Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
Cute property of Pascal hexagon config
Miquel-point   1
N 2 hours ago by FarrukhBurzu
Source: KoMaL B. 5444
In cyclic hexagon $ABCDEF$ let $P$ denote the intersection of diagonals $AD$ and $CF$, and let $Q$ denote the intersection of diagonals $AE$ and $BF$. Prove that if $BC=CP$ and $DP=DE$, then $PQ$ bisects angle $BQE$.

Proposed by Géza Kós, Budapest
1 reply
Miquel-point
3 hours ago
FarrukhBurzu
2 hours ago
another n x n table problem.
pohoatza   3
N 2 hours ago by reni_wee
Source: Romanian JBTST III 2007, problem 3
Consider a $n$x$n$ table such that the unit squares are colored arbitrary in black and white, such that exactly three of the squares placed in the corners of the table are white, and the other one is black. Prove that there exists a $2$x$2$ square which contains an odd number of unit squares white colored.
3 replies
1 viewing
pohoatza
May 13, 2007
reni_wee
2 hours ago
Concurrency from isogonal Mittenpunkt configuration
MarkBcc168   18
N 2 hours ago by ihategeo_1969
Source: Fake USAMO 2020 P3
Let $\triangle ABC$ be a scalene triangle with circumcenter $O$, incenter $I$, and incircle $\omega$. Let $\omega$ touch the sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ at points $D$, $E$, and $F$ respectively. Let $T$ be the projection of $D$ to $\overline{EF}$. The line $AT$ intersects the circumcircle of $\triangle ABC$ again at point $X\ne A$. The circumcircles of $\triangle AEX$ and $\triangle AFX$ intersect $\omega$ again at points $P\ne E$ and $Q\ne F$ respectively. Prove that the lines $EQ$, $FP$, and $OI$ are concurrent.

Proposed by MarkBcc168.
18 replies
MarkBcc168
Apr 28, 2020
ihategeo_1969
2 hours ago
Anything real in this system must be integer
Assassino9931   8
N 2 hours ago by Abdulaziz_Radjabov
Source: Al-Khwarizmi International Junior Olympiad 2025 P1
Determine the largest integer $c$ for which the following statement holds: there exists at least one triple $(x,y,z)$ of integers such that
\begin{align*} x^2 + 4(y + z) = y^2 + 4(z + x) = z^2 + 4(x + y) = c \end{align*}and all triples $(x,y,z)$ of real numbers, satisfying the equations, are such that $x,y,z$ are integers.

Marek Maruin, Slovakia
8 replies
Assassino9931
May 9, 2025
Abdulaziz_Radjabov
2 hours ago
Good Permutations in Modulo n
swynca   10
N 2 hours ago by MR.1
Source: BMO 2025 P1
An integer $n > 1$ is called $\emph{good}$ if there exists a permutation $a_1, a_2, a_3, \dots, a_n$ of the numbers $1, 2, 3, \dots, n$, such that:
$(i)$ $a_i$ and $a_{i+1}$ have different parities for every $1 \leq i \leq n-1$;
$(ii)$ the sum $a_1 + a_2 + \cdots + a_k$ is a quadratic residue modulo $n$ for every $1 \leq k \leq n$.
Prove that there exist infinitely many good numbers, as well as infinitely many positive integers which are not good.
10 replies
swynca
Apr 27, 2025
MR.1
2 hours ago
Quadratic + cubic residue => 6th power residue?
Miquel-point   0
3 hours ago
Source: KoMaL B. 5445
Decide whether the following statement is true: if an infinite arithmetic sequence of positive integers includes both a perfect square and a perfect cube, then it also includes a perfect $6$th power.

Proposed by Sándor Róka, Nyíregyháza
0 replies
Miquel-point
3 hours ago
0 replies
II_a - r_a = R - r implies A = 60
Miquel-point   0
3 hours ago
Source: KoMaL B. 5421
The incenter and the inradius of the acute triangle $ABC$ are $I$ and $r$, respectively. The excenter and exradius relative to vertex $A$ is $I_a$ and $r_a$, respectively. Let $R$ denote the circumradius. Prove that if $II_a=r_a+R-r$, then $\angle BAC=60^\circ$.

Proposed by Class 2024C of Fazekas M. Gyak. Ált. Isk. és Gimn., Budapest
0 replies
Miquel-point
3 hours ago
0 replies
Cheating effectively in game of luck
Miquel-point   0
3 hours ago
Source: KoMaL B. 5420
Ádám, the famous conman signed up for the following game of luck. There is a rotating table with a shape of a regular $13$-gon, and at each vertex there is a black or a white cap. (Caps of the same colour are indistinguishable from each other.) Under one of the caps $1000$ dollars are hidden, and there is nothing under the other caps. The host rotates the table, and then Ádám chooses a cap, and take what is underneath. Ádám's accomplice, Béla is working at the company behind this game. Béla is responsible for the placement of the $1000$ dollars under the caps, however, the colors of the caps are chosen by a different collegaue. After placing the money under a cap, Béla
[list=a]
[*] has to change the color of the cap,
[*] is allowed to change the color of the cap, but he is not allowed to touch any other cap.
[/list]
Can Ádám and Béla find a strategy in part a. and in part b., respectively, so that Ádám can surely find the money? (After entering the casino, Béla cannot communicate with Ádám, and he also cannot influence his colleague choosing the colors of the caps on the table.)

Proposed by Gábor Damásdi, Budapest
0 replies
Miquel-point
3 hours ago
0 replies
IMO Genre Predictions
ohiorizzler1434   68
N 3 hours ago by Koko11
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
68 replies
ohiorizzler1434
May 3, 2025
Koko11
3 hours ago
Gcd(m,n) and Lcm(m,n)&F.E.
Jackson0423   1
N 3 hours ago by WallyWalrus
Source: 2012 KMO Second Round

Find all functions \( f : \mathbb{N} \to \mathbb{N} \) such that for all positive integers \( m, n \),
\[
f(mn) = \mathrm{lcm}(m, n) \cdot \gcd(f(m), f(n)),
\]where \( \mathrm{lcm}(m, n) \) and \( \gcd(m, n) \) denote the least common multiple and the greatest common divisor of \( m \) and \( n \), respectively.
1 reply
Jackson0423
May 13, 2025
WallyWalrus
3 hours ago
Trigonometric Product
Henryfamz   3
N 3 hours ago by Aiden-1089
Compute $$\prod_{n=1}^{45}\sin(2n-1)$$
3 replies
Henryfamz
May 13, 2025
Aiden-1089
3 hours ago
isogonal geometry
Tuguldur   7
N Apr 7, 2025 by whwlqkd
Let $P$ and $Q$ be isogonal conjugates with respect to $\triangle ABC$. Let $\triangle P_1P_2P_3$ and $\triangle Q_1Q_2Q_3$ be their respective pedal triangles. Let\[ X_1=P_2Q_3\cap P_3Q_2,\quad X_2=P_1Q_3\cap P_3Q_1,\quad X_3=P_1Q_2\cap P_2Q_1 \]Prove that the points $X_1$, $X_2$ and $X_3$ lie on the line $PQ$.
7 replies
Tuguldur
Apr 7, 2025
whwlqkd
Apr 7, 2025
isogonal geometry
G H J
G H BBookmark kLocked kLocked NReply
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Tuguldur
19 posts
#1
Y by
Let $P$ and $Q$ be isogonal conjugates with respect to $\triangle ABC$. Let $\triangle P_1P_2P_3$ and $\triangle Q_1Q_2Q_3$ be their respective pedal triangles. Let\[ X_1=P_2Q_3\cap P_3Q_2,\quad X_2=P_1Q_3\cap P_3Q_1,\quad X_3=P_1Q_2\cap P_2Q_1 \]Prove that the points $X_1$, $X_2$ and $X_3$ lie on the line $PQ$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
whwlqkd
106 posts
#2
Y by
Let $M$ be the midpoint of $PQ$. It is well known that $P_1, P_2, P_3, Q_1, Q_2, Q_3$ are cyclic, and call this circle $\Gamma$. Note that $M$ is the center of $\Gamma$. Let $R_i$ be the Miquel point of quadrilateral $P_{i+1}Q_{i+1}Q_{i+2}P_{i+2}$. ($P_{i+3}=P_{i}$) And rename $A,B,C$ as $A_1, A_2, A_3$. Note that $\angle PR_{i}A_{i}= \angle QR_{i}A_{i}=90°$ by cyclic, $R_{i}$ lies on $PQ$. And by Miquel’s theorem, $X_{i}$ lies on $R_{i}M$, and we are done.
note
This post has been edited 5 times. Last edited by whwlqkd, Apr 7, 2025, 1:25 PM
Reason: Misspelling
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Tuguldur
19 posts
#3
Y by
Note that $\angle PR_{i}A_{i}= \angle QR{i}A{i}=90°$ by cyclic, $R_{i}$ lies on $PQ$. And by Miquel’s theorem, $X_{i}$ lies on $R_{i}M$, and we are done.
explain this part
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
whwlqkd
106 posts
#4
Y by
Miquel’s theorem:
Let $ABCD$ be cyclic quadrilateral inscribed on circle $O$. And let $X$ be the Miquel point of quadrilateral $ABCD$. Then $AC$, $BD$, $OX$ are concurrent
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Tuguldur
19 posts
#5
Y by
Shouldnt $R_{i}$ exist in the line$X_{i}A_{i}$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
whwlqkd
106 posts
#6
Y by
whwlqkd wrote:
Miquel’s theorem:
Let $ABCD$ be cyclic quadrilateral inscribed on circle $O$. And let $X$ be the Miquel point of quadrilateral $ABCD$. Then $AC$, $BD$, $OX$ are concurrent

In here, $R_{i}=X$, $A_{i}=AB\cap CD$, $X_{i}=AC\cap BD$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
aidenkim119
33 posts
#8 • 1 Y
Y by whwlqkd
I have nice solution with cool construction.

Obviously, $P_1Q_1P_2Q_2P_3Q_3$ is cyclic, let its center be $O$
So let $S$ be the antipode of $Q_{2}$, and $R$ be the antipode of $Q_{3}$

Pascal on $Q_{2}P_{3}RQ_{3}P_{2}S$ gives that $X_{1} = Q_{2}P_{3} \cap Q_{3}P_{2}$ lies on $PO = PQ$, so problem solved
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
whwlqkd
106 posts
#9
Y by
aidenkim119 wrote:
I have nice solution with cool construction.

Obviously, $P_1Q_1P_2Q_2P_3Q_3$ is cyclic, let its center be $O$
So let $S$ be the antipode of $Q_{2}$, and $R$ be the antipode of $Q_{3}$

Pascal on $Q_{2}P_{3}RQ_{3}P_{2}S$ gives that $X_{1} = Q_{2}P_{3} \cap Q_{3}P_{2}$ lies on $PO = PQ$, so problem solved

Very nice solution with Pascal!!!!!!!!!
Z K Y
N Quick Reply
G
H
=
a