $x^{256}-2021x^2-3=0$, evaluate $\sum^{256}_{i=1}r_i^{256}$.
1 reply
aops-g5-gethsemanea2
6 hours ago
aops-g5-gethsemanea2
6 hours ago
Original Problem
wonderboy807   3
N Today at 1:08 PM by reyaansh_agrawal
A non-constant polynomial function f : \mathbb{R} \to \mathbb{R} satisfies f(f(x)) = f(3x) + f(x) + 3. Also, f(0) = 1. Find f(2025).

Answer: Click to reveal hidden text

Solution: Click to reveal hidden text
3 replies
wonderboy807
Today at 1:02 AM
reyaansh_agrawal
Today at 1:08 PM
Funny easy transcendental geo
qwerty123456asdfgzxcvb   1
N Apr 21, 2025 by golue3120
Let $\mathcal{S}$ be a logarithmic spiral centered at the origin (ie curve satisfying for any point $X$ on it, line $OX$ makes a fixed angle with the tangent to $\mathcal{S}$ at $X$). Let $\mathcal{H}$ be a rectangular hyperbola centered at the origin, scaled such that it is tangent to the logarithmic spiral at some point.

Prove that for a point $P$ on the spiral, the polar of $P$ wrt. $\mathcal{H}$ is tangent to the spiral.
1 reply
qwerty123456asdfgzxcvb
Apr 21, 2025
golue3120
Apr 21, 2025
Funny easy transcendental geo
G H J
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qwerty123456asdfgzxcvb
1088 posts
#1 • 8 Y
Y by CyclicISLscelesTrapezoid, OronSH, pineapply, kiyoras_2001, golue3120, KevinYang2.71, EpicBird08, Ciobi_
Let $\mathcal{S}$ be a logarithmic spiral centered at the origin (ie curve satisfying for any point $X$ on it, line $OX$ makes a fixed angle with the tangent to $\mathcal{S}$ at $X$). Let $\mathcal{H}$ be a rectangular hyperbola centered at the origin, scaled such that it is tangent to the logarithmic spiral at some point.

Prove that for a point $P$ on the spiral, the polar of $P$ wrt. $\mathcal{H}$ is tangent to the spiral.
This post has been edited 3 times. Last edited by qwerty123456asdfgzxcvb, Apr 21, 2025, 7:33 PM
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golue3120
58 posts
#2 • 1 Y
Y by qwerty123456asdfgzxcvb
Let $\mathfrak I$ be inverflection centered at $O$ fixing $P$, and let $\mathfrak p$ be polarity with respect to $\mathcal H$. Then $\mathcal S$ is preserved by $\mathfrak I$. Let $P_0$ be the tangency point of $\mathcal H$ and $\mathcal S$. Because $\mathfrak I$ is an isoconjugation in $\triangle O\infty_{+i}\infty_{-i}$ fixing $P$ and $\mathcal H$ is a diagonal conic through $P$ in $\triangle O\infty_{+i}\infty_{-i}$, the points $P$ and $\mathfrak I(P)$ are always conjugate in $\mathcal H$. Now we claim the polar of a point $P\in\mathcal S$ is the tangent at $\mathcal S$ to $\mathfrak I(P)$. If $T_{\mathfrak I(P)}$ is the tangent, then $\arg O\mathfrak I(P)-\arg T_{\mathfrak I(P)}$ is constant, $\arg O\mathfrak I(P)+\arg OP$ is constant, and $\arg OP+\arg \mathfrak p(P)$ is constant, so the angle between $T_{\mathfrak I(P)}$ and $\mathfrak p(P)$ is constant. This angle is $0$ when $P=P_0$, and hence $T_{\mathfrak I(P)}$ and $\mathfrak p(P)$ are parallel. They both pass through $\mathfrak I(P)$, and hence coincide.
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