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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
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IMO ShortList 2002, geometry problem 2
orl   27
N 4 minutes ago by ZZzzyy
Source: IMO ShortList 2002, geometry problem 2
Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \]
27 replies
orl
Sep 28, 2004
ZZzzyy
4 minutes ago
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FE based on (x+1)(y+1)
CrazyInMath   4
N 19 minutes ago by jasperE3
Source: 2023 CK Summer MSG I-A
Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that \[f(xf(y)+f(x+y)+1)=(y+1)f(x+1)\]holds for all $x,y\in\mathbb{R}$.

Proposed by owoovo.shih and CrazyInMath
4 replies
CrazyInMath
Aug 14, 2023
jasperE3
19 minutes ago
J
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Multiplicative polynomial exactly 2025 times
Assassino9931   1
N 31 minutes ago by sami1618
Source: Bulgaria Balkan MO TST 2025
Does there exist a polynomial $P$ on one variable with real coefficients such that the equation $P(xy) = P(x)P(y)$ has exactly $2025$ ordered pairs $(x,y)$ as solutions?
1 reply
Assassino9931
Yesterday at 10:14 PM
sami1618
31 minutes ago
J
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Geometric inequality problem
mathlover1231   1
N 34 minutes ago by Double07
Given an acute triangle ABC, where H and O are the orthocenter and circumcenter, respectively. Point K is the midpoint of segment AH, and ℓ is a line through O. Points P and Q are the projections of B and C onto ℓ. Prove that KP + KQ ≥BC
1 reply
mathlover1231
2 hours ago
Double07
34 minutes ago
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No more topics!
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Inspired by old results
sqing   2
N Apr 2, 2025 by sqing
Source: Own
Let $ a,b $ be reals such that $ a^2+b^2 +ab =1.$ Prove that $$9\geq (a^2-a+1)(b^2-b+1) (a^2-ab+b^2) \geq \frac{19-8\sqrt 3}{27}$$Let $ a,b $ be reals such that $ a^3+b^3 -2ab =1.$ Prove that $$ (a^2-a+1)(b^2-b+1) (a^2-ab+b^2) \geq 1$$
2 replies
sqing
Apr 2, 2025
sqing
Apr 2, 2025
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Inspired by old results
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Reveal topic tags
inequalities proposed
algebra
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Source: Own
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sqing
41518 posts
sqing
#1 h  Apr 2, 2025, 2:32 AM
Y by
Let $ a,b $ be reals such that $ a^2+b^2 +ab =1.$ Prove that $$9\geq (a^2-a+1)(b^2-b+1) (a^2-ab+b^2) \geq \frac{19-8\sqrt 3}{27}$$Let $ a,b $ be reals such that $ a^3+b^3 -2ab =1.$ Prove that $$ (a^2-a+1)(b^2-b+1) (a^2-ab+b^2) \geq 1$$
This post has been edited 1 time. Last edited by sqing, Apr 2, 2025, 2:47 AM
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lbh_qys
506 posts
lbh_qys
#2 h  Apr 2, 2025, 2:36 AM • 1 Y
Y by aidan0626
sqing wrote:
Let $ a,b $ be reals such that $ a+b-ab =1.$ Prove that $$(a^2-a+1)(b^2-b+1) (a^2-ab+b^2) \geq \frac{9}{16}$$

$$a+b-ab=1 \Rightarrow a = 1 \vee b = 1$$
Z K Y
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sqing
41518 posts
sqing
#3 h  Apr 2, 2025, 2:40 AM
Y by
Sorry .Thank lbh_qys.
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