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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
[*]April 8th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS State Discussion
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]
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0 replies
jlacosta
Apr 2, 2025
0 replies
AMC 10/AIME Study Forum
PatTheKing806   140
N an hour ago by PikaPika999
[center]

Me (PatTheKing806) and EaZ_Shadow have created a AMC 10/AIME Study Forum! Hopefully, this forum wont die quickly. To signup, do /join or \join.

Click here to join! (or do some pushups) :P

People should join this forum if they are wanting to do well on the AMC 10 next year, trying get into AIME, or loves math!
140 replies
PatTheKing806
Mar 27, 2025
PikaPika999
an hour ago
memorize your 60 120 degree triangles
OronSH   14
N an hour ago by Apple_maths60
Source: 2024 AMC 12A #19
Cyclic quadrilateral $ABCD$ has lengths $BC=CD=3$ and $DA=5$ with $\angle CDA=120^\circ$. What is the length of the shorter diagonal of $ABCD$?

$
\textbf{(A) }\frac{31}7 \qquad
\textbf{(B) }\frac{33}7 \qquad
\textbf{(C) }5 \qquad
\textbf{(D) }\frac{39}7 \qquad
\textbf{(E) }\frac{41}7 \qquad
$
14 replies
OronSH
Nov 7, 2024
Apple_maths60
an hour ago
2025 Math and AI 4 Girls Competition: Win Up To $1,000!!!
audio-on   25
N an hour ago by mathandai
Join the 2025 Math and AI 4 Girls Competition for a chance to win up to $1,000!

Hey Everyone, I'm pleased to announce the dates for the 2025 MA4G Competition are set!
Applications will open on March 22nd, 2025, and they will close on April 26th, 2025 (@ 11:59pm PST).

Applicants will have one month to fill out an application with prizes for the top 50 contestants & cash prizes for the top 20 contestants (including $1,000 for the winner!). More details below!

Eligibility:
The competition is free to enter, and open to middle school female students living in the US (5th-8th grade).
Award recipients are selected based on their aptitude, activities and aspirations in STEM.

Event dates:
Applications will open on March 22nd, 2025, and they will close on April 26th, 2025 (by 11:59pm PST)
Winners will be announced on June 28, 2025 during an online award ceremony.

Application requirements:
Complete a 12 question problem set on math and computer science/AI related topics
Write 2 short essays

Prizes:
1st place: $1,000 Cash prize
2nd place: $500 Cash prize
3rd place: $300 Cash prize
4th-10th: $100 Cash prize each
11th-20th: $50 Cash prize each
Top 50 contestants: Over $50 worth of gadgets and stationary


Many thanks to our current and past sponsors and partners: Hudson River Trading, MATHCOUNTS, Hewlett Packard Enterprise, Automation Anywhere, JP Morgan Chase, D.E. Shaw, and AI4ALL.

Math and AI 4 Girls is a nonprofit organization aiming to encourage young girls to develop an interest in math and AI by taking part in STEM competitions and activities at an early age. The organization will be hosting an inaugural Math and AI 4 Girls competition to identify talent and encourage long-term planning of academic and career goals in STEM.

Contact:
mathandAI4girls@yahoo.com

For more information on the competition:
https://www.mathandai4girls.org/math-and-ai-4-girls-competition

More information on how to register will be posted on the website. If you have any questions, please ask here!


25 replies
2 viewing
audio-on
Jan 26, 2025
mathandai
an hour ago
Those who know
centslordm   29
N 2 hours ago by Apple_maths60
Source: 2024 AMC 12A #15
The roots of $x^3 + 2x^2 - x + 3$ are $p, q,$ and $r.$ What is the value of \[(p^2 + 4)(q^2 + 4)(r^2 + 4)?\]
$\textbf{(A) } 64 \qquad \textbf{(B) } 75 \qquad \textbf{(C) } 100 \qquad \textbf{(D) } 125 \qquad \textbf{(E) } 144$
29 replies
centslordm
Nov 7, 2024
Apple_maths60
2 hours ago
No more topics!
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
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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 $m+n$.
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scannose
996 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 $r=34(1+\tfrac{14}{x})=34(1+\tfrac{11}{85})=34(\tfrac{96}{85})=\tfrac{192}{5}\implies\boxed{197}$.
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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
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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
2438 posts
#9
Y by
ugh this problem was confusing how many people put 014 over/under 999.5
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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 \[ \frac{r - 34}{476} = \frac{r - 1}{4046} \to r = \frac{192}{5} \to \boxed{197}.\]
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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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