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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
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Tam giác nội tiếp
chunchun.math.2010   1
N 2 hours ago by giratina3
Bài 1:Cho tam giác abc nội tiếp đường tròn (o), đường cao ad (d ∈ bc). qua a kẻ đường song song với bc cắt (o) tại t. chứng minh rằng dt đi qua trọng tâm của tam giác abc.
Bài 2: Cho tứ giác ngoại tiếp abcd. p là một điểm bất kì trên cd. j, k, l lần lượt là tâm đường tròn nội tiếp của các tam giác apb, apd, bpc. chứng minh rằng ∠ajk + ∠bjl = 180°.
1 reply
chunchun.math.2010
5 hours ago
giratina3
2 hours ago
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Discord Server
mathprodigy2011   2
N 3 hours ago by Lhaj3
Theres a server where we are all like discussing problems+helping each other practice. Hopefully you guys can join.

https://discord.gg/6hN3w4eK
2 replies
mathprodigy2011
Yesterday at 11:00 PM
Lhaj3
3 hours ago
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Expected Vakue of coins around table
ehz2701   0
4 hours ago
There are $n$ people around in a circle, who at $t=0\text{ sec}$ each have one penny. After each second every person gives deals penny they have to the left or right, or keeps that penny with equal probability. What is the expected value of seconds such that one person has all $n$ pennies? (Express answer in terms of $n$).

I think the answer is $3^{n-1}$ but I’m not sure. Purely guessing.

For $n=3$ the answer is $9$ because from any arrangement there is exactly $1$ way to get to any of the other arrangements (consider the problem from the coin’s perspective as they switch humans. Label the humans $0,1,2$. Therefore the initial arrangement is $0,1,2$. Every time we add or subtract $1$ or leave alone each of the numbers. We can only obtain, say, $2, 2, 2$ (where every coin is in the second person’s hand), in one way: that is $-1, +1, 0$. This is true for every possible arrangement of coins from any possible arrangement of coins. Hence the probability of landing on $1, 1, 1$ or $2, 2, 2$ or $3, 3, 3$ is $\frac{1}{9}$ and the answer is $9$.
0 replies
ehz2701
4 hours ago
0 replies
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Assam Mathematics Olympiad 2022 Category III Q9
SomeonecoolLovesMaths   2
N 5 hours ago by ParthivCalculus
What is the number formed by the last three digits of $1201^{1202}$?
2 replies
SomeonecoolLovesMaths
Sep 12, 2024
ParthivCalculus
5 hours ago
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No more topics!
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Nesbitt in triangle
Dynic   1
N Mar 18, 2025 by bhontu
Let $a,b,c$ be the lengths of the three sides of a triangle, respectively. Let $\mathbb{X}$ as a set of positive real number and the function $D:\mathbb{X}^3\to\mathbb{X}$ determined by $$D(x,y,z)=\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y},\forall x,y,z >0$$Find all values that $D(a,b,c)$ can take.
1 reply
Dynic
Mar 18, 2025
bhontu
Mar 18, 2025
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Nesbitt in triangle
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Dynic
#1 h  Mar 18, 2025, 9:51 AM
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Let $a,b,c$ be the lengths of the three sides of a triangle, respectively. Let $\mathbb{X}$ as a set of positive real number and the function $D:\mathbb{X}^3\to\mathbb{X}$ determined by $$D(x,y,z)=\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y},\forall x,y,z >0$$Find all values that $D(a,b,c)$ can take.
This post has been edited 1 time. Last edited by Dynic, Mar 18, 2025, 9:52 AM
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bhontu
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bhontu
#2 h  Mar 18, 2025, 1:49 PM • 1 Y
Y by Dynic
https://artofproblemsolving.com/community/c6h295764p1601527
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