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k a June Highlights and 2025 AoPS Online Class Information
jlacosta   0
Jun 2, 2025
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!

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0 replies
jlacosta
Jun 2, 2025
0 replies
two subsets with no fewer than four common elements.
micliva   42
N an hour ago by math_holmes15
Source: All-Russian Olympiad 1996, Grade 9, First Day, Problem 4
In the Duma there are 1600 delegates, who have formed 16000 committees of 80 persons each. Prove that one can find two committees having no fewer than four common members.

A. Skopenkov
42 replies
micliva
Apr 18, 2013
math_holmes15
an hour ago
Prove the inequality
Butterfly   1
N an hour ago by Butterfly

Let $x,y,z>0$ such that $xyz=1$. Prove $(1+x)(1+x+y)(1+x+y+z)\ge16.$
1 reply
Butterfly
2 hours ago
Butterfly
an hour ago
Easy diophantine equation
Arne   15
N an hour ago by Chikiratori
Source: AMO 2006
Find all positive integers $m$ and $n$ such that $1 + 5 \cdot 2^m = n^2$.
15 replies
Arne
May 6, 2006
Chikiratori
an hour ago
Point on circumcircle with angle condition
Taco12   10
N an hour ago by dragoon
Source: OMMC POTM 12/2022
Let $\triangle ABC$ be such that the midpoint of $BC$ is $D$. Let $E$ be the point on the opposite side of $AC$ as $B$ on the circumcircle of $\triangle ABC$ such that $\angle DEA = \angle DEC$ and let $\omega$ be the circumcircle of $\triangle CED$. If $\omega$ intersects $AE$ at $X$ and the tangent to $\omega$ at $D$ intersects $AB$ at $Y$, show that $XY$ is parallel to $BC$.

Proposed by Taco12
10 replies
Taco12
Dec 3, 2022
dragoon
an hour ago
Daily Problem Writing Practice
KSH31415   17
N 2 hours ago by KSH31415
I'm trying to get better at writing problems so I decided to challenge myself to write one problem for every day this month (June 2025). I will post them in this thread as well as edit this post with all of them in hide tags. If I can, I'll include a difficulty level in the form of AIME placement. If anybody wants to solve them and give feedback on the problem and/or my difficulty rating, please do!

June 1 (AIME P4) - Solved
June 2 (AIME P5) - Solved
June 3 (AIME P12) - Solved
June 4 (AIME P7)
June 5 (AIME P5)
June 6 (AIME P4)

June 1 (AIME P4)
A bag contains $6$ red balls and $6$ blue balls. A draw consists of randomly selecting $2$ of the remaining balls from bag without replacement, and then setting them aside. A draw is called a match if the two balls have the same color. Compute the expected number of draws until either a match occurs or the bag is empty.
17 replies
KSH31415
Jun 2, 2025
KSH31415
2 hours ago
Solve the equation in integers
Math2030   0
2 hours ago
Solve the equation in integers
& 5^x - 3^y = 2&
0 replies
Math2030
2 hours ago
0 replies
An identity about floor function.
Tony_Yin_Math   0
3 hours ago
How to prove this identity in wikipedia?It's about floor function.
https://en.wikipedia.org/wiki/Floor_and_ceiling_functions#Quotients
0 replies
Tony_Yin_Math
3 hours ago
0 replies
Polynomials
P162008   0
4 hours ago
P1. Given that $p$ is a monic polynomial of smallest degree satisfying $p(4 - \sqrt{3}) = p(2 + \sqrt{5}) = 0.$
If coefficients of $p$ are rational then which of the following option(s) is/are true

A) $p(2) = -5$

B) $p(0) = -13$

C) $p'(0) = 46$

D) $p(1) = -24$

P2. Which of the following option(s) is/are correct?

A) There exists a $4$ degree polynomial $p$ with rational coefficients satisfying $p(\sqrt{2} + \sqrt{3}) = \sqrt{3}.$

B) There exists a $3$ degree polynomial $p$ with rational coefficients satisfying $p(\sqrt{2} + \sqrt{3}) = \sqrt{2}.$

C) If $\{x \in C : x^3 + x^2 = 1\} = \{\alpha,\beta,\gamma\}, \text{then} \{x \in C x^3 = x + 1\} = \{\alpha\beta,\beta\gamma,\gamma\alpha\}.$

D) If $a + b + c + d = a^3 + b^3 + c^3 + d^3 = 0,\text{then} 0 \in \{a + b,b + c,c + a,a + d,b + d,c + d\}.$

P3. If $p$ denotes a cyclic permutation of $(a,b,c)$. Use the idea of quadratic equation as an identity to prove that

(i) $\sum_{p} \frac{2c^2 + 3c + 4}{(c - a)(c - b)} = 2$

(ii) $\sum_{p} \frac{(2c^2 + 3c + 4)(a + b)}{(c - a)(c - b)} = -3$

P4. Let $p$ denote a cyclic permutation of $(a,b,c,d)$. Then find the value of $ \lambda = \sum_{p} \left(\frac{(2a^3 + 3a^2 + 4a + 5)(bc + cd + bd)}{(a - b)(a - c)(a - d)}\right).$
0 replies
P162008
4 hours ago
0 replies
Greatest Integer function and Fractional Part Function
P162008   2
N 4 hours ago by pingpongmerrily
P1. Find the value of $S$ defined as ($\lfloor . \rfloor$ denotes greatest integer function, $\{.\}$ denotes the fractional part function)

$S = \sum_{k=1}^{7} \sum_{r=k^2 + 1}^{k^2 + 2k} \lfloor \frac{1}{\{\sqrt{r}\}} \rfloor.$

P2. Prove that $\sum_{r=0}^{n-1} \lfloor \frac{n + 2^r}{2^{r + 1}} \rfloor = n \forall n \in N.$

P3. Find $\lambda = \lfloor \sum_{r=1}^{200} (\sqrt{100^2 + r}) \rfloor.$

P4. Find $\lambda = \sum_{k=1}^{7} \sum_{r=1}^{2k}\lfloor \frac{2k}{r} \rfloor.$
2 replies
P162008
5 hours ago
pingpongmerrily
4 hours ago
Exponents and Logarithms
wpdnjs   3
N 5 hours ago by trangbui
Suppose A and B are positive real numbers for which log base A of B equals log base B of A. If neither A nor B is 1 and A does not equal B, find the value of AB.
3 replies
wpdnjs
Yesterday at 6:34 PM
trangbui
5 hours ago
1919 water square
NicoN9   2
N 6 hours ago by abbominable_sn0wman
We call a positive integer $c$ bigger than $10^4$ water if the last four digit is $1919$. Does there exists a water perfect square?
2 replies
NicoN9
Yesterday at 10:32 PM
abbominable_sn0wman
6 hours ago
99...99400...009
NicoN9   2
N Yesterday at 11:44 PM by NicoN9
Let $k>2$ be a positive integer. Prove that\[
\underbrace{99\dots 99}_{k-1}4\underbrace{00\dots 00}_{k-1}9
\]can't be a prime number.
2 replies
NicoN9
Yesterday at 10:35 PM
NicoN9
Yesterday at 11:44 PM
quadratic equation solving with 399999999...
NicoN9   1
N Yesterday at 10:41 PM by LearnMath_105
Let $m$ be a positive integer. Solve over $\mathbb{R}$, the equation\[
x^2+2x-3\underbrace{99\dots 99}_{2m}.
\]
1 reply
NicoN9
Yesterday at 10:26 PM
LearnMath_105
Yesterday at 10:41 PM
Extended Wilson's?
NamelyOrange   2
N Yesterday at 9:46 PM by rhydon516
Let $\mathbb{Z}^*_n$ be the set of positive integers less than $n$ relatively prime to it. Is there a nice pattern for $\left(\prod_{k\in \mathbb{Z}^*_n} k\right) \text{ mod }n$? I know from a Wilson's theorem-style argument that it's either $1$ or $-1$, but when is it which?
2 replies
NamelyOrange
Yesterday at 5:41 PM
rhydon516
Yesterday at 9:46 PM
Marking vertices in splitted triangle
mathisreal   2
N May 18, 2025 by sopaconk
Source: Mexico
Let $n$ be a positive integer. Consider a figure of a equilateral triangle of side $n$ and splitted in $n^2$ small equilateral triangles of side $1$. One will mark some of the $1+2+\dots+(n+1)$ vertices of the small triangles, such that for every integer $k\geq 1$, there is not any trapezoid(trapezium), whose the sides are $(1,k,1,k+1)$, with all the vertices marked. Furthermore, there are no small triangle(side $1$) have your three vertices marked. Determine the greatest quantity of marked vertices.
2 replies
mathisreal
Feb 7, 2022
sopaconk
May 18, 2025
Marking vertices in splitted triangle
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Source: Mexico
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mathisreal
647 posts
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Let $n$ be a positive integer. Consider a figure of a equilateral triangle of side $n$ and splitted in $n^2$ small equilateral triangles of side $1$. One will mark some of the $1+2+\dots+(n+1)$ vertices of the small triangles, such that for every integer $k\geq 1$, there is not any trapezoid(trapezium), whose the sides are $(1,k,1,k+1)$, with all the vertices marked. Furthermore, there are no small triangle(side $1$) have your three vertices marked. Determine the greatest quantity of marked vertices.
This post has been edited 1 time. Last edited by mathisreal, Feb 7, 2022, 1:54 PM
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parmenides51
30653 posts
#2
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Let $n$ be a positive integer. Consider a figure of a equilateral triangle of side $n$ and splitted in $n^2$ small equilateral triangles of side $1$. One will mark some of the $1+2+\dots+(n+1)$ vertices of the small triangles, such that for every integer $k\geq 1$, there is not any trapezoid(trapezium), whose the sides are $(1,k,1,k+1)$, with all the vertices marked. Furthermore, there are no small triangle(side $1$) have your three vertices marked. Determine the greatest quantity of marked vertices.
https://cdn.artofproblemsolving.com/attachments/7/d/a31dd041fcafa9c90b451b16efe3592ec9d721.png
Note: The figure shows an example of a board when $n = 4$. and an example of one of the mentioned trapezoids, with three of its vertices marked and with $k = 2$.
This post has been edited 3 times. Last edited by parmenides51, Sep 17, 2022, 8:25 PM
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sopaconk
19 posts
#3 • 1 Y
Y by plagueis
The answer is $(k+1)^2$ if $n=2k$ or $(k+1)(k+2)$ if $n=2k+1$.

First, we are going to prove for $n\geq 2$ that if $f(n)$ is the answer for $n$, then
\[f(n) \leq f(n-2)+n+1\]
To prove this.
https://cdn.artofproblemsolving.com/attachments/2/2/416721cf806f2f8f5f00066f9986ee61558929.png
In the pink triangle of length $n-2$ we have maximum $f(n-2)$ marked vertices, and we have $2n+1$ more points, we consider these points in a path (the red path in the picture).

We cannot have 3 consecutive marked points because that implies a short triangle with their three vertices marked.

A packet of marked points is the maximum consecutive marked points.
Let's say we have $a$ packets of size 2, and $b$ packets of size 1.
Now, between two packets of size 2, it should be an even distance otherwise it will form a trapezoid with all the vertices marked.

Then between two packets of size 2 it's at least one extra not marked vertex, otherwise the distance is $2k+1$ where $k$ is the number of packets of size 1 between the packets of size 2.

Then we use at least
\[2a+b+ (a+b-1)+(a-1) = 2(2a+b)-2 \leq 2n+1\]The marked points, the spaces between each packet and the "extra" not marked point between the packets of size 2.

Then $2a+b \leq \frac{2n+3}{2}$ but $2a+b$ is integer then $2a+b \leq n+1$, then the marked points in the path are maximum $n+1$, and the total points are maximum $f(n-2)+n+1$.

We have $f(0)=1, f(1)=2$ and then if $n=2k+1$ is odd the maximum is
\[2+4+\ldots+(2k+2) = 2 \left(1+2+\ldots+k+1\right)=(k+1)(k+2) \]and if $n=2k$ even the maximum is
\[1+3+\ldots+ 2k+1 = (k+1)^2\]
And both are possible, if you enumerate the rows from 1 to $n+1$ and taking the rows with the same parity as $n+1$ (It takes just the sums $2+4+\ldots$ and $1+3+\ldots$ depending of the parity of $n$) we have the values that we want, and it works because each trapezium or triangle has a segment of size 1 that it's not horizontal and it has one vertex not marked.
This post has been edited 1 time. Last edited by sopaconk, May 18, 2025, 1:37 AM
Reason: I don't know how to put images
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