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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)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]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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Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.

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0 replies
jwelsh
Jul 1, 2025
0 replies
Inequalitis
sqing   12
N 2 minutes ago by sqing
Let $ a,b,c\geq  0 , a^2+b^2+c^2 =3.$ Prove that
$$a^3 +b^3 +c^3 +\frac{11}{5}abc  \leq \frac{26}{5}$$
12 replies
1 viewing
sqing
May 31, 2025
sqing
2 minutes ago
[PMO26 Qualifying III.3] Another Quadrilateral Unlocked
kae_3   1
N 7 minutes ago by BinariouslyRandom
Let $ABCD$ be a square and $P$ be a point on segment $AB$. Segments $CP$ and $BD$ intersect at $Q$, and $BD$ is extended beyond $B$ to a point $S$. Finally, $T$ is the intersection of line $CP$ and the line through $S$ parallel to $AB$. If $AB=18$ and $PT=2CQ$, what is the area of quadrilateral $PBST$?

Answer Confirmation
1 reply
kae_3
Feb 23, 2025
BinariouslyRandom
7 minutes ago
[PMO27 Areas] I.3 Locus around a rectangle
aops-g5-gethsemanea2   2
N 24 minutes ago by BinariouslyRandom
Let $a$ and $b$ be positive integers. Let $A$ be the region that consists of all points on a $4\times15$ rectangle and its interior, and also the points in the exterior which are at most $a$ units away from the boundary of this rectangle. Similarly, let $B$ be the region that consists of all points on a $3\times20$ rectangle and its interior, and also the points in the exterior which are at most $b$ units away from the boundary of this rectangle. Suppose that the area of $B$ is $44\%$ larger than the area of $A$. What is $a+b$?
2 replies
aops-g5-gethsemanea2
Jan 25, 2025
BinariouslyRandom
24 minutes ago
Inequality from CTPCM
Gaussian0000   4
N 27 minutes ago by Not__Infinity
For a fixed positive integer $n$ compute the minimum value of the sum:
$x_1 + \frac{(x_2)^2}{2} + \frac{(x_3)^3}{3} +...+\frac{(x_n)^n}{n}$ given that
$\frac{1}{x_1} + \frac{1}{x_2}+ \frac{1}{x_3}+...+ \frac{1}{x_n}= n$
4 replies
Gaussian0000
Jul 23, 2025
Not__Infinity
27 minutes ago
[PMO24 Qualifying I.11] Odd Divisor Sum
kae_3   1
N 39 minutes ago by BinariouslyRandom
How many positive integers $n<2022$ are there for which the sum of the odd positive divisors of $n$ is $24$?

$\text{(a) }7\qquad\text{(b) }8\qquad\text{(c) }14\qquad\text{(d) }15$

Answer Confirmation
1 reply
kae_3
Feb 16, 2025
BinariouslyRandom
39 minutes ago
Complex numbers
preatsreard   6
N 4 hours ago by TedBot
How many complex z exist, such that z,z²,z³,....,z²⁰²¹ form a perfect 2021-gon (in arbitary order) in a complex plane.
6 replies
preatsreard
Jul 20, 2025
TedBot
4 hours ago
Quadratic equations
Streit31415   7
N 5 hours ago by Ox.pi
Let (p) and (q) be integers. Knowing that x² + px + q is positive for all integer (x), prove that the equation x² + px + q = 0 has no real solution.
7 replies
Streit31415
Yesterday at 5:45 PM
Ox.pi
5 hours ago
My first proof problem
OWOW   10
N Yesterday at 7:52 PM by TedBot
In the equation $\sum_{i=0}^n i$ prove (or disprove) that there exists infinitely many positive integers n in which $\sum_{i=0}^n i$ sums to k! where k is a positive integer. (examples, 1+2+3=3! and 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15=5!

Also this is my first proof problem so don't get mad at me if it's really bad. (Technically I'm a middle schooler so I should post this in the middle school threads but I'm thinking proofs are more high school level so I'm posting it here.)
10 replies
OWOW
Friday at 11:44 PM
TedBot
Yesterday at 7:52 PM
Angle bisector, medians
xeroxia   0
Yesterday at 6:58 PM
Let $AM$ be a median and $AN$ be an angle bisector of $\triangle ABC$. Let $BK$ be a median and $BH$ be an altitude of $\triangle ABN$. If $AN=6$ and $MK=4$, what is $MH$?
0 replies
xeroxia
Yesterday at 6:58 PM
0 replies
Trigonometry equation practice
ehz2701   26
N Yesterday at 6:34 PM by vanstraelen
There is a lack of trigonometric bash practice, and I want to see techniques to do these problems. So here are 10 kinds of problems that are usually out in the wild. How do you tackle these problems? (I had ChatGPT write a code for this.). Please give me some general techniques to solve these kinds of problems, especially set 2b. I’ll add more later.

Leaderboard and Solved Problems

problem set 1a (1-10)

problem set 2a (1-20)

problem set 2b (1-20)
answers 2b

General techniques so far:

Trick 1: one thing to keep in mind is

[center] $\frac{1}{2} = \cos 36 - \sin 18$. [/center]

Many of these seem to be reducible to this. The half can be written as $\cos 60 = \sin 30$, and $\cos 36 = \sin 54$, $\sin 18 = \cos 72$. This is proven in solution 1a-1. We will refer to this as Trick 1.
26 replies
ehz2701
Jul 12, 2025
vanstraelen
Yesterday at 6:34 PM
Find the value of angle C
markosa   12
N Yesterday at 6:29 PM by sunken rock
Given a triangle ABC with base BC

angle B = 3x
angle C = x
AP is the bisector of base BC (i.e.) BP = PC
angle APB = 45 degrees

Find x

I know there are multiple methods to solve this problem using cosine law, coord geo
But is there any pure geometrical solution?
12 replies
markosa
Friday at 12:45 PM
sunken rock
Yesterday at 6:29 PM
10 Problems
Sedro   57
N Yesterday at 3:25 PM 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 (solved by Math-lover1): 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 (solved by CubeAlgo15): 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 (solved by maromex): 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 (solved by Mathsll-enjoy): 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 (solved by sami1618): 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 (solved by Math-lover1, sami1618): 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 (solved by aaravdodhia, sami1618): 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 $995$.
57 replies
Sedro
Jul 10, 2025
Sedro
Yesterday at 3:25 PM
Trigonometry prove
Studying_geometry   2
N Yesterday at 2:39 PM by CuriousMathBoy72
Prove that $ sin36^\circ - cos18^\circ = \frac{1}{2} $
2 replies
Studying_geometry
Yesterday at 2:18 PM
CuriousMathBoy72
Yesterday at 2:39 PM
PROVINCIAL MATHEMATICS 9 MATH QUESTIONS FOR REFERENCE
tuananh_vvvbb   0
Yesterday at 1:33 PM
Hello friends, I would like to share with you a reference to an HSG question for me to get a score of 70% or more, which is quite difficult. Readers, please refer to me for an explanation. Thank you all very much. Good health.
0 replies
tuananh_vvvbb
Yesterday at 1:33 PM
0 replies
Minimum number of points
Ecrin_eren   8
N Jun 16, 2025 by HAL9000sk
There are 18 teams in a football league. Each team plays against every other team twice in a season—once at home and once away. A win gives 3 points, a draw gives 1 point, and a loss gives 0 points. One team became the champion by earning more points than every other team. What is the minimum number of points this team could have?

8 replies
Ecrin_eren
May 15, 2025
HAL9000sk
Jun 16, 2025
Minimum number of points
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Ecrin_eren
122 posts
#1 • 1 Y
Y by PikaPika999
There are 18 teams in a football league. Each team plays against every other team twice in a season—once at home and once away. A win gives 3 points, a draw gives 1 point, and a loss gives 0 points. One team became the champion by earning more points than every other team. What is the minimum number of points this team could have?
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rachelcassano
17 posts
#2 • 1 Y
Y by PikaPika999
The minimum number the team could have is 103 points.
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Shan3t
548 posts
#3 • 1 Y
Y by PikaPika999
this wrong @1below is right, bc of ties, except 1
This post has been edited 1 time. Last edited by Shan3t, May 16, 2025, 5:58 PM
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Ecrin_eren
122 posts
#4
Y by
Answer is 36
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aidan0626
2083 posts
#5
Y by
Ecrin_eren wrote:
Answer is 36

achieved by every game being a draw except for one
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Shan3t
548 posts
#6
Y by
aidan0626 wrote:
Ecrin_eren wrote:
Answer is 36

achieved by every game being a draw except for one

ohh shoot i forgot abt ties :skull:
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Ecrin_eren
122 posts
#7
Y by
Yes draws and win works but why it cant be less
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Ecrin_eren
122 posts
#8
Y by
Bumpingg
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HAL9000sk
113 posts
#9 • 1 Y
Y by aidan0626
The minimum point sum over all 18 teams is $\binom{18}{2}\cdot 2\cdot 2 = 612$, if all games are draw. Then every team has 34 points. But the champion in your scenario earns more than every other team, so there is at least one win game in the championship and therefore $\geq 613$ points overall sum. (*)

Is it possible, that the champion has exactly 35 points? Then all other seventeen teams must have $\leq 34$ points, in combination with (*) that means exactly 34 points and exactly one win in the championship. The latter fact contradicts the 35 points of the champion: One win and 33 draws means 36 points!

And 36 points for the champion are possible: Sixteen teams have 34 points and one team (the loser of the only non-draw game) has 33 points.


A champion with 35 points is only possible, if at least two other teams have also 35 points:

Each of the three teams wins and loses one game against the other two. All other games draw, as above.
This post has been edited 2 times. Last edited by HAL9000sk, Jun 16, 2025, 6:12 PM
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