Stay ahead of learning milestones! Enroll in a class over the summer!

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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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All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
1 viewing
jlacosta
Apr 2, 2025
0 replies
Inequalities
sqing   4
N an hour ago by SunnyEvan
Let $ a,b,c,d\geq 0 ,a-b+d=21 $ and $ a+3b+4c=101 $. Prove that
$$ - \frac{1681}{3}\leq   ab - cd \leq 820$$$$ - \frac{16564}{9}\leq   ac -bd \leq 420$$$$ - \frac{10201}{48}\leq ad- bc \leq\frac{1681}{3}$$
4 replies
sqing
Yesterday at 3:53 AM
SunnyEvan
an hour ago
JEE Related ig?
mikkymini2   10
N 5 hours ago by Idiot_of_the64squares
Hey everyone,

Just wanted to see if there are any other JEE aspirants on this forum currently prepping for it[mention year if you can]

I am actually entering 10th this year and have decided to try for it...So this year is just going to go in me strengthening my math (IOQM level (heard its enough till Mains part, so will start from there) for the problem solving part, and learn some topics from 11th and 12th as well)

It would be great to connect with others who are going through the same thing - share study strategies, tips, resources, discuss, and maybe even form study groups(not sure how to tho :maybe: ) and motivate each other ig?. :D
So yea, cya later
10 replies
mikkymini2
Thursday at 2:54 PM
Idiot_of_the64squares
5 hours ago
Inequalities
sqing   0
5 hours ago
Let $ a,b,c\in [0,1] $ . Prove that
$$(a+b+c)\left(\frac{1}{a^2+3}+\frac{2}{b^2+2}+\frac{2}{c^2+2}\right)\leq   \frac{19}{4}$$$$(a+b+c)\left(\frac{1}{a^2+ 4}+\frac{2}{b^2+2}+\frac{2}{c^2+2}\right)\leq   \frac{23}{5}$$$$(a+b+c)\left(\frac{1}{a^2+ \frac{5}{2}}+\frac{2}{b^2+2}+\frac{2}{c^2+2}\right)\leq   \frac{34}{7}$$$$(a+b+c)\left(\frac{1}{a^2+ \frac{7}{2}}+\frac{2}{b^2+2}+\frac{2}{c^2+2}\right)\leq   \frac{14}{3}$$
0 replies
sqing
5 hours ago
0 replies
lcm(1,2,3,...,n)
lgx57   5
N 6 hours ago by Kempu33334
Let $M=\operatorname{lcm}(1,2,3,\cdots,n)$.Estimate the range of $M$.
5 replies
lgx57
Apr 9, 2025
Kempu33334
6 hours ago
No more topics!
area chasing, two cevians in a triangle (2007 Euler Teacher's Olympiad II p5)
parmenides51   4
N Jul 6, 2020 by Bole
Points $A_1$ and $B_1$ lie on the sides $BC$ and $AC$ of the triangle $ABC$, respectively, segments $AA_1$ and $BB_1$ intersect at $K$ (see picture). Find the area of the quadrangle $A_1CB_1K$, if it is known that the area of triangle $A_1BK$ is $5$, the area of the triangle $AB_1K$ is $8$, and the area of the triangle $ABK$ is $10$.
IMAGE
4 replies
parmenides51
Jul 6, 2020
Bole
Jul 6, 2020
area chasing, two cevians in a triangle (2007 Euler Teacher's Olympiad II p5)
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G H BBookmark kLocked kLocked NReply
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parmenides51
30630 posts
#1 • 2 Y
Y by Mango247, Mango247
Points $A_1$ and $B_1$ lie on the sides $BC$ and $AC$ of the triangle $ABC$, respectively, segments $AA_1$ and $BB_1$ intersect at $K$ (see picture). Find the area of the quadrangle $A_1CB_1K$, if it is known that the area of triangle $A_1BK$ is $5$, the area of the triangle $AB_1K$ is $8$, and the area of the triangle $ABK$ is $10$.
https://cdn.artofproblemsolving.com/attachments/3/c/8016edf9ac0da56f7d4c0f4a669da739b80c4c.png
This post has been edited 2 times. Last edited by parmenides51, Jul 6, 2020, 1:44 PM
Reason: typos, there is no K_1
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parmenides51
30630 posts
#2 • 2 Y
Y by Mango247, Mango247
posted for the image link
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HumanCalculator9
6231 posts
#3 • 1 Y
Y by Mango247
What is this $K_1$ you speak of

Nevermind
This post has been edited 1 time. Last edited by HumanCalculator9, Jul 6, 2020, 1:47 PM
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HumanCalculator9
6231 posts
#4
Y by
It is this right?
Z K Y
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Bole
576 posts
#5
Y by
Sol: We can draw a line from C to K. We call the area of triangle CB_1K x and the area of triangle CA_1K y. Since we know the ratio of the area of AKB to A_1KB is 2:1, we know that the ratio of AKC to A_1KC is also 2:1 so we get the equation (8+x)/2=y then the ratio of AKB_1 to AKB is 4:5 we then know the ratio of KB_1C to CKB is also 4:5 so then we get the equation x/4=(5+y)/5. Now we have a system of 2 equations. Solving this we get x=10, y=12 so the area is 10+12=22.
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