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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)!

Be sure to mark your calendars for the following events:
[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]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
jlacosta
Apr 2, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Incenter and concurrency
jenishmalla   4
N 30 minutes ago by Double07
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)
4 replies
jenishmalla
Mar 15, 2025
Double07
30 minutes ago
inequalities
pennypc123456789   0
34 minutes ago
Let $a,b,c$ be positive real numbers . Prove that :
$$\dfrac{(a+b+c)^2}{ab+bc +ac } \ge \dfrac{2ab}{a^2+b^2} + \dfrac{2bc}{b^2+c^2} + \dfrac{2ac}{a^2+c^2} $$
0 replies
pennypc123456789
34 minutes ago
0 replies
Ratio of lengths in right-angled triangle
DylanN   1
N 39 minutes ago by Mathzeus1024
Source: South African Mathematics Olympiad 2021, Problem 2
Let $PAB$ and $PBC$ be two similar right-angled triangles (in the same plane) with $\angle PAB = \angle PBC = 90^\circ$ such that $A$ and $C$ lie on opposite sides of the line $PB$. If $PC = AC$, calculate the ratio $\frac{PA}{AB}$.
1 reply
DylanN
Aug 11, 2021
Mathzeus1024
39 minutes ago
Pythagorean new journey
XAN4   4
N an hour ago by XAN4
Source: Inspired by sarjinius
The number $4$ is written on the blackboard. Every time, Carmela can erase the number $n$ on the black board and replace it with a new number $m$, if and only if $|n^2-m^2|$ is a perfect square. Prove or disprove that all positive integers $n\geq4$ can be written exactly once on the blackboard.
4 replies
XAN4
Yesterday at 3:41 AM
XAN4
an hour ago
wu2481632 Mock Geometry Olympiad problems
wu2481632   14
N an hour ago by bin_sherlo
To avoid clogging the fora with a horde of geometry problems, I'll post them all here.

Day I

Day II

Enjoy the problems!
14 replies
wu2481632
Mar 13, 2017
bin_sherlo
an hour ago
bashing out 3^7 terms
i_equal_tan_90   43
N an hour ago by akliu
Source: 2023 AIME II/8
Let $\omega=\cos\frac{2\pi}{7}+i\cdot\sin\frac{2\pi}{7}$, where $i=\sqrt{-1}$. Find
$$\prod_{k=0}^{6}(\omega^{3k}+\omega^k+1).$$
43 replies
i_equal_tan_90
Feb 16, 2023
akliu
an hour ago
Straight line
uTOPi_a   19
N an hour ago by NerdyNashville
Source: 41-st Vietnamese Mathematical Olympiad 2003
The circles $ C_{1}$ and $ C_{2}$ touch externally at $ M$ and the radius of $ C_{2}$ is larger than that of $ C_{1}$. $ A$ is any point on $ C_{2}$ which does not lie on the line joining the centers of the circles. $ B$ and $ C$ are points on $ C_{1}$ such that $ AB$ and $ AC$ are tangent to $ C_{1}$. The lines $ BM$, $ CM$ intersect $ C_{2}$ again at $ E$, $ F$ respectively. $ D$ is the intersection of the tangent at $ A$ and the line $ EF$. Show that the locus of $ D$ as $ A$ varies is a straight line.
19 replies
uTOPi_a
Aug 28, 2004
NerdyNashville
an hour ago
inequalities
pennypc123456789   1
N an hour ago by Double07
Let \( x,y \) be non-negative real numbers.Prove that :
\[ \sqrt{x^4+y^4 } +(2+\sqrt{2})xy \geq x^2+y^2 \]
1 reply
pennypc123456789
Today at 3:28 AM
Double07
an hour ago
Inspired by old results
sqing   2
N an hour ago by sqing
Source: Own
Let $a,b$ be real numbers such that $  a^3 +b^3+6ab=8 . $ Prove that
$$a+b \leq 2$$Let $a,b$ be real numbers such that $ a^2+b^2+a^3 +b^3+8ab=12 . $ Prove that
$$a+b \leq 2$$Let $a,b$ be real numbers such that $a+b + a^2+b^2+a^3 +b^3+8ab=14 . $ Prove that
$$a+b \leq 2$$
2 replies
sqing
Today at 4:53 AM
sqing
an hour ago
2 var inquality
sqing   2
N an hour ago by sqing
Source: Own
Let $ a,b>0 . $ Prove that
$$\dfrac{1}{a^2+b^2}+\dfrac{1}{4ab} \ge \dfrac{\dfrac{3}{2} +\sqrt 2}{(a+b)^2} $$$$\dfrac{1}{a^3+b^3}+\dfrac{1}{4ab(a+b)}\ge \dfrac{\dfrac{7}{4} +\sqrt 3}{(a+b)^3} $$
2 replies
sqing
4 hours ago
sqing
an hour ago
Correlation between USAMO and CMO qualification
rantaccountcs   7
N an hour ago by elasticwealth
What is the correlation between cmo qualification and usamo qualification? What proportion of Canadian usamo qualifiers are cmo qualifiers and what proportion of cmo qualifiers are usamo qualifiers?
7 replies
1 viewing
rantaccountcs
Nov 2, 2023
elasticwealth
an hour ago
Problem 1
blug   7
N 2 hours ago by kokcio
Source: Polish Math Olympiad 2025 Finals P1
Find all $(a, b, c, d)\in \mathbb{R}$ satisfying
\[\begin{aligned}
\begin{cases}
    a+b+c+d=0,\\
    a^2+b^2+c^2+d^2=12,\\
    abcd=-3.\\
\end{cases}
\end{aligned}\]
7 replies
blug
Apr 4, 2025
kokcio
2 hours ago
Olympiad NT, Olympiad Geometry, and Math Modeling: Math4All 24-25 Classes
Geometry285   2
N 2 hours ago by Pengu14
Hi everyone!

[center]IMAGE[/center]

I am excited to announce that Math4All, a 501(c)3 nonprofit organization dedicated to making mathematics more accessible through problem-solving, will be holding Olympiad Number Theory, Olympiad Geometry, and Math Modeling for M3/MTFC classes in the 2024-2025 school year. Our instructors are highly experienced and have qualified for AIME, USA(J)MO, and/or MOP.

In particular, this year we are opening an exclusive one-of-a-kind Math Modeling for M3/Modeling the Future Challenge (MTFC) class, where students will learn how to apply mathematics and computer to solve problems in the real-world. If you’re interested in these topics and/or want to conduct productive research in the future, this class is for you!

To register, we ask that all prospective students fill out the corresponding registration form(s) for each of their classes at https://math4all.org/classes. You MUST fill your corresponding form(s) to be considered registered for the class. Once you have filled the Google Form, please react with an appropriate role in the ⁠classes-announcements channel so our instructors can contact you accordingly.

Our instructors have invested lots of time and effort into making this year's classes. We hope you won't miss out!
2 replies
1 viewing
Geometry285
Oct 21, 2024
Pengu14
2 hours ago
usamOOK geometry
KevinYang2.71   94
N 2 hours ago by SatisfiedMagma
Source: USAMO 2025/4, USAJMO 2025/5
Let $H$ be the orthocenter of acute triangle $ABC$, let $F$ be the foot of the altitude from $C$ to $AB$, and let $P$ be the reflection of $H$ across $BC$. Suppose that the circumcircle of triangle $AFP$ intersects line $BC$ at two distinct points $X$ and $Y$. Prove that $C$ is the midpoint of $XY$.
94 replies
KevinYang2.71
Mar 21, 2025
SatisfiedMagma
2 hours ago
Incircles
r00tsOfUnity   15
N Mar 29, 2025 by Mathgloggers
Source: 2024 AIME I #8
Eight circles of radius $34$ can be placed tangent to side $\overline{BC}$ of $\triangle ABC$ such that the first circle is tangent to $\overline{AB}$, subsequent circles are externally tangent to each other, and the last is tangent to $\overline{AC}$. Similarly, $2024$ circles of radius $1$ can also be placed along $\overline{BC}$ in this manner. The inradius of $\triangle ABC$ is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
15 replies
r00tsOfUnity
Feb 2, 2024
Mathgloggers
Mar 29, 2025
Incircles
G H J
G H BBookmark kLocked kLocked NReply
Source: 2024 AIME I #8
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r00tsOfUnity
695 posts
#1
Y by
Eight circles of radius $34$ can be placed tangent to side $\overline{BC}$ of $\triangle ABC$ such that the first circle is tangent to $\overline{AB}$, subsequent circles are externally tangent to each other, and the last is tangent to $\overline{AC}$. Similarly, $2024$ circles of radius $1$ can also be placed along $\overline{BC}$ in this manner. The inradius of $\triangle ABC$ is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Z K Y
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scannose
998 posts
#2 • 4 Y
Y by mahaler, amogussussybingchilling, r00tsOfUnity, crazyeyemoody907
alternative title: incircles make maa lose the game

$478 + 34x = 4046 + x = k$, solve for $\frac{k}{x}$
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r00tsOfUnity
695 posts
#3
Y by
Let $x=\cot(B/2)+\cot(C/2)$. Then $BC=x+2023\cdot 2=34x+14\cdot 34=rx$, where $r$ is the inradius of $\triangle ABC$. Dividing by $34$ gives $\tfrac{BC}{34}=\tfrac{x}{34}+119=x+14$, so $\tfrac{33}{34}x=105$ and $x=\tfrac{1190}{11}$. Then $r=34(1+\tfrac{14}{x})=34(1+\tfrac{11}{85})=34(\tfrac{96}{85})=\tfrac{192}{5}\implies\boxed{197}$.
Z K Y
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think4l
344 posts
#5
Y by
I already posted it here before -- https://artofproblemsolving.com/community/c5h3248098_tangent_circles_on_bc
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shendrew7
793 posts
#6
Y by
Denote the center of the leftmost circle as $I_B$, the center of the rightmost circle as $I_C$, and the incenter of $\triangle ABC$ as $I$. The key step is notice that $\triangle II_BI_C \sim \triangle IBC$.

Then, if we let the inradius be $r$, similarity ratios with the bases and corresponding heights of these two triangles, we get the two equations
\[\frac{r-34}{r} = \frac{14 \cdot 34}{BC}, \quad \frac{r-1}{r} = \frac{4046 \cdot 1}{BC},\]
from which solving gives $r = \frac{192}{5} \implies \boxed{197}$.
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OlympusHero
17019 posts
#7
Y by
scannose wrote:
alternative title: incircles make maa lose the game

$478 + 34x = 4046 + x = k$, solve for $\frac{k}{x}$

Yep exactly what I did
Z K Y
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r00tsOfUnity
695 posts
#8
Y by
think4l wrote:
I already posted it here before -- https://artofproblemsolving.com/community/c5h3248098_tangent_circles_on_bc

me losing the game (literally)
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fruitmonster97
2439 posts
#9
Y by
ugh this problem was confusing how many people put 014 over/under 999.5
Z K Y
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bluelinfish
1446 posts
#10
Y by
Observe that as $k$ varies linearly, the distance $d$ between the centers of circles of two circles of radius $k$, one tangent to $BC$ and $AB$ and one tangent to $BC$ and $AC$, also varies linearly. When $k = 1$, $d = 4046$, and when $k = 34$, $d = 476$. When $k = r$, $d = 0$. Therefore \[ \frac{r - 34}{476} = \frac{r - 1}{4046} \to r = \frac{192}{5} \to \boxed{197}.\]
Z K Y
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cinnamon_e
703 posts
#11
Y by
The triangles similar to $\triangle ABC$ with one side connecting the centers of the leftmost and rightmost circles satisfy
\[\frac{14\cdot 34}{BC}=\frac{r-34}{r}\qquad\text{and}\qquad\frac{4046}{BC}=\frac{r-1}{r},\]so
\[\frac{r-34}{r-1}=\frac{2}{17}\implies r=\frac{192}{5}\implies\boxed{197}.\]
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JD9
50 posts
#12
Y by
I don't remember the problem well but you can also just let $B$ be a right angle and it works. Makes calculations a little easier I guess?
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bachkieu
131 posts
#13
Y by
Similarity with a bunch of incenters works
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pi_is_3.14
1437 posts
#14
Y by
Video Solution

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Mr.Sharkman
496 posts
#16
Y by
Bruh this problem is literally trivial, but somehow I messed it up :wallbash_red:
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bomberdoodles
67 posts
#17
Y by
Let the incenter of $ABC$ be $I$, let the incircle be tangent to $BC$ at $H$, and let $IH$ = $r$. From similar triangles, $\frac {r-34}{34*14} = \frac {r-1}{1*4046}$ which yields $r= \frac {192}{5}$ $\implies m+n=197$ where we have used the fact that the ratio of heights in similar triangles is the same scale factor k as the ratio of sides.
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Mathgloggers
57 posts
#18
Y by
yea the key observation here was to note centers of extreme circles lies on the bisector of the main triangle.
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