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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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Inequalities
sqing   6
N 17 minutes ago by sqing
Let $a,b,c> 0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1.$ Prove that
$$ (1-abc) (1-a)(1-b)(1-c) \ge 208 $$$$ (1+abc) (1-a)(1-b)(1-c) \le -224 $$$$(1+a^2b^2c^2) (1-a)(1-b)(1-c) \le -5840 $$
6 replies
sqing
Jul 12, 2024
sqing
17 minutes ago
J
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A problem with a rectangle
Raul_S_Baz   11
N 20 minutes ago by Raul_S_Baz
On the sides AB and AD of the rectangle ABCD, points M and N are taken such that MB = ND. Let P be the intersection of BN and CD, and Q be the intersection of DM and CB. How can we prove that PQ || MN?
IMAGE
11 replies
Raul_S_Baz
Apr 26, 2025
Raul_S_Baz
20 minutes ago
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(14n+25)/(2n+1) 'is a perfect square - Portugal OPM 2017 p1
parmenides51   3
N an hour 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
an hour 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 2 hours 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
2 hours ago
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No more topics!
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a A+b B+cC >= aB+b C+cA (1989 Lomonosov Tournament )
parmenides51   3
N May 15, 2021 by khanhnx
Let $a,b,c$ be the lengths of the sides of the triangle, $A,B,C$ are the values of opposite angles.
Prove that $a A+b B+cC \ge aB+b C+cA$.
3 replies
parmenides51
Apr 29, 2021
khanhnx
May 15, 2021
J
a A+b B+cC >= aB+b C+cA (1989 Lomonosov Tournament )
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Reveal topic tags
geometry
geometric inequality
Lomonosov Tournament
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parmenides51
30650 posts
parmenides51
#1 h  Apr 29, 2021, 7:36 PM
Y by
Let $a,b,c$ be the lengths of the sides of the triangle, $A,B,C$ are the values of opposite angles.
Prove that $a A+b B+cC \ge aB+b C+cA$.
This post has been edited 2 times. Last edited by parmenides51, May 20, 2022, 7:47 AM
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khanhnx
1618 posts
khanhnx
#2 h  May 15, 2021, 2:28 AM
Y by
Suppose $a \ge b \ge c$ then $A \ge B \ge C$. So by rearrange inequality, we have $aA + bB + cC \ge aB + bC + cA$. The equality hold when $\triangle ABC$ is equilateral
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Quantum_fluctuations
1282 posts
Quantum_fluctuations
#3 h  May 15, 2021, 5:08 AM
Y by
khanhnx wrote:
Suppose $a \ge b \ge c$ then $A \ge B \ge C$.

You can't suppose that without loss of generality because inequality is not symmetric in $a,b,c$ or $A,B,C$.
This post has been edited 1 time. Last edited by Quantum_fluctuations, May 15, 2021, 5:08 AM
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khanhnx
1618 posts
khanhnx
#4 h  May 15, 2021, 9:17 AM
Y by
Quantum_fluctuations wrote:
khanhnx wrote:
Suppose $a \ge b \ge c$ then $A \ge B \ge C$.

You can't suppose that without loss of generality because inequality is not symmetric in $a,b,c$ or $A,B,C$.

Why ? Can you explain more ?
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