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
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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
New geometry problem
titaniumfalcon   4
N an hour ago by titaniumfalcon
Post any solutions you have, with explanation or proof if possible, good luck!
4 replies
1 viewing
titaniumfalcon
Apr 3, 2025
titaniumfalcon
an hour ago
A rather difficult question
BeautifulMath0926   1
N an hour ago by jasperE3
I got a difficult equation for users to solve:
Find all functions f: R to R, so that to all real numbers x and y,
1+f(x)f(y)=f(x+y)+f(xy)+xy(x+y-2) holds.
1 reply
BeautifulMath0926
Today at 11:10 AM
jasperE3
an hour ago
Vietnam Mock Test
imnotgoodatmathsorry   0
an hour ago
Second Entrance Mock test for grade 10 specialized in Mathematics at High School for Gifted Students, HNUE, Vietnam
13/4/2025

Problem 1:
1) Let $a,b$ be positive reals. Prove that: $\frac{a}{a+1} + \frac{b}{b+2} < \frac{\sqrt{a} + \sqrt{b}}{2}$
2) In a small garden there are $3$ rabbits and $3$ carrots. Each rabbit will choose randomly a carrot to eat. Find the probability of a carrot was chose by less than $2$ rabbit.
Problem 2:
1) Solve the equation system: $(x+y)(x^2+y^2)=567$ and $\sqrt{xy}(x+y)^2=243$
2) Let $a,b,c$ be positive rational numbers such that: $a+b+c=2\sqrt{abc}$
Problem 3:
Let triangle $ABC$ ($\angle A$, $\angle B$, $\angle C < 90$) with excircle $(O)$ and incircle $(I)$. Incircle $(I)$ touches $BC,CA,AB$ at $D,E,F$. The excircle with the diameter of $AI$ cuts excircle $(O)$ at $K$ ($K \neq A$). $KD$ cuts the excircle with the diameter of $AI$ at $P$ ($P \neq K$) and $AK$ cuts $BC$ at $Q$. Prove that:
1) $\Delta KEC$ ~ $\Delta KFB$ and $KD$ is the bisector of $\angle BKC$
2) $AP \bot  BC$
3) $IQ$ is the tangent line of the excircle of $\Delta IBC$
Problem 4,5: (will type tomorrow)
0 replies
imnotgoodatmathsorry
an hour ago
0 replies
Combinatorics.
NeileshB   0
an hour ago
An odd integer is written in each cell of a 2009  2009 table. For 1  i  2009 let Ri be
the sum of the numbers in the ith row, and for 1  j  2009 let Cj be the sum of the
numbers in the jth column. Finally, let A be the product of the Ri, and B the product of
the Cj . Prove that A + B is different from zero.

I really need help on this. Can people give me hints? I don’t know where to start.
0 replies
NeileshB
an hour ago
0 replies
No more topics!
2014 preRMO p10, computational with ratios and areas
parmenides51   10
N Aug 16, 2023 by Serengeti22
In a triangle $ABC, X$ and $Y$ are points on the segments $AB$ and $AC$, respectively, such that $AX : XB = 1 : 2$ and $AY :YC = 2:1$. If the area of triangle $AXY$ is $10$, then what is the area of triangle $ABC$?
10 replies
parmenides51
Aug 9, 2019
Serengeti22
Aug 16, 2023
2014 preRMO p10, computational with ratios and areas
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G H BBookmark kLocked kLocked NReply
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parmenides51
30630 posts
#1 • 3 Y
Y by Samujjal101, Adventure10, Mango247
In a triangle $ABC, X$ and $Y$ are points on the segments $AB$ and $AC$, respectively, such that $AX : XB = 1 : 2$ and $AY :YC = 2:1$. If the area of triangle $AXY$ is $10$, then what is the area of triangle $ABC$?
Z K Y
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a87321dc0
917 posts
#2 • 1 Y
Y by Adventure10
Solution
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Wiskunder
2 posts
#3 • 1 Y
Y by Samujjal101
a87321dc0 wrote:
Solution

I did this problem by the following way which gave me the answer 45
$\frac{[AXY]}{[BXY]}=\frac{AX}{BX}=\frac12$ (Collinear Bases with common vertex (since their altitude is the same, the ratio of area is equal to the ratio of bases))
Therefore, $[BXY]=20,[BYA]=30$
$\frac{[BYA]}{[BYC]}=\frac{AY}{YC}=\frac21$
Therefore, $[BYC]=15$
$[ABC]=10+20+15 = 45$
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NTistrulove
183 posts
#4
Y by
a87321dc0 wrote:
Solution

Actually you did a mis calculation at the end.

@above you are right!!!
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stretfordend
13 posts
#5
Y by
yo @a97321dc0 blud you forgot to multiply with 1/2 before finding the area innit as area = 1/2*b*hsinA. So the answer should be 45 blud
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WildSquid
1915 posts
#6
Y by
$\boxed{90}$
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stretfordend
13 posts
#7
Y by
@[quotehttps://latex.artofproblemsolving.com/3/5/2/3522f4b289672dfc0894a6d26a239cb91efd7f0b.png][/quote] bruv its 45 innit
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Johan_Liebert
4 posts
#8
Y by
Draw median XD to AY in AXY.
From this, [AXD] = 5
Notice AX/AB=AD/CA = 1/3. AXD is similar to ABC
(1/3)^2 = 5/[ABC]
[ABC]=45
This post has been edited 1 time. Last edited by Johan_Liebert, Aug 15, 2023, 5:08 PM
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torch
1001 posts
#9
Y by
Wiskunder wrote:
a87321dc0 wrote:
Solution

I did this problem by the following way which gave me the answer 45
$\frac{[AXY]}{[BXY]}=\frac{AX}{BX}=\frac12$ (Collinear Bases with common vertex (since their altitude is the same, the ratio of area is equal to the ratio of bases))
Therefore, $[BXY]=20,[BYA]=30$
$\frac{[BYA]}{[BYC]}=\frac{AY}{YC}=\frac21$
Therefore, $[BYC]=15$
$[ABC]=10+20+15 = 45$

congrats on first post!
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am07
316 posts
#10
Y by
I got $45$ as the answer
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Serengeti22
1133 posts
#11
Y by
I got 45 to
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