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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
+1 w
jlacosta
May 1, 2025
0 replies
Popular children at camp with algebra and geometry
Assassino9931   2
N a few seconds ago by cj13609517288
Source: RMM Shortlist 2024 C3
Fix an odd integer $n\geq 3$. At a maths camp, there are $n^2$ children, each of whom selects
either algebra or geometry as their favourite topic. At lunch, they sit at $n$ tables, with $n$ children
on each table, and start talking about mathematics. A child is said to be popular if their favourite
topic has a majority at their table. For dinner, the students again sit at $n$ tables, with $n$ children
on each table, such that no two children share a table at both lunch and dinner. Determine the
minimal number of young mathematicians who are popular at both mealtimes. (The minimum is across all sets of topic preferences and seating arrangements.)
2 replies
Assassino9931
Friday at 11:07 PM
cj13609517288
a few seconds ago
Interesting inequalities
sqing   1
N 7 minutes ago by sqing
Source: Own
Let $ a,b> 0 $ and $  a^2+ab+b^2=a+b   $. Prove that
$$   \frac{a }{2b^2+1}+ \frac{b }{2a^2+1}+ \frac{1}{2ab+1} \geq \frac{21}{17}$$Let $ a,b> 0 $ and $ a^2+ab+b^2=a+b+1   $. Prove that
$$   \frac{a }{2b^2+1}+ \frac{b }{2a^2+1}+ \frac{1}{2ab+1} \geq1$$
1 reply
1 viewing
sqing
28 minutes ago
sqing
7 minutes ago
Interesting inequalities
sqing   0
18 minutes ago
Source: Own
Let $ a,b> 0 $ and $  a^2+ab+b^2=k(a+b)   $. Prove that
$$   \frac{a }{2b^2+1}+ \frac{b }{2a^2+1}+ \frac{1}{2ab+1} \geq \frac{12k+9}{8k^2+9}$$Where $ k\in N^+.$
Let $ a,b> 0 $ and $  a^2+ab+b^2=3(a+b)   $. Prove that
$$   \frac{a }{2b^2+1}+ \frac{b }{2a^2+1}+ \frac{1}{2ab+1} \geq \frac{5}{9}$$
0 replies
sqing
18 minutes ago
0 replies
Queue geo
vincentwant   6
N 25 minutes ago by ihategeo_1969
Let $ABC$ be an acute scalene triangle with circumcenter $O$. Let $Y, Z$ be the feet of the altitudes from $B, C$ to $AC, AB$ respectively. Let $D$ be the midpoint of $BC$. Let $\omega_1$ be the circle with diameter $AD$. Let $Q\neq A$ be the intersection of $(ABC)$ and $\omega$. Let $H$ be the orthocenter of $ABC$. Let $K$ be the intersection of $AQ$ and $BC$. Let $l_1,l_2$ be the lines through $Q$ tangent to $\omega,(AYZ)$ respectively. Let $I$ be the intersection of $l_1$ and $KH$. Let $P$ be the intersection of $l_2$ and $YZ$. Let $l$ be the line through $I$ parallel to $HD$ and let $O'$ be the reflection of $O$ across $l$. Prove that $O'P$ is tangent to $(KPQ)$.
6 replies
vincentwant
Apr 30, 2025
ihategeo_1969
25 minutes ago
No more topics!
trolling geometry problem
iStud   4
N Apr 22, 2025 by GreenTea2593
Source: Monthly Contest KTOM April P3 Essay
Given a cyclic quadrilateral $ABCD$ with $BC<AD$ and $CD<AB$. Lines $BC$ and $AD$ intersect at $X$, and lines $CD$ and $AB$ intersect at $Y$. Let $E,F,G,H$ be the midpoints of sides $AB,BC,CD,DA$, respectively. Let $S$ and $T$ be points on segment $EG$ and $FH$, respectively, so that $XS$ is the angle bisector of $\angle{DXA}$ and $YT$ is the angle bisector of $\angle{DYA}$. Prove that $TS$ is parallel to $BD$ if and only if $AC$ divides $ABCD$ into two triangles with equal area.
4 replies
iStud
Apr 21, 2025
GreenTea2593
Apr 22, 2025
trolling geometry problem
G H J
Source: Monthly Contest KTOM April P3 Essay
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iStud
268 posts
#1
Y by
Given a cyclic quadrilateral $ABCD$ with $BC<AD$ and $CD<AB$. Lines $BC$ and $AD$ intersect at $X$, and lines $CD$ and $AB$ intersect at $Y$. Let $E,F,G,H$ be the midpoints of sides $AB,BC,CD,DA$, respectively. Let $S$ and $T$ be points on segment $EG$ and $FH$, respectively, so that $XS$ is the angle bisector of $\angle{DXA}$ and $YT$ is the angle bisector of $\angle{DYA}$. Prove that $TS$ is parallel to $BD$ if and only if $AC$ divides $ABCD$ into two triangles with equal area.
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iStud
268 posts
#2
Y by
i can prove the $\Longleftarrow$ direction but not pretty sure with the $\Longrightarrow$ direction, can anyone help me?
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mathuz
1520 posts
#3
Y by
Note that $EFGH$ is a parallelogram and $FG\parallel EH \parallel BD$. Therefore, $TS\parallel BD$ iff $ES:SG = HT:TF$.

Assume now $TS\parallel BD$. Then $ES:SG = HT:TF$.
We have $\triangle XCD \sim \triangle XAB$ and $\triangle YBC\sim \triangle YDA$. This implies $XS$ is the angle bisector of $\angle EXG$ and $YT$ is the angle bisector of $\angle HYF$. Therefore, we have \[ \frac{EX}{GX} = \frac{ES}{SG} = \frac{HT}{TF} = \frac{HY}{FY}. \]Again, by the similarity this means \[ \frac{AB}{CD} = \frac{DA}{BC} \quad \Rightarrow \quad AB\cdot BC = CD\cdot DA, \]which essentially means the areas of $ABC$ and $ADC$ are equal.

This gives $\Longrightarrow $ direction.
This post has been edited 1 time. Last edited by mathuz, Apr 22, 2025, 1:35 AM
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iStud
268 posts
#4
Y by
mathuz wrote:
Note that $EFGH$ is a parallelogram and $FG\parallel EH \parallel BD$. Therefore, $TS\parallel BD$ iff $ES:SG = HT:TF$.

Assume now $TS\parallel BD$. Then $ES:SG = HT:TF$.
We have $\triangle XCD \sim \triangle XAB$ and $\triangle YBC\sim \triangle YDA$. This implies $XS$ is the angle bisector of $\angle EXG$ and $YT$ is the angle bisector of $\angle HYF$. Therefore, we have \[ \frac{EX}{GX} = \frac{ES}{SG} = \frac{HT}{TF} = \frac{HY}{FY}. \]Again, by the similarity this means \[ \frac{AB}{CD} = \frac{DA}{BC} \quad \Rightarrow \quad AB\cdot BC = CD\cdot DA, \]which essentially means the areas of $ABC$ and $ADC$ are equal.

This gives $\Longrightarrow $ direction.

demn, all of my Menelause bash worth nothing in front of this dude's solution
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GreenTea2593
236 posts
#5
Y by
Oops, I took this from 2016 Japan MO Finals Problem 2
This post has been edited 1 time. Last edited by GreenTea2593, Apr 22, 2025, 3:12 AM
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