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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Thursday at 11:16 PM
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
Thursday at 11:16 PM
0 replies
IMO Genre Predictions
ohiorizzler1434   4
N 2 minutes ago by whwlqkd
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
4 replies
ohiorizzler1434
3 hours ago
whwlqkd
2 minutes ago
Find the product
sqing   0
11 minutes ago
Source: Ecrin_eren
The roots of $ x^3 - 2x^2 - 11x + k=0 $ are $r_1, r_2,  r_3 $ and $ r_1+2 r_2+3 r_3= 0.$ Find the product of all possible values of $ k .$
0 replies
sqing
11 minutes ago
0 replies
Problem 6
SlovEcience   0
28 minutes ago
Given two points A and B on the unit circle. The tangents to the circle at A and B intersect at point P. Then:
\[ p = \frac{2ab}{a + b} \], \[ p, a, b \in \mathbb{C} \]
0 replies
SlovEcience
28 minutes ago
0 replies
cubefree divisibility
DottedCaculator   59
N 31 minutes ago by SimplisticFormulas
Source: 2021 ISL N1
Find all positive integers $n\geq1$ such that there exists a pair $(a,b)$ of positive integers, such that $a^2+b+3$ is not divisible by the cube of any prime, and $$n=\frac{ab+3b+8}{a^2+b+3}.$$
59 replies
DottedCaculator
Jul 12, 2022
SimplisticFormulas
31 minutes ago
No more topics!
Concurrency from exterior equilateral triangles
WakeUp   2
N Apr 12, 2025 by Nari_Tom
Source: Baltic Way 2003
Equilateral triangles $AMB,BNC,CKA$ are constructed on the exterior of a triangle $ABC$. The perpendiculars from the midpoints of $MN, NK, KM$ to the respective lines $CA, AB, BC$ are constructed. Prove that these three perpendiculars pass through a single point.
2 replies
WakeUp
Nov 7, 2010
Nari_Tom
Apr 12, 2025
Concurrency from exterior equilateral triangles
G H J
Source: Baltic Way 2003
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WakeUp
1347 posts
#1 • 2 Y
Y by Adventure10, Mango247
Equilateral triangles $AMB,BNC,CKA$ are constructed on the exterior of a triangle $ABC$. The perpendiculars from the midpoints of $MN, NK, KM$ to the respective lines $CA, AB, BC$ are constructed. Prove that these three perpendiculars pass through a single point.
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Luis González
4148 posts
#2 • 2 Y
Y by Adventure10, Mango247
Let $L$ be the reflection of $N$ across line $BC.$ Since $\angle LBC=\angle MBA=60^{\circ},$ we deduce that $\angle MBL=\angle ABC$ and because of $MB=MA$ and $BL=BC,$ it follows that $\triangle ABC \cong \triangle MBL$ by SAS criterion $\Longrightarrow$ $ML=AC=AK.$ Similar reasoning yields $KL=AM$ $\Longrightarrow$ $AMLK$ is a parallelogram, thus $AL$ and $MK$ bisect each other at $U.$ If $H_a,M_a$ denote the foot of the A-altitude and midpoint of $BC,$ then perpendicular $\ell_a$ from $U$ to $BC$ is the midline of the trapezoid $ALM_aH_a$ $\Longrightarrow$ $\ell_a$ passes through the 9-point center of $\triangle ABC.$ Likewise, perpendiculars from midpoints of $MN,NK$ to $CA,AB$ pass through the 9-point center of $\triangle ABC.$

P.S. The result is still true for similar isosceles triangles AMB,BNC,CKA and the proof is similar.
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Nari_Tom
117 posts
#3
Y by
I have different proof using Carnot's theorem.
First let $a,b,c$ denotes the sides of triangle as usual. Let $D,E,F$ be the midpoints of $MK,KN,NM$, respectively. In $\triangle BMK$, $BD$ is a median $\implies$ $BD^2=\frac{2c^2+2BK^2-MK^2}{4}$. Similarly we have $CD^2=\frac{2CM^2+2b^2-MK^2}{4}$ $\implies$ $BD^2-CD^2=\frac{c^2-b^2+BK^2-CM^2}{2}$.

So $\sum {BD^2-CD^2}=\sum_{cyc}{\frac{c^2-b^2}{2}}+\sum_{cyc} {\frac{BK^2-CM^2}{2}}=0$, and we are done by Carnot's theorem.
This post has been edited 1 time. Last edited by Nari_Tom, Apr 12, 2025, 7:49 AM
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