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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!

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0 replies
jlacosta
Yesterday at 3:57 PM
0 replies
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Center lies on altitude
plagueis   17
N 5 minutes ago by bin_sherlo
Source: Mexico National Olympiad 2018 Problem 6
Let $ABC$ be an acute-angled triangle with circumference $\Omega$. Let the angle bisectors of $\angle B$ and $\angle C$ intersect $\Omega$ again at $M$ and $N$. Let $I$ be the intersection point of these angle bisectors. Let $M'$ and $N'$ be the respective reflections of $M$ and $N$ in $AC$ and $AB$. Prove that the center of the circle passing through $I$, $M'$, $N'$ lies on the altitude of triangle $ABC$ from $A$.

Proposed by Victor Domínguez and Ariel García
17 replies
plagueis
Nov 6, 2018
bin_sherlo
5 minutes ago
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IMO Shortlist 2014 C6
hajimbrak   22
N 11 minutes ago by awesomeming327.
We are given an infinite deck of cards, each with a real number on it. For every real number $x$, there is exactly one card in the deck that has $x$ written on it. Now two players draw disjoint sets $A$ and $B$ of $100$ cards each from this deck. We would like to define a rule that declares one of them a winner. This rule should satisfy the following conditions:
1. The winner only depends on the relative order of the $200$ cards: if the cards are laid down in increasing order face down and we are told which card belongs to which player, but not what numbers are written on them, we can still decide the winner.
2. If we write the elements of both sets in increasing order as $A =\{ a_1 , a_2 , \ldots, a_{100} \}$ and $B= \{ b_1 , b_2 , \ldots , b_{100} \}$, and $a_i > b_i$ for all $i$, then $A$ beats $B$.
3. If three players draw three disjoint sets $A, B, C$ from the deck, $A$ beats $B$ and $B$ beats $C$ then $A$ also beats $C$.
How many ways are there to define such a rule? Here, we consider two rules as different if there exist two sets $A$ and $B$ such that $A$ beats $B$ according to one rule, but $B$ beats $A$ according to the other.

Proposed by Ilya Bogdanov, Russia
22 replies
2 viewing
hajimbrak
Jul 11, 2015
awesomeming327.
11 minutes ago
J
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annoying algebra with sequence :/
tabel   1
N 17 minutes ago by L_.
Source: random 9th grade text book (section meant for contests)
Let \( a_1 = 1 \) and \( a_{n+1} = 1 + \frac{n}{a_n} \) for \( n \geq 1 \). Prove that the sequence \( (a_n)_{n \geq 1} \) is increasing.
1 reply
tabel
5 hours ago
L_.
17 minutes ago
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The Return of Triangle Geometry
peace09   16
N 28 minutes ago by NO_SQUARES
Source: 2023 ISL A7
Let $N$ be a positive integer. Prove that there exist three permutations $a_1,\dots,a_N$, $b_1,\dots,b_N$, and $c_1,\dots,c_N$ of $1,\dots,N$ such that \[\left|\sqrt{a_k}+\sqrt{b_k}+\sqrt{c_k}-2\sqrt{N}\right|<2023\]for every $k=1,2,\dots,N$.
16 replies
peace09
Jul 17, 2024
NO_SQUARES
28 minutes 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
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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
J
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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
J
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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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Find all natural numbers $n$
ItsBesi   7
N Apr 26, 2025 by justaguy_69
Source: Kosovo Math Olympiad 2025, Grade 9, Problem 2
Find all natural numbers $n$ such that $\frac{\sqrt{n}}{2}+\frac{10}{\sqrt{n}}$ is a natural number.
7 replies
ItsBesi
Nov 17, 2024
justaguy_69
Apr 26, 2025
J
Find all natural numbers $n$
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Reveal topic tags
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Source: Kosovo Math Olympiad 2025, Grade 9, Problem 2
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ItsBesi
148 posts
ItsBesi
#1 h  Nov 17, 2024, 4:55 PM
Y by
Find all natural numbers $n$ such that $\frac{\sqrt{n}}{2}+\frac{10}{\sqrt{n}}$ is a natural number.
Z K Y
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Mhremath
65 posts
Mhremath
#3 h  Nov 17, 2024, 5:17 PM
Y by
For $n \in \mathbb{N}$, $n \neq 0$, if $\frac{\sqrt{n}}{2} + \frac{10}{\sqrt{n}} \in \mathbb{N}$, find all possible values of $n$.

We know that $n$ is a natural number and the result is also a natural number.

Now, we must search for the divisors of 10: $b | 10$, so $b \in \{1, 2, 5, 10\}$.

Therefore, from this part, $n \in \{1, 4, 25, 100\}$.

Trying each one:

1. For $n = 1$, $\frac{1}{2}$ is not a natural number, so the final result will not be natural either. Therefore, $n = 1$ is a contradiction.
2. $n = 4$ is true.
3. $n = 25$ is not true because dividing by 2 will not give a natural number.
4. $n = 100$ is true.

Therefore, the true answers are $n \in \{4, 100\}$.
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Mhremath
65 posts
Mhremath
#4 h  Nov 17, 2024, 5:19 PM
Y by
is that really olympiad problem?
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grupyorum
1435 posts
grupyorum
#5 h  Nov 17, 2024, 5:49 PM • 3 Y
Y by ItsBesi, bsf714, Mhremath
Mhremath wrote:
is that really olympiad problem?

I mean, it's a Grade 9 problem. Higher grades (e.g., 12th) have actually quite nice problems, take a look ;)
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DensSv
62 posts
DensSv
#6 h  Nov 17, 2024, 6:28 PM
Y by
Sol.
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AshAuktober
1014 posts
AshAuktober
#7 h  Nov 18, 2024, 6:58 AM
Y by
Let $n = k^2$ (this can be checked to be necessary), then $2k \mid k^2 + 20 \implies 2k \mid 80 \implies k \mid 40$.
Now manually check to see which work.
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lksb
183 posts
lksb
#8 h  Nov 19, 2024, 2:20 PM
Y by
DensSv wrote:
2)$k=2 \Leftrightarrow 6\in \mathbb{N}$ which is true.
5)$k=10\Leftrightarrow 6\in\mathbb{N}$, which is true.

Nice that when it is an integer, it is equal to six
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justaguy_69
17 posts
justaguy_69
#9 h  Apr 26, 2025, 10:45 AM
Y by
we get $$2\sqrt{n}\vert n+20 $$$$2\sqrt{n}\vert 2n+40$$$$2\sqrt{n}\vert 2n+40-2n$$$$\sqrt{n}\vert20$$$$ we\ get \ \sqrt{n}={1,2,4,5,10,20}$$
we try the values and get $$n={{2^2,10^2}}$$
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