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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)!

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[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
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[*]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
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(14n+25)/(2n+1) 'is a perfect square - Portugal OPM 2017 p1
parmenides51   3
N 27 minutes ago by justaguy_69
Determine all integer values of n for which the number $\frac{14n+25}{2n+1}$ 'is a perfect square.
3 replies
parmenides51
May 15, 2024
justaguy_69
27 minutes ago
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(3x-1)^2/x+(3y-1)^2/y >=1, for x,y>0, x+y=1 Austria Beginners' 2010 p3
parmenides51   22
N an hour ago by justaguy_69
Let $x$ and $y$ be positive real numbers with $x + y =1 $. Prove that
$$\frac{(3x-1)^2}{x}+ \frac{(3y-1)^2}{y} \ge1.$$For which $x$ and $y$ equality holds?

(K. Czakler, GRG 21, Vienna)
22 replies
parmenides51
Oct 3, 2021
justaguy_69
an hour ago
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New but easy
ZETA_in_olympiad   2
N an hour ago by jasperE3
Find all functions $f:\mathbb R\to \mathbb R$ such that $$f(f(x)+f(y))=xf(y)+yf(x)$$for all $x,y\in \mathbb R.$
2 replies
ZETA_in_olympiad
Oct 1, 2022
jasperE3
an hour ago
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2025 CMIMC team p7, rephrased
scannose   4
N 2 hours ago by scannose
In the expansion of $(x^2 + x + 1)^{2024}$, find the number of terms with coefficient divisible by $3$.
4 replies
scannose
Apr 18, 2025
scannose
2 hours ago
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angle bisector of <BAD _|_ KL, AB+BK = AD+DL, #ABCD (2021 Polish JMO R2 p4)
parmenides51   2
N Jul 13, 2023 by Ramanujanmath
Points $K$ and $L$ are on the sides $BC$ and $CD$, respectively of the parallelogram $ABCD$, such that $AB + BK = AD + DL$. Prove that the bisector of angle $BAD$ is perpendicular to the line $KL$.
2 replies
parmenides51
Apr 8, 2021
Ramanujanmath
Jul 13, 2023
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angle bisector of <BAD _|_ KL, AB+BK = AD+DL, #ABCD (2021 Polish JMO R2 p4)
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Reveal topic tags
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parallelogram
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parmenides51
30650 posts
parmenides51
#1 h  Apr 8, 2021, 5:30 PM
Y by
Points $K$ and $L$ are on the sides $BC$ and $CD$, respectively of the parallelogram $ABCD$, such that $AB + BK = AD + DL$. Prove that the bisector of angle $BAD$ is perpendicular to the line $KL$.
This post has been edited 1 time. Last edited by parmenides51, Nov 28, 2022, 8:54 PM
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math_maniac
19 posts
math_maniac
#2 h  Apr 8, 2021, 5:40 PM
Y by
AB + BK + CK = AD + DL + LC
therefore CL = CK
giving angle KLC = angle LKC
angle DLK = angle BKL
quadrilateral ADLI and ABKI(where I is the foot of bisector) have three angles equal so fourth must be equal and they form a line so it is 90 degree

P.S. not written in code so hope u can understand
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Ramanujanmath
18 posts
Ramanujanmath
#3 h  Jul 13, 2023, 12:38 PM
Y by
Continue the bisector to meet $DC$ at $M$ and $KL$ at $O$. Now ,
$AB + BK = AD + DL$
$AB + AD - KC = AD + AB - LC$
$KC = LC$ so now simple angle chasing
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