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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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0 replies
jlacosta
Mar 2, 2025
0 replies
J
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Prime for square numbers
giangtruong13   0
3 minutes ago
Source: City’s Specialized Math Examination
Given that $a,b$ are natural numbers satisfy that: $\frac{a^3}{a+b}$ and $\frac{b^3}{a+b}$ are prime numbers. Prove that $$a^2+3ab+3a+b+1$$is a perfect squared number
0 replies
giangtruong13
3 minutes ago
0 replies
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euler-totient function
Laan   1
N 17 minutes ago by top1vien
Proof that there are infinitely many positive integers $n$ such that
$\varphi(n)<\varphi(n+1)<\varphi(n+2)$
1 reply
Laan
Today at 7:13 AM
top1vien
17 minutes ago
J
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2 var inquality
sqing   5
N 21 minutes ago by sqing
Source: Own
Let $ a,b $ be nonnegative real numbers such that $ a^2+ab+b^2+a+b=1. $ Prove that
$$ (ab+1)(a+b)\leq \frac{ 20}{27} $$$$ (ab+1)(a+b-1)\leq - \frac{ 10}{27} $$Let $ a,b $ be nonnegative real numbers such that $ a^2+b^2+a+b=1. $ Prove that
$$ (ab+1)(a+b)\leq \frac{ 5\sqrt 3-7}{2} $$$$ (ab+1)(a+b-1)\leq 3\sqrt 3- \frac{ 11}{2} $$
5 replies
sqing
Yesterday at 3:00 PM
sqing
21 minutes ago
J
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Is it fake? how can someone score 12 in AIME but can&#039;t qualify RMO
Bruce_wayne123   3
N 27 minutes ago by Bruce_wayne123
Source: https://www.reddit.com/r/JEENEETards/comments/1jgduci/op_qualified_for_usamo/#lightbox
He also claims to have scored 93 percentile in JEEM maths another thing which makes it more doubtful and also he didn't got any letter from MAA
3 replies
Bruce_wayne123
an hour ago
Bruce_wayne123
27 minutes ago
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No more topics!
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geometry in a isosceles
AlanLG   4
N Nov 1, 2023 by gnoka
Source: 2nd National Women´s Contest of Mexican Mathematics Olympiad 2023 , problem 1 teams
Let $\triangle ABC$ such that $AB=AC$, $D$ and $E$ points on $AB$ and $BC$, respectively, with $DE\parallel AC$. Let $F$ on line $DE$ such that $CADF$ it´s a parallelogram. If $O$ is the circumcenter of $\triangle BDE$, prove that $O,F,A$ and $D$ lie on a circle.
4 replies
AlanLG
Jul 23, 2023
gnoka
Nov 1, 2023
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geometry in a isosceles
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Reveal topic tags
geometry
isosceles
Mexico
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G H BBookmark kLocked kLocked NReply
Source: 2nd National Women´s Contest of Mexican Mathematics Olympiad 2023 , problem 1 teams
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AlanLG
241 posts
AlanLG
#1 h  Jul 23, 2023, 4:38 AM
Y by
Let $\triangle ABC$ such that $AB=AC$, $D$ and $E$ points on $AB$ and $BC$, respectively, with $DE\parallel AC$. Let $F$ on line $DE$ such that $CADF$ it´s a parallelogram. If $O$ is the circumcenter of $\triangle BDE$, prove that $O,F,A$ and $D$ lie on a circle.
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SQTHUSH
154 posts
SQTHUSH
#2 h  Jul 23, 2023, 5:05 AM
Y by
Consider that $AD=EF,\angle ODA=\angle OEF,OD=OE$
Hence $\triangle ODA\cong \triangle OEF$
Which means that $\triangle ODE\sim\triangle OAF$
So $\angle ODF=\angle OAF\Rightarrow O,F,A,D$ are cyclic
Z K Y
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jayme
9767 posts
jayme
#3 h  Jul 23, 2023, 2:50 PM
Y by
Dear Mathlinkers,

another point of view by considerind the pivot theorem...

Sncerely
Jean-Louis
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X.Allaberdiyev
101 posts
X.Allaberdiyev
#4 h  Sep 2, 2023, 1:35 PM
Y by
Since $OB=OD$, $OF=AC=AB$ and $\angle ABO=90-\angle DEB=90-\angle DBE=\angle ODF$ we have $\triangle OBA\cong \triangle ODF$, which gives that $\angle DFO=\angle BAO=\angle DAO$, which completes the solution.
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gnoka
245 posts
gnoka
#5 h  Nov 1, 2023, 7:07 AM
Y by
Since $AB=AC, DE//AC$ we have $BD=DE$,then $OD$ is angle bisector of $\angle BDE$
from parallelogram ADFC we have $DF=AC=AB=AD+DB$
which means $DF=AD+2OD\cos\angle ODB$
then $DF\sin\angle ODB=AD\sin\angle ODB+OD\sin2\angle ODB$
So $DF\sin\angle ADO=AD\sin\angle ODE+OD\sin2\angle ADE$
According to Ptolemy's theorem, $O,F,A$ and $D$ lie on a circle.
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