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k a June Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 3:57 PM
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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Summer camps are starting this month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have a transformative summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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0 replies
jlacosta
Yesterday at 3:57 PM
0 replies
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Daily Problem Writing Practice
KSH31415   10
N an hour 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)

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.
10 replies
KSH31415
Yesterday at 3:13 PM
KSH31415
an hour ago
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[PMO17 Qualifying II.4] A Consecutive Telescoping Product Sum
Shinfu   1
N 2 hours ago by Shinfu
Simplify

$$\dfrac{1+2+3}{1+2+3+4}\times\dfrac{1+2+3+4+5}{1+2+3+4+5+6}\times\dots\times\dfrac{1+2+3+\dots+19}{1+2+3+\dots+20}$$
$\text{(a) }\frac{1}{5}\qquad\text{(b) }\frac{3}{20}\qquad\text{(c) }\dfrac{1}{7}\qquad\text{(d) }\frac{1}{35}$

Answer Confirmation
1 reply
Shinfu
2 hours ago
Shinfu
2 hours ago
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13th Philippine Mathematical Olympiad Area Stage #8
qrxz17   1
N 2 hours ago by Mathelets
Problem: Find all complex numbers \(x\) satisfying \(x^3+x^2+x+1=0\).

Answer: \(-1, i, -i\)

Solution: Notice that \(x^3+x^2+x+1\) is a geometric series that can be expressed as
\begin{align*} \frac{x^4-1}{x-1}. \end{align*}
We have

\begin{align*} \frac{x^4-1}{x-1} &= 0 \\ x^4-1&=0 \\ x^4 &= 1. \end{align*}
The solutions to this are the fourth roots of unity. So,
\begin{align*} x^4 = 1 &= \cos (2\pi)+i\sin(2\pi) \\ &= (\cos \theta+i \sin\theta)^4 \\ &= \cos 4\theta+i \sin4\theta. \end{align*}
We then have \(\theta = \frac{2\pi k}{4}\) for \(k=0,1,2,3\). The complex numbers \(x\) that satisfy the equation are \(\boxed{-1, i, -i}\), with \(1\) excluded as a solution. This is because the equation \(\frac{x^4 - 1}{x - 1} = 0,\) is valid only when \(x \ne 1\).
1 reply
qrxz17
Today at 4:15 PM
Mathelets
2 hours ago
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a problem about cups
pzzd   3
N 3 hours ago by vincentwant
Hello! I’m new to the AoPS forums, so apologies in advance if I’ve posted in the wrong place :-)

Here’s an interesting counting problem I’ve come up with that’s been bothering me for a while:

Say we have a tower of cups such that the top row has one cup, the row directly under that has two cups, the row under that has 3 cups, and so on such that every cup is supported by exactly two cups in the row under it. Each cup in the tower is labelled with a number or letter so that every cup is distinct. If a cup tower has $n$ rows, how many different ways can we pick up all the cups, starting at the top? (Assume that you can’t pick up a cup that’s under another cup, the tower would fall over.)
3 replies
pzzd
3 hours ago
vincentwant
3 hours ago
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Min max của pt bậc 2
chunchun.math.2010   2
N 5 hours ago by HS12309
Cho x,y là hai số thực thoa mãn x^2+ y^2+6x-8y+21 =0.Tìm min,max P=x^2+y^2
2 replies
chunchun.math.2010
Today at 3:41 PM
HS12309
5 hours ago
J
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Original Problem (Roots of Unity)
qrxz17   1
N 6 hours ago by xHypotenuse
Problem: A regular hexagon in the complex plane has its center located at the complex number \(C=338+169i\). Its vertices are given by the complex numbers \(z_0,z_1,z_2,z_3,z_4,z_5\).

What is \(z_0+z_1+z_2+z_3+z_4+z_5\)?

Answer: Click to reveal hidden text

Solution: A regular hexagon centered at the origin has the sum of its coordinates equal to \(0\) since the coordinates correspond to the 6th roots of unity scaled by the radius. Since the sum of all n-th roots of unity is zero, the coordinates add up to zero.

The center was translated to \(338 + 169i\), so all corresponding vertices of the hexagon are also translated by \(338 + 169i\).

We have
\begin{align*} & (z_0 + 338 + 169i) + (z_1 + 338 + 169i) + (z_2 + 338 + 169i) \\ & + (z_3 + 338 + 169i) + (z_4 + 338 + 169i) + (z_5 + 338 + 169i) \\ &= (z_0 + z_1 + z_2 + z_3 + z_4 + z_5) + 6(338 + 169i) \\ &= 0 + 6(338 + 169i) = \boxed{2028 + 1014i}. \end{align*}
1 reply
qrxz17
6 hours ago
xHypotenuse
6 hours ago
J
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Geometry
MathsII-enjoy   6
N 6 hours ago by aaravdodhia
Given triangle $ABC$ inscribed in $(O)$, $M, N$ are respectively the midpoints of the major and minor arcs $BC$. Let $I$ be the center of the inscribed circle, $R$ be $A-mix$, $D$ is the intersection point of the line through $A$ parallel to $BC$ with $(O)$. $DI$ intersects $AR$ at $K$, take $L$ on $AK$ so that $LI//BC$. $NL$ intersects $(O)$ at $G$, $AG$ intersects $LI$ at $Z$. Prove that: $ZM$ is perpendicular to $KN$.
6 replies
MathsII-enjoy
May 23, 2025
aaravdodhia
6 hours ago
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Adapted from MATHirang MATHibay 2013 Eliminations Difficult E1
qrxz17   0
Today at 4:20 PM
Problem: Janina was able to “extend” the complex number \( \mathbb{C} \) into what she calls the \(\textit{Kompurekkusu}\) numbers, \( \mathbb{K} \), by including her favorite letter, \( j \). This \( j \) behaves much like \( i \) except that \( j^2 = 1 \), but \( j \) is not any complex number (particularly not 1 nor \(-1\)).

For \( r, s, t, u \in \mathbb{R} \), what is the sum of all solutions \( \kappa = (r + si) + (t + ui)j \in \mathbb{K} \) such that \( \kappa^4 = 4 \) and both \( i \) and \( j \) appear with nonzero coefficients?

Answer: Click to reveal hidden text

Solution: Let \(A=r+si\) and \(B=t+ui\). We have
\begin{align*} k^4&=(A+Bj)^4 \\ &= (A^2+B^2+2ABj)^2 \\ &= (A^4+B^4+6A^2B^2) + 4AB(A^2+B^2)j. \end{align*}For this to be equal to \(4\), which is a real number, \(4AB(A^2+B^2)\) must be equal to \(0\).

If \( B = 0 \), then \( Bj = 0 \), which fails to meet the condition that \( j \) must appear with nonzero coefficients.

If \( A = 0 \), then \( r + si = 0 \), which can only occur if both \( r = 0 \) and \( s = 0 \). So,
\begin{align*} 4&=[(t+ui)j]^4 \\ 4&=(t+ui)^4 \\ 4&=r^4(\cos4\theta+i\sin4\theta) \end{align*}
We are then going to find the roots of unity. But we are only asked for the sum, which we know is equal to \(0\).

If, \(A^2+B^2=0\). We have any of the following cases:
\begin{align*} A &= \pm iB \\ B &= \pm iA \end{align*}
So,
\begin{align*} 4&=(\pm iB+ Bj)^4 \\ 4&=[B(\pm i+j)]^4 \\ 4&=B^4(-4)\\ -1&=B^4 \end{align*}and
\begin{align*} 4&=(A \pm Aij)^4 \\ 4&=[A(1\pm ij)]^4 \\ 4&=A^4(-4)\\ -1&=A^4 \end{align*}
Thus, the solutions \(\kappa\) are the fourth roots of unity of \(-1\). We know that the sum of all \( n \)th roots of unity is zero, so the sum of all the solutions is \(\boxed{0}\).

Comment. The original problem asks for how many solutions we have for \(\kappa\).
0 replies
qrxz17
Today at 4:20 PM
0 replies
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Find f(1)+f(3)+f(5)+...+f(7999)
Darealzolt   1
N Today at 4:19 PM by aaravdodhia
It is known that
\[ f(x) = \frac{1}{\sqrt[3]{x^2+2x+1}+\sqrt[3]{x^2-1}+\sqrt[3]{x^2-2x+1}} \]for all positive integers \(x\), hence find the value of
\[ f(1)+f(3)+f(5)+\dots +f(7999) \]
1 reply
Darealzolt
Today at 3:33 PM
aaravdodhia
Today at 4:19 PM
J
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Geometry Trigonometry Olympiads
Foxellar   1
N Today at 3:46 PM by vanstraelen
Let \( \triangle ABC \) be a triangle such that \( \angle ABC = 120^\circ \). Points \( X, Y, Z \) lie on segments \( BC, CA, AB \), respectively, such that lines \( AX, BY, \) and \( CZ \) are the angle bisectors of triangle \( ABC \). Find the measure of angle \( \angle XYZ \).
1 reply
Foxellar
May 23, 2025
vanstraelen
Today at 3:46 PM
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Min,maã của pt bậc 2
chunchun.math.2010   0
Today at 3:37 PM
Cho x,y là hai số thực thoa mãn x^2+ y^2+6x-8y+21 =0.Tìm min,maã của P=x^2+y^2
0 replies
chunchun.math.2010
Today at 3:37 PM
0 replies
J
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Alternating colors in chairs
tapilyoca   1
N Today at 3:30 PM by tapilyoca
Six individuals, wearing shirts that alternate in color between black and white, are seated around a circular table. Three times in succession, a pair of adjacent seats is chosen uniformly at random (with repetition allowed), and the occupants of those seats exchange places. What is the probability that, after these three adjacent swaps, the six shirts still alternate in color around the table?

Answer!!
1 reply
tapilyoca
Today at 3:23 PM
tapilyoca
Today at 3:30 PM
J
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Find the number of elements in S
Darealzolt   7
N Today at 3:09 PM by iniffur
It is given the set \(S\), such that
\[ S=\left\{ x \in \mathbb{Z} | \frac{x^2 - 2x +7}{2x-1} \in \mathbb{Z} \right\} \]Hence find the number of elements in \(S \).
7 replies
Darealzolt
Today at 1:45 AM
iniffur
Today at 3:09 PM
J
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Monotonically increasing dice rolls
arcticfox009   1
N Today at 3:03 PM by arcticfox009
Alice and Bob take turns rolling a regular 6-sided die. They repeatedly do so until either a player rolls a number less than the previous roll or a total of $4$ rolls have been made. If Alice makes the first roll, the probability that Bob makes the last roll can be written as $p/q$, where $p$ and $q$ are relatively prime positive integers. What is $p + q$?

answer!
1 reply
arcticfox009
Today at 3:01 PM
arcticfox009
Today at 3:03 PM
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J
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100th post
MathJedi108   1
N Apr 20, 2025 by mdk2013
Well I guess this is my 100th post, it would be really funny if it isn't can yall share your favorite experience on AoPS here?
1 reply
MathJedi108
Apr 20, 2025
mdk2013
Apr 20, 2025
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100th post
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Reveal topic tags
100th Post
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MathJedi108
100 posts
MathJedi108
#1 h  Apr 20, 2025, 10:59 PM
Y by
Well I guess this is my 100th post, it would be really funny if it isn't can yall share your favorite experience on AoPS here?
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650 posts
mdk2013
#2 h  Apr 20, 2025, 11:10 PM • 1 Y
Y by ChristianYoo
why is this post in hsm though
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