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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.

Summer camps are starting next 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 an enriching 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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[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
Divisibilty...
Sadigly   0
7 minutes ago
Source: Azerbaijan Junior NMO 2025 P2
Find all $4$ consecutive even numbers, such that the square of their product divides the sum of their squares.
0 replies
Sadigly
7 minutes ago
0 replies
Quadratic system
juckter   35
N an hour ago by shendrew7
Source: Mexico National Olympiad 2011 Problem 3
Let $n$ be a positive integer. Find all real solutions $(a_1, a_2, \dots, a_n)$ to the system:

\[a_1^2 + a_1 - 1 = a_2\]\[ a_2^2 + a_2 - 1 = a_3\]\[\hspace*{3.3em} \vdots \]\[a_{n}^2 + a_n - 1 = a_1\]
35 replies
juckter
Jun 22, 2014
shendrew7
an hour ago
IMO Shortlist 2012, Geometry 3
lyukhson   75
N 2 hours ago by numbertheory97
Source: IMO Shortlist 2012, Geometry 3
In an acute triangle $ABC$ the points $D,E$ and $F$ are the feet of the altitudes through $A,B$ and $C$ respectively. The incenters of the triangles $AEF$ and $BDF$ are $I_1$ and $I_2$ respectively; the circumcenters of the triangles $ACI_1$ and $BCI_2$ are $O_1$ and $O_2$ respectively. Prove that $I_1I_2$ and $O_1O_2$ are parallel.
75 replies
lyukhson
Jul 29, 2013
numbertheory97
2 hours ago
Diophantine
TheUltimate123   31
N 2 hours ago by SomeonecoolLovesMaths
Source: CJMO 2023/1 (https://aops.com/community/c594864h3031323p27271877)
Find all triples of positive integers \((a,b,p)\) with \(p\) prime and \[a^p+b^p=p!.\]
Proposed by IndoMathXdZ
31 replies
TheUltimate123
Mar 29, 2023
SomeonecoolLovesMaths
2 hours ago
No more topics!
<DNF=90
sinajackson   5
N Apr 5, 2007 by Virgil Nicula
$E$ and $F$ are points on sides $AB$ and $BC$ of square $ABCD$ respectively such that: $BE=BF$. If $BN$ is the altitude of triangle $BEC$, Prove that $DNF$ is right angle.
5 replies
sinajackson
Apr 4, 2007
Virgil Nicula
Apr 5, 2007
<DNF=90
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sinajackson
594 posts
#1 • 1 Y
Y by Adventure10
$E$ and $F$ are points on sides $AB$ and $BC$ of square $ABCD$ respectively such that: $BE=BF$. If $BN$ is the altitude of triangle $BEC$, Prove that $DNF$ is right angle.
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vittasko
1327 posts
#2 • 2 Y
Y by Adventure10, Mango247
We denote as $M,$ the intersection point of $BD,$ $EF$ and so, we have that the quadrilateral $BENM$ is cyclic, because of $BD\perp EF.$

Hence, we conclude that $\angle NEF = \angle NBD$ $,(1)$

From the similar right triangles $\bigtriangleup BNE,$ $\bigtriangleup BNC,$ we have that

$\frac{EN}{BN}= \frac{BE}{BC}=\frac{EF}{AC}= \frac{EF}{BD}$ $\Longrightarrow$ $\frac{EN}{BN}= \frac{EF}{BD}$ $,(2)$

From $(1),$ $(2)$ we conclude that the triangles $\bigtriangleup ENF,$ $\bigtriangleup BND,$ are similar and so, we have that $\angle NDB = \angle NFE$ $,(3)$

From $(3),$ $\Longrightarrow$ the quadrilateral $DNMF,$ is cyclic.

Hence, we have that $\angle DNF = \angle DMF = 90^{o}$ and the pproof is completed.

Kostas vittas.
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stergiu
1648 posts
#3 • 1 Y
Y by Adventure10
sinajackson wrote:
$E$ and $F$ are points on sides $AB$ and $BC$ of square $ABCD$ respectively such that: $BE=BF$. If $BN$ is the altitude of triangle $BEC$, Prove that $DNF$ is right angle.

It is a nice problem from Leningrad or All Russian olympiad..

After Vitta's nice (and new to me !) solution , I send you one more solution, not metric.

If $BN$ meets $AD$ at $Z$ , then triangles $ABZ, BEC$ are equal (since $\angle ABZ= \angle BCE$ . Thus $AZ=BE=BF$ and $DZ=CF$. If $DF$ meets $CZ$ at $O$ , then :
$ON= \frac{ZC}{2}= \frac{DF}{2}$, because in right triangle $ZNC$ segment $NO$ is median and $ZC=DF$.

This means that triangle $DNF$ is right , since $NO$ is a median in this triangle.
This post has been edited 1 time. Last edited by stergiu, Apr 5, 2007, 8:14 AM
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Virgil Nicula
7054 posts
#4 • 1 Y
Y by Adventure10
A nice and easy problem !A short proof.Stergiu, are only four minutes between the our messages !
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stergiu
1648 posts
#5 • 1 Y
Y by Adventure10
Virgil Nicula wrote:
Stergiu[/u][/b], are only four minutes between the our messages !

It seems , we solve problems about the same time !!!

Greetings.-
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Virgil Nicula
7054 posts
#6 • 1 Y
Y by Adventure10
A metrical proof with the following simple remark applied to the right triangle NBC :Indeed, $\tan\widehat{CNF}=\frac{CF}{CB}=\frac{CF}{CD}=\tan\widehat{CDF}$, i.e. $\widehat{CNF}\equiv\widehat{CDF}$ a.s.o.

Stergiu, (again) now is only one minute between our messages ! Greetings !
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