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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)!

Be sure to mark your calendars for the following upcoming events:
[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.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]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
everyone will get zero marx on this
Th3Numb3rThr33   48
N 18 minutes ago by Blast_S1
Source: JMO 2018 Problem 6
Karl starts with $n$ cards labeled $1,2,3,\dots,n$ lined up in a random order on his desk. He calls a pair $(a,b)$ of these cards swapped if $a>b$ and the card labeled $a$ is to the left of the card labeled $b$. For instance, in the sequence of cards $3,1,4,2$, there are three swapped pairs of cards, $(3,1)$, $(3,2)$, and $(4,2)$.

He picks up the card labeled 1 and inserts it back into the sequence in the opposite position: if the card labeled 1 had $i$ card to its left, then it now has $i$ cards to its right. He then picks up the card labeled $2$ and reinserts it in the same manner, and so on until he has picked up and put back each of the cards $1,2,\dots,n$ exactly once in that order. (For example, the process starting at $3,1,4,2$ would be $3,1,4,2\to 3,4,1,2\to 2,3,4,1\to 2,4,3,1\to 2,3,4,1$.)

Show that, no matter what lineup of cards Karl started with, his final lineup has the same number of swapped pairs as the starting lineup.
48 replies
+1 w
Th3Numb3rThr33
Apr 19, 2018
Blast_S1
18 minutes ago
USAMO Medals
YauYauFilter   7
N an hour ago by pingpongmerrily
YauYauFilter
Apr 24, 2025
pingpongmerrily
an hour ago
Find the radius of circle O
TheMaskedMagician   3
N 2 hours ago by fruitmonster97
Source: 1976 AHSME Problem 18
IMAGE

In the adjoining figure, $AB$ is tangent at $A$ to the circle with center $O$; point $D$ is interior to the circle; and $DB$ intersects the circle at $C$. If $BC=DC=3$, $OD=2$, and $AB=6$, then the radius of the circle is

$\textbf{(A) }3+\sqrt{3}\qquad\textbf{(B) }15/\pi\qquad\textbf{(C) }9/2\qquad\textbf{(D) }2\sqrt{6}\qquad \textbf{(E) }\sqrt{22}$
3 replies
TheMaskedMagician
May 18, 2014
fruitmonster97
2 hours ago
MathPath
PatTheKing806   13
N 2 hours ago by Nora2021
Is anybody else going to MathPath?

I haven't gotten in. its been 3+ weeks since they said my application was done.
13 replies
PatTheKing806
Mar 24, 2025
Nora2021
2 hours ago
GCD of consecutive terms
nsato   33
N 3 hours ago by reni_wee
The numbers in the sequence 101, 104, 109, 116, $\dots$ are of the form $a_n = 100 + n^2$, where $n = 1$, 2, 3, $\dots$. For each $n$, let $d_n$ be the greatest common divisor of $a_n$ and $a_{n + 1}$. Find the maximum value of $d_n$ as $n$ ranges through the positive integers.
33 replies
nsato
Mar 14, 2006
reni_wee
3 hours ago
Largest Divisor
4everwise   19
N 3 hours ago by reni_wee
What is that largest positive integer $n$ for which $n^3+100$ is divisible by $n+10$?
19 replies
4everwise
Dec 22, 2005
reni_wee
3 hours ago
Can anyone solve this binomial identity
sasu1ke   0
4 hours ago


\[
\sum_{0 \leq k \leq l} (l - k) \binom{m}{k} \binom{q + k}{n}
= \binom{l + q + 1}{m + n + 1},
\]\[
\text{integers } l, m \geq 0,\quad \text{integers } n \geq q \geq 0.
\]
0 replies
sasu1ke
4 hours ago
0 replies
[ABCD] = n [CDE], areas in trapezoid - IOQM 2020-21 p1
parmenides51   3
N 4 hours ago by iamahana008
Let $ABCD$ be a trapezium in which $AB \parallel CD$ and $AB = 3CD$. Let $E$ be then midpoint of the diagonal $BD$. If $[ABCD] = n \times  [CDE]$, what is the value of $n$?

(Here $[t]$ denotes the area of the geometrical figure$ t$.)
3 replies
parmenides51
Jan 18, 2021
iamahana008
4 hours ago
Looking for users and developers
derekli   5
N Today at 3:59 PM by musicalpenguin
Guys I've been working on a web app that lets you grind high school lvl math. There's AMCs, AIME, BMT, HMMT, SMT etc. Also, it's infinite practice so you can keep grinding without worrying about finding new problems. Please consider helping me out by testing and also consider joining our developer team! :P :blush:

Link: https://stellarlearning.app/competitive
5 replies
derekli
Today at 12:57 AM
musicalpenguin
Today at 3:59 PM
Sequences and GCD problem
BBNoDollar   0
Today at 3:23 PM
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
0 replies
BBNoDollar
Today at 3:23 PM
0 replies
Sum of digits is 18
Ecrin_eren   13
N Today at 3:20 PM by NamelyOrange
How many 5 digit numbers are there such that sum of its digits is 18
13 replies
Ecrin_eren
Yesterday at 1:10 PM
NamelyOrange
Today at 3:20 PM
Inequalities
sqing   1
N Today at 3:18 PM by DAVROS
Let $ a,b,c>0 $ and $ a+b\leq 16abc. $ Prove that
$$ a+b+kc^3\geq\sqrt[4]{\frac{4k} {27}}$$$$ a+b+kc^4\geq\frac{5} {8}\sqrt[5]{\frac{k} {2}}$$Where $ k>0. $
$$ a+b+3c^3\geq\sqrt{\frac{2} {3}}$$$$ a+b+2c^4\geq \frac{5} {8}$$
1 reply
sqing
Today at 12:46 PM
DAVROS
Today at 3:18 PM
an algebra problem
Asyrafr09   2
N Today at 12:03 PM by pooh123
Determine all real number($x,y,z$) that satisfy
$$x=1+\sqrt{y-z^2}$$$$y=1+\sqrt{z-x^2}$$$$z=1+\sqrt{x-y^2}$$
2 replies
Asyrafr09
Today at 10:05 AM
pooh123
Today at 12:03 PM
Inequalities
sqing   1
N Today at 11:51 AM by sqing
Let $ a,b,c\geq 0 ,   2a +ab + 12a bc \geq 8. $ Prove that
$$  a+  (b+c)(a+1)+\frac{4}{5}  bc \geq 4$$$$  a+  (b+c)(a+0.9996)+ 0.77  bc \geq 4$$
1 reply
sqing
Today at 5:23 AM
sqing
Today at 11:51 AM
Segment has Length Equal to Circumradius
djmathman   72
N Mar 16, 2025 by Zhaom
Source: 2014 USAMO Problem 5
Let $ABC$ be a triangle with orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $AHC$ with the internal bisector of the angle $\angle BAC$. Let $X$ be the circumcenter of triangle $APB$ and $Y$ the orthocenter of triangle $APC$. Prove that the length of segment $XY$ is equal to the circumradius of triangle $ABC$.
72 replies
djmathman
Apr 30, 2014
Zhaom
Mar 16, 2025
Segment has Length Equal to Circumradius
G H J
Source: 2014 USAMO Problem 5
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djmathman
7938 posts
#1 • 8 Y
Y by narutomath96, spacekid4, Davi-8191, centslordm, megarnie, Adventure10, Mango247, Rounak_iitr
Let $ABC$ be a triangle with orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $AHC$ with the internal bisector of the angle $\angle BAC$. Let $X$ be the circumcenter of triangle $APB$ and $Y$ the orthocenter of triangle $APC$. Prove that the length of segment $XY$ is equal to the circumradius of triangle $ABC$.
This post has been edited 1 time. Last edited by djmathman, Apr 30, 2014, 9:56 PM
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BOGTRO
5818 posts
#2 • 3 Y
Y by Condorcet, Adventure10, Mango247
I managed to reduce this to proving XY was perpendicular to BC, but i couldn't manage to prove that. Is this even true?
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AceOfDiamonds
1017 posts
#3 • 4 Y
Y by opptoinfinity, Imayormaynotknowcalculus, Adventure10, Mango247
Complex
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pi37
2079 posts
#4 • 9 Y
Y by mathandyou, narutomath96, r31415, ProbaBillity, yds, mathtiger6, Adventure10, Mango247, and 1 other user
Synthetic solution
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soy_un_chemisto
927 posts
#5 • 2 Y
Y by Adventure10, Mango247
djmathman wrote:
Let $ABC$ be a triangle with orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $AHC$ with the internal bisector of the angle $\angle BAC$. Let $X$ be the circumcenter of triangle $AHC$ and $Y$ the orthocenter of triangle $APC$. Prove that the length of segment $XY$ is equal to the circumradius of triangle $ABC$.
It should be:
let $P$ be the second intersection of the circumcircle of triangle $ABC$ with the internal bisector of the angle $\angle BAC$

EDIT: oops this is not the original problem but this version also has $XY$ equaling the circumradius.
This post has been edited 1 time. Last edited by soy_un_chemisto, Apr 30, 2014, 9:57 PM
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msinghal
725 posts
#6 • 2 Y
Y by Adventure10, Mango247
Reflecting the whole thing over line AC gives a nice solution
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v_Enhance
6877 posts
#7 • 12 Y
Y by dantx5, zschess, Ultroid999OCPN, Imayormaynotknowcalculus, HamstPan38825, megarnie, Kingsbane2139, nguyenducmanh2705, Adventure10, Mango247, Rounak_iitr, and 1 other user
msinghal wrote:
Reflecting the whole thing over line AC gives a nice solution
It also gives a nice complex bash :P took me around 20 minutes, thankfully... I finished the test with five to spare so any more synthetic effort would have finished me.
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pi37
2079 posts
#8 • 3 Y
Y by megarnie, Adventure10, and 1 other user
Question: Do you think not addressing possible configuration issues will lose points? (i.e. if I said assume the configuration is as shown, other cases are handled similarly)
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msinghal
725 posts
#9 • 1 Y
Y by Adventure10
pi37 wrote:
Question: Do you think not addressing possible configuration issues will lose points? (i.e. if I said assume the configuration is as shown, other cases are handled similarly)

I was thinking about that but I really didn't want to take care of that. I wasn't sure whether to add at the end that directed angles/lengths could resolve this, but decided against it.
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XmL
552 posts
#10 • 2 Y
Y by Adventure10, Mango247
Outline of mine
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JuanOrtiz
366 posts
#11 • 2 Y
Y by Adventure10, Mango247
$Y$ is on circumcircle by angle chasing. Let $O_1$ be the reflection of $O$ through $AC$. Then look at the rotohomothety with center $O$ that sends $AOY$ to $O_1OX$ (these triangles are similar). It turns out $AO_1=R$ and so $XY=R$, so we're done.

hmm
This post has been edited 1 time. Last edited by JuanOrtiz, May 1, 2014, 1:17 AM
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henrypickle
238 posts
#12 • 1 Y
Y by Adventure10
Starting to worry that reducing to BC perpendicular to XY was not actually useful for finding a solution...
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JuanOrtiz
366 posts
#13 • 2 Y
Y by Adventure10, Mango247
Well, it could be used to find the spiral similarity in my solution... I'm not sure.
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mathandyou
77 posts
#14 • 2 Y
Y by Adventure10, Mango247
pi37 wrote:
Synthetic solution
Why $(O_1)$ is the reflection of $(O)$ across $AC$?
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pohoatza
1145 posts
#15 • 3 Y
Y by HoRI_DA_GRe8, lrjr24, Adventure10
pi37 wrote:
Question: Do you think not addressing possible configuration issues will lose points? (i.e. if I said assume the configuration is as shown, other cases are handled similarly)
Your solution was the same as the official solution, so it is safe to assume that you won't lose any points :).
This post has been edited 1 time. Last edited by pohoatza, May 1, 2014, 6:16 PM
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