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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
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[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
Prove that lines parallel in triangle
jasperE3   6
N a few seconds ago by Retemoeg
Source: Mongolian MO 2007 Grade 11 P1
Let $M$ be the midpoint of the side $BC$ of triangle $ABC$. The bisector of the exterior angle of point $A$ intersects the side $BC$ in $D$. Let the circumcircle of triangle $ADM$ intersect the lines $AB$ and $AC$ in $E$ and $F$ respectively. If the midpoint of $EF$ is $N$, prove that $MN\parallel AD$.
6 replies
jasperE3
Apr 8, 2021
Retemoeg
a few seconds ago
Check upper bound
Sadigly   1
N 8 minutes ago by Sadigly
Source: Azerbaijan Senior MO 2025 P5
A 9-digit number $N$ is given, whose digits are non-zero and all different.The sums of all consecutive three-digit segments in the decimal representation of number $N$ are calculated and arranged in increasing order.Is it possible to obtain the following sequences as a result of this operation?

$\text{a)}$ $11,15,16,18,19,21,22$

$\text{b)}$ $11,15,16,18,19,21,23$
1 reply
Sadigly
37 minutes ago
Sadigly
8 minutes ago
Number Theory Marathon!!!
starchan   435
N 13 minutes ago by Primeniyazidayi
Source: Possibly Mercury??
Number theory Marathon
Let us begin
P1
435 replies
starchan
May 28, 2020
Primeniyazidayi
13 minutes ago
one cyclic formed by two cyclic
CrazyInMath   39
N 15 minutes ago by trigadd123
Source: EGMO 2025/3
Let $ABC$ be an acute triangle. Points $B, D, E$, and $C$ lie on a line in this order and satisfy $BD = DE = EC$. Let $M$ and $N$ be the midpoints of $AD$ and $AE$, respectively. Suppose triangle $ADE$ is acute, and let $H$ be its orthocentre. Points $P$ and $Q$ lie on lines $BM$ and $CN$, respectively, such that $D, H, M,$ and $P$ are concyclic and pairwise different, and $E, H, N,$ and $Q$ are concyclic and pairwise different. Prove that $P, Q, N,$ and $M$ are concyclic.
39 replies
CrazyInMath
Apr 13, 2025
trigadd123
15 minutes ago
No more topics!
Circumference with diameter HG
Seventh   6
N Feb 14, 2025 by Saucepan_man02
Source: Problem 6, Brazilian MO 2015
Let $\triangle ABC$ be a scalene triangle and $X$, $Y$ and $Z$ be points on the lines $BC$, $AC$ and $AB$, respectively, such that $\measuredangle AXB = \measuredangle BYC = \measuredangle CZA$. The circumcircles of $BXZ$ and $CXY$ intersect at $P$. Prove that $P$ is on the circumference which diameter has ends in the ortocenter $H$ and in the baricenter $G$ of $\triangle ABC$.
6 replies
Seventh
Oct 20, 2015
Saucepan_man02
Feb 14, 2025
Circumference with diameter HG
G H J
G H BBookmark kLocked kLocked NReply
Source: Problem 6, Brazilian MO 2015
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Seventh
82 posts
#1 • 3 Y
Y by Davi-8191, Adventure10, Mango247
Let $\triangle ABC$ be a scalene triangle and $X$, $Y$ and $Z$ be points on the lines $BC$, $AC$ and $AB$, respectively, such that $\measuredangle AXB = \measuredangle BYC = \measuredangle CZA$. The circumcircles of $BXZ$ and $CXY$ intersect at $P$. Prove that $P$ is on the circumference which diameter has ends in the ortocenter $H$ and in the baricenter $G$ of $\triangle ABC$.
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pi37
2079 posts
#2 • 4 Y
Y by don2001, edfearay123, ThisNameIsNotAvailable, Adventure10
Solution
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parmenides51
30651 posts
#3 • 1 Y
Y by Adventure10
also Serbia RMM TST 2019 P4
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william122
1576 posts
#5 • 1 Y
Y by Adventure10
Solution with MathStudent2002 and Muriatic.

Denote the HM points of the triangle as $A',B',C'$. We first claim that $(AYZ)$ always passes through $A'$. Note that $Y\to Z$ and $Y\to (AYA')\cap AB$ are projective, so we only need to check 3 cases. When $BY\perp AC$, the claim is obvious. When $\angle BYC=\angle B$, $(AYZ)$ is the circle passing through $A$ tangent to $BC$, so $A'$ lies on $(AYZ)$ by the definition of the HM point. As $\angle BYC=180-\angle C$ follows similarly, we have the desired result.

Now, note that $X\to (BB'X)\cap (HG)$ is projective, as well as $X\to Y\to (CC'Y)\cap (HG)$. So, to show that $(BXZ)$ and $(CXY)$ intersect on $(HG)$, we only need to check three points. However, by choosing $X=(BB'C')\cap BC$, $X=(BB'A')\cap BC$, $X=(CC'B')\cap BC$, we can make $P$ the three HM points, which do indeed lie on $(HG)$. Thus, we are done.
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NiltonCesar
163 posts
#7
Y by
pi37 wrote:
Solution

Sorry, but if you draw in Geogebra, you'll not have $\angle BCX_A=\angle X_AAB$.
Attachments:
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VicKmath7
1389 posts
#8 • 1 Y
Y by Maths_Girl
Here is a sketch of a bit different solution.
Sketch
This post has been edited 1 time. Last edited by VicKmath7, Apr 18, 2023, 2:07 PM
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Saucepan_man02
1337 posts
#9
Y by
Note that $AX \cap CZ \in (BXZ)$, which implies the $B$ HM point lies on $(BXZ)$ (due to ELMO 2013 SL G3). Denote $B$ HM point to be $X_B$, and similarly define $X_C$.
Note that $B, X_B, G$ are collinear along with $H X_B \perp X_B G$ implies $X_B$ lies on the circle with diameter $HG$. Similarly $X_c$ lies on the circle with diameter $HG$.
Notice that, by Miquel theorem on $\triangle BGC$, $(G X_B X_C), (C X_C X), (B X_B X)$ are concurrent at $P$. Thus, $P \in (G X_b X_c)$ and we are done.
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