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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

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[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Jul 1, 2025
0 replies
Grade 10 Inequality
BomboaneMentos316   1
N 12 minutes ago by aaravdodhia
Prove that for any x,y,z real and positive the following is true:
(PS: It is written correctly)

\begin{align*}
\frac{x^3}{y^2 + y z + z^2}
\;+\;
\frac{y^3}{2 z^2 + y z x}
\;+\;
\frac{z^3}{x^2 + x y + y^2}
\;\ge\;
\frac{x + y + z}{3}.
\end{align*}
1 reply
BomboaneMentos316
3 hours ago
aaravdodhia
12 minutes ago
10 Problems
Sedro   14
N 13 minutes ago by Sedro
Title says most of it. I've been meaning to post a problem set on HSM since at least a few months ago, but since I proposed the most recent problems I made to the 2025 SSMO, I had to wait for that happen. (Hence, most of these problems will probably be familiar if you participated in that contest, though numbers and wording may be changed.) The problems are very roughly arranged by difficulty. Enjoy!

Problem 1: An sequence of positive integers $u_1, u_2, \dots, u_8$ has the property for every positive integer $n\le 8$, its $n^\text{th}$ term is greater than the mean of the first $n-1$ terms and the sum of its first $n$ terms is a multiple of $n$. Let $S$ be the number of such sequences satisfying $u_1+u_2+\cdots + u_8 = 144$. Compute the remainder when $S$ is divided by $1000$.

Problem 2 (solved by fruitmonster97): Rhombus $PQRS$ has side length $3$. Point $X$ lies on segment $PR$ such that line $QX$ is perpendicular to line $PS$. Given that $QX=2$, the area of $PQRS$ can be expressed as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

Problem 3: Positive integers $a$ and $b$ satisfy $a\mid b^2$, $b\mid a^3$, and $a^3b^2 \mid 2025^{36}$. If the number of possible ordered pairs $(a,b)$ is equal to $N$, compute the remainder when $N$ is divided by $1000$.

Problem 4: Let $ABC$ be a triangle. Point $P$ lies on side $BC$, point $Q$ lies on side $AB$, and point $R$ lies on side $AC$ such that $PQ=BQ$, $CR=PR$, and $\angle APB<90^\circ$. Let $H$ be the foot of the altitude from $A$ to $BC$. Given that $BP=3$, $CP=5$, and $[AQPR] = \tfrac{3}{7} \cdot [ABC]$, the value of $BH\cdot CH$ can be expressed as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

Problem 5: Anna has a three-term arithmetic sequence of integers. She divides each term of her sequence by a positive integer $n>1$ and tells Bob that the three resulting remainders are $20$, $52$, and $R$, in some order. For how many values of $R$ is it possible for Bob to uniquely determine $n$?

Problem 6: There is a unique ordered triple of positive reals $(x,y,z)$ satisfying the system of equations \begin{align*} x^2 + 9 &= (y-\sqrt{192})^2 + 4 \\ y^2 + 4 &= (z-\sqrt{192})^2 + 49 \\ z^2 + 49 &= (x-\sqrt{192})^2 + 9. \end{align*}The value of $100x+10y+z$ can be expressed as $p\sqrt{q}$, where $p$ and $q$ are positive integers such that $q$ is square-free. Compute $p+q$.

Problem 7: Let $S$ be the set of all monotonically increasing six-term sequences whose terms are all integers between $0$ and $6$ inclusive. We say a sequence $s=n_1, n_2, \dots, n_6$ in $S$ is symmetric if for every integer $1\le i \le 6$, the number of terms of $s$ that are at least $i$ is $n_{7-i}$. The probability that a randomly chosen element of $S$ is symmetric is $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Compute $p+q$.

Problem 8: For a positive integer $n$, let $r(n)$ denote the value of the binary number obtained by reading the binary representation of $n$ from right to left. Find the smallest positive integer $k$ such that the equation $n+r(n)=2k$ has at least ten positive integer solutions $n$.

Problem 9: Let $p$ be a quadratic polynomial with a positive leading coefficient. There exists a positive real number $r$ such that $r < 1 < \tfrac{5}{2r} < 5$ and $p(p(x)) = x$ for $x \in \{ r,1,  \tfrac{5}{2r} , 5\}$. Compute $p(20)$.

Problem 10: Find the number of ordered triples of positive integers $(a,b,c)$ such that $a+b+c=995$ and $ab+bc+ca$ is a multiple of $995$.
14 replies
Sedro
Jul 10, 2025
Sedro
13 minutes ago
Find the min value of function f(x,y)
PhysicsIsChad   2
N 36 minutes ago by maromex
Let
\[
f(x, y) = \sqrt{x^2 + y^2} + \sqrt{x^2 + y^2 - 2x + 1} + \sqrt{x^2 + y^2 - 2y + 1} + \sqrt{x^2 + y^2 - 6x - 8y + 25}
\]for all real \( x, y \in \mathbb{R} \).
Find the min value of this function . What are your insights on this problem ?
My insights :
All these quadratics under the square root can be converted into \[
f(x, y) = \sqrt{(x - 0)^2 + (y - 0)^2} 
+ \sqrt{(x - 1)^2 + (y - 0)^2} 
+ \sqrt{(x - 0)^2 + (y - 1)^2} 
+ \sqrt{(x - 3)^2 + (y - 4)^2}
\]This function can be interpreted as the sum of the distances of points (0,0) , (1,0) , (0,1) , (3,4) from (x,y) .
We can prove that for minimising this distance (x,y) must be the mean of the x and y coordinates of the points respectively . Thus we get (x,y) = (1,1.25) . \[
\begin{aligned}
f(x, y) &= \sqrt{(x - 0)^2 + (y - 0)^2} 
+ \sqrt{(x - 1)^2 + (y - 0)^2} \\
&\quad + \sqrt{(x - 0)^2 + (y - 1)^2} 
+ \sqrt{(x - 3)^2 + (y - 4)^2} \\
\\
f(1, 1.25) &= \sqrt{1^2 + 1.25^2} 
+ \sqrt{(1 - 1)^2 + 1.25^2} \\
&\quad + \sqrt{1^2 + (1.25 - 1)^2} 
+ \sqrt{(1 - 3)^2 + (1.25 - 4)^2} \\
\\
&= \sqrt{2.5625} + \sqrt{1.5625} + \sqrt{1.0625} + \sqrt{11.5625} \\
\\
&\approx 1.6 + 1.25 + 1.03078 + 3.40184 \\
\\
&= \boxed{7.2826}
\end{aligned}
\]But apparently this is wrong . I will post the proof for the intermediate i used in the replies .





2 replies
PhysicsIsChad
5 hours ago
maromex
36 minutes ago
Number Theory (Divisibility)
AbdulWaheed   2
N an hour ago by NamelyOrange
Find all triples (a, b, c) of natural numbers such that the numbers $a^2$ + bc, $b^2$ + ac,
and $c^2$ + ab are powers of 2. (source, Sozopol 2023, Grades 8-9)
2 replies
AbdulWaheed
Today at 7:14 AM
NamelyOrange
an hour ago
Some Calculating Rules to handle Exponents and their bases
Klaus-Anton   0
Dec 8, 2023
In Exponentialfunktion (german wiki) there are given some calculating rules, which i want to note. They say this makes it possible to "convert" multiplikation into addition (wiki/Exponentialfunktion#Rechenregeln).

IMAGE
[center]Table: Some Calculating Rules to handle Exponents and their bases[/center]

They there give further on the link to Rechenregeln für Logarithmus (wiki). And there in they send you to Potenz_(Mathematik)#Potenzgesetze (wiki). The main wiki article of this is wiki/Potenz_(Mathematik). And from that german article the branching to the english version is wiki/Exponentiation.
0 replies
Klaus-Anton
Dec 8, 2023
0 replies
No more topics!
Recursion
Sid-darth-vater   6
N Apr 20, 2025 by vanstraelen
Help, I can't characterize ts and I dunno what to do
6 replies
Sid-darth-vater
Apr 20, 2025
vanstraelen
Apr 20, 2025
Recursion
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Sid-darth-vater
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Help, I can't characterize ts and I dunno what to do
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aidan0626
2069 posts
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well by testing small cases i got the general form to be Click to reveal hidden text, which can probably be proven by induction
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Sid-darth-vater
67 posts
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Ahh yeah, it works out. tyty. also, can you kinda explain how you thought of the numerator part? like I had figured out $a_2 = \frac{7}{24}, a_3 = \frac{13}{96},$ and $a_4 = \frac{25}{384}$ but I couldn't find the $3 \cdot 2^{n-1} + 1$ portion.
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vanstraelen
9164 posts
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$a_{1}=\frac{4}{6},a_{2}=\frac{7}{24},a_{3}=\frac{13}{96},a_{4}=\frac{25}{384},a_{5}=\frac{49}{1536},a_{6}=\frac{97}{6144},\cdots$

$a_{n}=\frac{3 \cdot 2^{n-1} +1}{3 \cdot 2^{2n-1}}$.
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aidan0626
2069 posts
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Sid-darth-vater wrote:
Ahh yeah, it works out. tyty. also, can you kinda explain how you thought of the numerator part? like I had figured out $a_2 = \frac{7}{24}, a_3 = \frac{13}{96},$ and $a_4 = \frac{25}{384}$ but I couldn't find the $3 \cdot 2^{n-1} + 1$ portion.
well I noticed that the differences between consecutive numerators was 3, 6, 12, etc.
and so I got a geometric series which resulted in that
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Sid-darth-vater
67 posts
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mmmm i see, thanks!
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vanstraelen
9164 posts
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$a_{k}=\frac{3 \cdot 2^{k-1} +1}{3 \cdot 2^{2k-1}}=2^{-k}+\frac{1}{3} \cdot 2^{1-2k}$.

$S=\lim_{n \to \infty} \sum_{k=1}^{n}a_{k}=\lim_{n \to \infty} \sum_{k=1}^{n}\left[2^{-k}+\frac{1}{3} \cdot 2^{1-2k}\right]$,
$S=\lim_{n \to \infty} \left[1-2^{-n}+\frac{1}{3}(\frac{2}{3}-\frac{2^{1-2n}}{2})\right]=1+\frac{2}{9}=\frac{11}{9}$.
This post has been edited 1 time. Last edited by vanstraelen, Apr 20, 2025, 6:00 PM
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