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

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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. 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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[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
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April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
[*]April 10th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MathILy and MathILy-Er Math Jam: Multibackwards Numbers
[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
1 viewing
jlacosta
Apr 2, 2025
0 replies
Serbian selection contest for the BMO 2025 - P2
OgnjenTesic   1
N 3 minutes ago by grupyorum
Find all functions $f : \mathbb{N} \to \mathbb{N}$ such that $f(a) + f(b) \mid af(a) - bf(b)$, for all $a, b \in \mathbb{N}$.
(Here, $\mathbb{N}$ is a set of positive integers.)

Proposed by Vukašin Pantelić
1 reply
OgnjenTesic
2 hours ago
grupyorum
3 minutes ago
Incenter and concurrency
jenishmalla   6
N 10 minutes ago by ihategeo_1969
Source: 2025 Nepal ptst p3 of 4
Let the incircle of $\triangle ABC$ touch sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Let $D'$ be the diametrically opposite point of $D$ with respect to the incircle. Let lines $AD'$ and $AD$ intersect the incircle again at $X$ and $Y$, respectively. Prove that the lines $DX$, $D'Y$, and $EF$ are concurrent, i.e., the lines intersect at the same point.

(Kritesh Dhakal, Nepal)
6 replies
jenishmalla
Mar 15, 2025
ihategeo_1969
10 minutes ago
Nine-point centers in triangle with a cevian
Talmon   0
28 minutes ago
A point $D$ is on side $BC$ of triangle $ABC$. Let $N_1$, $N_2$ and $N$ be nine-point centers of triangles $ABD$, $ACD$ and $ABC$ respectively. Prove that perpendicular lines from $N_1$ to $AC$, from $N_2$ to $AB$, from $N$ to $AD$ and from $A$ to $BC$ are concurrent.
0 replies
Talmon
28 minutes ago
0 replies
Third degree and three variable system of equations
MellowMelon   56
N 31 minutes ago by Marcus_Zhang
Source: USA TST 2009 #7
Find all triples $ (x,y,z)$ of real numbers that satisfy the system of equations
\[ \begin{cases}x^3 = 3x-12y+50, \\ y^3 = 12y+3z-2, \\ z^3 = 27z + 27x. \end{cases}\]

Razvan Gelca.
56 replies
MellowMelon
Jul 18, 2009
Marcus_Zhang
31 minutes ago
No more topics!
The reflections of the vertices and the intersections of circumcircle with HaHb
bigant146   5
N Feb 23, 2022 by lazizbek42
Source: VI Caucasus Mathematical Olympiad
In an acute triangle $ABC$ let $AH_a$ and $BH_b$ be altitudes. Let $H_aH_b$ intersect the circumcircle of $ABC$ at $P$ and $Q$. Let $A'$ be the reflection of $A$ in $BC$, and let $B'$ be the reflection of $B$ in $CA$. Prove that $A', B'$, $P$, $Q$ are concyclic.
5 replies
bigant146
Mar 14, 2021
lazizbek42
Feb 23, 2022
The reflections of the vertices and the intersections of circumcircle with HaHb
G H J
Source: VI Caucasus Mathematical Olympiad
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bigant146
111 posts
#1 • 2 Y
Y by samrocksnature, jhu08
In an acute triangle $ABC$ let $AH_a$ and $BH_b$ be altitudes. Let $H_aH_b$ intersect the circumcircle of $ABC$ at $P$ and $Q$. Let $A'$ be the reflection of $A$ in $BC$, and let $B'$ be the reflection of $B$ in $CA$. Prove that $A', B'$, $P$, $Q$ are concyclic.
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L567
1184 posts
#2 • 3 Y
Y by samrocksnature, jhu08, Timmy456
Let the orthocenter be $H$. It is well known that $(AHC)$ is just the reflection of $(ABC)$ across $AC$ and so $A'$ is just $(AHC) \cap BH_b$ and similarly, $B' = (BHC) \cap AH_a$.

Now, its just PoP, $A'H_a.H_aH = CH_a.H_aB = QH_a.H_a.P$ and so $A'B'PQ$ is cyclic
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GorgonMathDota
1063 posts
#3 • 2 Y
Y by samrocksnature, jhu08
bigant146 wrote:
In an acute triangle $ABC$ let $AH_a$ and $BH_b$ be altitudes. Let $H_aH_b$ intersect the circumcircle of $ABC$ at $P$ and $Q$. Let $A'$ be the reflection of $A$ in $BC$, and let $B'$ be the reflection of $B$ in $CA$. Prove that $A', B'$, $P$, $Q$ are concyclic.

What...
We claim that $A'QHPB'$ is cyclic. let $AH_A$ intersects $(ABC)$ at $A_1$. By POP, we have
\[ A'H_A \cdot H_A H = AH_A \cdot H_A A_1 = QH_A \cdot H_A P \]Hence, $(A' Q HP)$ is cyclic. Do this similarly, and we get $(BP'HQ)$ is cyclic, and we are done.
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Maths_geometryTPC2023
163 posts
#4 • 2 Y
Y by jhu08, Mango247
bigant146 wrote:
In an acute triangle $ABC$ let $AH_a$ and $BH_b$ be altitudes. Let $H_aH_b$ intersect the circumcircle of $ABC$ at $P$ and $Q$. Let $A'$ be the reflection of $A$ in $BC$, and let $B'$ be the reflection of $B$ in $CA$. Prove that $A', B'$, $P$, $Q$ are concyclic.

Note that the orthocenter $H$ lies on $(A'B'PQ)$ , then power of a point will finish it
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Mahdi_Mashayekhi
689 posts
#6 • 1 Y
Y by jhu08
Let AHa and BHb meet circumcircle at A1 and B1. H is orthocenter. We will prove B'PHQ and A'QHP are cyclic.
B'Hb = Bhb , B1Hb = HHb ---> B'Hb . HbH = B1Hb . HbB = PHb . HbQ ---> B'PHQ is cyclic.
We can prove A'QHP is cyclic the same way so we have proved A'QHPB' is cyclic and we're Done.
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lazizbek42
548 posts
#7 • 1 Y
Y by Jasurbek
$AH_a$ and $(ABC)$ intersection $X$.
$$BH_a \cdot CH_a = H_aX \cdot AH_a = HH_a \cdot H_a A'= PH_a \cdot QH_a$$$PHQA'$ - cyclic.Similarity $PQHB'$ cyclic.
$PQA'B'$ - cyclic quadrateral.$\blacksquare$.
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