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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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0 replies
jwelsh
Jul 1, 2025
0 replies
Easy Combinatorial Geometry
EthanWYX2009   1
N a minute ago by Cats_on_a_computer
Source: 2025 February 谜之竞赛-4
Given integer $n\ge 2$. Define set
\[V_n:=\{(a_1,\cdots ,a_n)\in\mathbb R^n\mid a_1,\cdots ,a_n\ge 0\text{ and }a_1+\cdots +a_n=1\}.\]For $\alpha =(a_1,\cdots ,a_n)$, $\beta =(b_1,\cdots ,b_n)\in V_n$, define $d(\alpha ,\beta):=\sum\limits_{i=1}^n|a_i-b_i|.$

Determine the minimum real number $\lambda$, such that for all $v_1,v_2,\cdots ,v_{n+2}\in V_n$, there exists $1\le i<j\le n+2$ such that $d(x_i,x_j)\le\lambda$.

Created by Cheng Jiang
1 reply
EthanWYX2009
2 hours ago
Cats_on_a_computer
a minute ago
Abusing surjectivity
Sadigly   8
N 25 minutes ago by alfonsoramires
Find all functions $f:\mathbb{Q}\rightarrow\mathbb{Q}$ and $g:\mathbb{Q}\rightarrow\mathbb{Q}$ such that

$$f(f(x)+yg(x))=(x+1)g(y)+f(y)$$
for any $x;y\in\mathbb{Q}$
8 replies
Sadigly
Apr 13, 2025
alfonsoramires
25 minutes ago
Problem 4 of the HMO
GreekIdiot   15
N 30 minutes ago by Cats_on_a_computer
Prove that no perfect cube is of the form $y^2+108$ where $y \in \mathbb{Z}$.
15 replies
GreekIdiot
Feb 22, 2025
Cats_on_a_computer
30 minutes ago
Every popular person is the best friend of a popular person?
yunxiu   9
N 36 minutes ago by AshAuktober
Source: 2012 European Girls’ Mathematical Olympiad P6
There are infinitely many people registered on the social network Mugbook. Some pairs of (different) users are registered as friends, but each person has only finitely many friends. Every user has at least one friend. (Friendship is symmetric; that is, if $A$ is a friend of $B$, then $B$ is a friend of $A$.)
Each person is required to designate one of their friends as their best friend. If $A$ designates $B$ as her best friend, then (unfortunately) it does not follow that $B$ necessarily designates $A$ as her best friend. Someone designated as a best friend is called a $1$-best friend. More generally, if $n> 1$ is a positive integer, then a user is an $n$-best friend provided that they have been designated the best friend of someone who is an $(n-1)$-best friend. Someone who is a $k$-best friend for every positive integer $k$ is called popular.
(a) Prove that every popular person is the best friend of a popular person.
(b) Show that if people can have infinitely many friends, then it is possible that a popular person is not the best friend of a popular person.

Romania (Dan Schwarz)
9 replies
yunxiu
Apr 13, 2012
AshAuktober
36 minutes ago
Digits problem
menseggerofgod   0
5 hours ago
Jean came up with a positive integer that is divisible by 13, whose digits are not zero and distinct two by two. He noticed that in this number two digits can be interchanged so that the result is also divisible by 13. What is the smallest number of digits that Jean's number could have?
0 replies
menseggerofgod
5 hours ago
0 replies
[PMO27 Areas] I.8 Radical equations
aops-g5-gethsemanea2   9
N 5 hours ago by Siopao_Enjoyer
Positive real numbers $x$ and $y$ satisfy $\sqrt x+\sqrt y=4$ and $\sqrt{x+2}+\sqrt{y+2}=5$. If $x+y=m/n$ where $m$ and $n$ are relatively prime positive integers, what is $m+n$?
9 replies
aops-g5-gethsemanea2
Jan 25, 2025
Siopao_Enjoyer
5 hours ago
Trigonometry equation practice
ehz2701   6
N Yesterday at 7:56 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

problem set 1a

problem set 2a

problem set 2b
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.
6 replies
ehz2701
Saturday at 8:48 AM
vanstraelen
Yesterday at 7:56 PM
Easy geometry problem
menseggerofgod   5
N Yesterday at 6:27 PM by ehz2701
ABC is a right triangle, right at B, in which the height BD is drawn. E is a point on side BC such that AE = EC = 8. If BD is 6 and DE = k , find k
5 replies
menseggerofgod
Yesterday at 2:47 AM
ehz2701
Yesterday at 6:27 PM
Geometry easy
AlexCenteno2007   4
N Yesterday at 5:49 PM by AlexCenteno2007
In triangle ABC, if angle B=120°, AB=5u and BC=15u. Draw the interior bisector BE. Calculate BE
4 replies
AlexCenteno2007
Jul 11, 2025
AlexCenteno2007
Yesterday at 5:49 PM
Select 3 frm {1,2,..,4n}, 4 divides their sum
Sayan   12
N Yesterday at 4:14 PM by ParthivCalculus
Find the number of ways in which three numbers can be selected from the set $\{1,2,\cdots ,4n\}$, such that the sum of the three selected numbers is divisible by $4$.
12 replies
Sayan
May 9, 2012
ParthivCalculus
Yesterday at 4:14 PM
[JBMO 2013/3]
arcticfox009   2
N Yesterday at 3:39 PM by DAVROS
Show that

\[ \left( a + 2b + \frac{2}{a + 1} \right) \left( b + 2a + \frac{2}{b + 1} \right) \geq 16 \]
for all positive real numbers $a$ and $b$ such that $ab \geq 1$.
2 replies
arcticfox009
Jul 11, 2025
DAVROS
Yesterday at 3:39 PM
10 Problems
Sedro   8
N Yesterday at 3:34 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 increasing sequence of positive integers $u_1, u_2, \dots, u_8$ has the property that the sum of its first $n$ terms is divisible by $n$ for every positive integer $n\le 8$. 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: 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 $995$.
8 replies
Sedro
Jul 10, 2025
Sedro
Yesterday at 3:34 PM
angle chasing question
mahi.314   6
N Yesterday at 3:29 PM by sunken rock
Hi! I'm not comfortable with latex yet so bear with me please.
Q. in triABC, BD and CE are the bisectors of angles B,C cutting CA, AB at D,E respectively. if angle BDE= 24deg and angle CED= 18deg, find the angles of triABC.
I did find out angle A which comes out to be Click to reveal hidden text
but i'm stuck on the other two. help would be appreciated.
thanks!
6 replies
mahi.314
Jul 10, 2025
sunken rock
Yesterday at 3:29 PM
AM-GM Problem
arcticfox009   13
N Yesterday at 3:06 PM by nudinhtien
Let $x, y$ be positive real numbers such that $xy \geq 1$. Find the minimum value of the expression

\[ \frac{(x^2 + y)(x + y^2)}{x + y}. \]
answer confirmation
13 replies
arcticfox009
Jul 11, 2025
nudinhtien
Yesterday at 3:06 PM
Polynomial approximation and intersections
egxa   2
N Apr 29, 2025 by iliya8788
Source: All Russian 2025 10.6
What is the smallest value of \( k \) such that for any polynomial \( f(x) \) of degree $100$ with real coefficients, there exists a polynomial \( g(x) \) of degree at most \( k \) with real coefficients such that the graphs of \( y = f(x) \) and \( y = g(x) \) intersect at exactly $100$ points?
2 replies
egxa
Apr 18, 2025
iliya8788
Apr 29, 2025
Polynomial approximation and intersections
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Source: All Russian 2025 10.6
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egxa
215 posts
#1
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What is the smallest value of \( k \) such that for any polynomial \( f(x) \) of degree $100$ with real coefficients, there exists a polynomial \( g(x) \) of degree at most \( k \) with real coefficients such that the graphs of \( y = f(x) \) and \( y = g(x) \) intersect at exactly $100$ points?
This post has been edited 1 time. Last edited by egxa, Apr 18, 2025, 5:21 PM
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MathLuis
1594 posts
#2
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Recall the well known analysis fact that if a polynomial has exactly $n$ real roots counting multiplicity then its derivative has exactly $n-1$ real roots counting multiplicity. So if $k \le 97$ the considering the $98$-th derivative of $f-g$ gives that for any three given reals $a,b,c$ we must have that $ax^2+bx+c$ always has two real root if $f-g$ had exactly $100$ different roots, thus a contradiction by picking suitable $a,b,c$ (my example would just be $a,c>0$ and $b=0$ lol).
Now we will show $k=98$ works, first clean all the polynomial of degree $98$ inside $f$ using $g$ and from there many a suitable pick, in fact $100$ does not matter so we can prove a slightly stronger claim using induction.
We show that it is possible to construct a polynomial of degree $n$ where only coefficients $x^{n}, x^{n-1}$ are not in our control, so that the polynomial has all real roots with no multiplicity and none of them are zero.
For $n=2$ we can obviously shift the parabola by a constant acording to what is needed due to discriminant criteria so we should be fine here, due to continuity in fact you can just avoid a root being zero or the resulting parabola being tangent to the $x$ axis, for $n=3$ you have something unfree of the form $x^2(ax+b)$ and so you pick some $c(ax+b)+d$ on the free part where $c,d$ have freedom acording to needs like if $b=0$ then $d \ne 0$ and could just be suficiently small given we put $c<0$ to have two roots and that in order to not have repeated roots or the cubic being tangent to the $x$ axis at any moment (continuity yay).
For $n=4$ you basically have control over a depressed cubic upon doing a similar setup as seen above just shift everything by a small constant at the end to satisfy all the given conditions needed and now we will show this is the kind of work you have to do.
Suppose it was true for $n=\ell$ then we prove it for $n=\ell+1$, notice the not free term is $x^{\ell}(ax+b)$ so we just use the free term to focusing on a polynomial of the kind $x^{\ell}+q(x)$ for $\deg q \le \ell-2$, notice here it might happen that $-b \cdot a^{-1}$ is a root twice then we just shift by a very small constant as needed whether is above or below, one of them should work by continuity and IVT, since have $0$ as a root is only a one thing thing on a small interval we can also use continuity to avoid this and still satisfy the no double roots or more condition, hence by using the inductive hypothesis this is also completed and thus our claim is proven.
Just throw the claim and use a suitable $g$ to win, and thus we are done :cool:.
This post has been edited 1 time. Last edited by MathLuis, Apr 19, 2025, 3:49 AM
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iliya8788
8 posts
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Y by fe.
We claim that the answer is $\boxed{k=98}$.
Claim 1: $k>97$
Assume the contrary. there exists a polynomial $g(x)$ with degree less than $98$ such that $f(x)-g(x)$ has 100 roots for $f(x)=x^100$. Define $k$ as the degree of $g(x)$
Define $a_{1},...,a_{k}$ such that $f(x)-g(x)=x^{100}+a_{k}x^k+...+a_{0}$. Define $r_{1},...,r_{100}$ as the real roots of $f(x)-g(x)$.
By vieta's formula since the coefficient of $x^{99}$ and $x^{98}$ is equal to $0$: $\sum_{i=1}^{100}r_{i} = 0$ and $\sum_{i<j}^{}r_{i}r_{j} = 0 \implies \sum_{i=1}^{100}r_{i}^2 = 0 \implies r_{i}=0 \implies f(x)-g(x)=0$ which is a contradiction. It follows that $k>97$.
Claim 2: $k=98$
We can basically alter all the coefficients of $f(x)-g(x)$ except the coefficients of $x^{100}$ and $x^{99}$ so by vieta's formula our only restriction is the sum of the roots of the polynomial. So we just need to pick 100 arbitrary pairwise different real numbers such that their sum is equal to $-\frac{a_{99}}{a_{100}}$ with $a_{99}$ being the coefficient of $x^{99}$ and $a_{100}$ being the leading coefficient. Obviously this is possible. From here we just alter the other coefficients so that each one of the coefficients becomes equal to the coefficients of the polynomial that has all these $100$ numbers as roots multiplied by $a_{n}$ and so we are done. $\blacksquare$.
This post has been edited 2 times. Last edited by iliya8788, May 14, 2025, 9:54 AM
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