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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. 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 upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
do not bring back downvotes
alcumusftwgrind   0
a minute ago
AOPS forums will not be the same with downvotes. When people are downvoted they will be discouraged to post more, thus reducing the size and activity of our community, as well as increase toxicity

It's not worth wrecking the morale and the culture of the AOPS community by bringing back downvotes.

Here's AOPS's reason why they removed downvotes:
https://artofproblemsolving.com/community/c10t68191f10h1224945_more_information_on_why_we_removed_downvotes
0 replies
alcumusftwgrind
a minute ago
0 replies
Close to JMO, but not close enough
isache   0
43 minutes ago
Im currently a freshman in hs, and i rlly wanna make jmo in sophmore yr. Ive been cooking at in-person competitions recently (ucsd hmc, scmc, smt, mathcounts) but I keep fumbling jmo. this yr i had a 133.5 on 10b and a 9 on aime. How do i get that up by 20 points to a 240?
0 replies
isache
43 minutes ago
0 replies
usamOOK geometry
KevinYang2.71   108
N an hour ago by ray66
Source: USAMO 2025/4, USAJMO 2025/5
Let $H$ be the orthocenter of acute triangle $ABC$, let $F$ be the foot of the altitude from $C$ to $AB$, and let $P$ be the reflection of $H$ across $BC$. Suppose that the circumcircle of triangle $AFP$ intersects line $BC$ at two distinct points $X$ and $Y$. Prove that $C$ is the midpoint of $XY$.
108 replies
KevinYang2.71
Mar 21, 2025
ray66
an hour ago
Bring Back Downvotes
heheman   18
N an hour ago by Evanlovemath
i would like to start a petition to bring back downvote, it you agree then write "bbd $    $" in threads
18 replies
heheman
5 hours ago
Evanlovemath
an hour ago
How do I prove this? - Sets and symmetric difference
smadadi1000   1
N 4 hours ago by KSH31415
For sets A, B and C, prove that (A Δ B) Δ C = (A Δ C) Δ (A \ B).

The textbook - Proofs by Jay Cummings - had this definition for symmetric difference:
A Δ B = ( A U B)/(A ∩ B)

this is exercise 3.37 (e).
1 reply
smadadi1000
Yesterday at 3:23 PM
KSH31415
4 hours ago
help me ..
exoticc   2
N Yesterday at 5:00 PM by sodiumaka
Find all pairs of functions (f,g) : R->R that satisfy:
f(1)=2025;
g(f(x+y))+2x+y-1=f(x)+(2x+y)g(y) , ∀x,y∈R
2 replies
exoticc
Yesterday at 9:26 AM
sodiumaka
Yesterday at 5:00 PM
Inequalities
sqing   20
N Yesterday at 3:32 PM by ytChen
Let $ a,b,c\geq 0 ,a+b+c\leq 3. $ Prove that
$$a^2+b^2+c^2+ab +2ca+2bc +  abc \leq \frac{251}{27}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{2}{5}abc  \leq \frac{4861}{540}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{7}{20}abc  \leq \frac{2381411}{26460}$$
20 replies
sqing
May 21, 2025
ytChen
Yesterday at 3:32 PM
21st PMO National Orals #9
yes45   0
Yesterday at 3:22 PM
In square $ABCD$, $P$ and $Q$ are points on sides $CD$ and $BC$, respectively, such that $\angle{APQ} = 90^\circ$. If $AP = 4$ and $PQ = 3$, find the area of $ABCD$.

Answer Confirmation
Solution
0 replies
yes45
Yesterday at 3:22 PM
0 replies
Inequalities
sqing   0
Yesterday at 2:31 PM
Let $ a,b> 0 $ and $2a+2b+ab=5. $ Prove that
$$ \frac{a^4}{b^4}+\frac{1}{a^4}+42ab-a^4\geq  43$$$$ \frac{a^5}{b^5}+\frac{1}{a^5}+64ab-a^5\geq  65$$$$ \frac{a^6}{b^6}+\frac{1}{a^6}+90ab-a^6\geq  91$$$$ \frac{a^7}{b^7}+\frac{1}{a^7}+121ab-a^7\geq  122$$
0 replies
sqing
Yesterday at 2:31 PM
0 replies
Original Problem (qrxz17)
qrxz17   0
Yesterday at 1:32 PM
Problem. Suppose that
\begin{align*}
    (a^2+b^2)^2 + (b^2+c^2)^2 +(c^2+a^2)^2 &= 138 \\
    (a^2+b^2+c^2)^2 &=100.
\end{align*}Find \(a^2(a^2-1) + b^2(b^2-1)+c^2(c^2-1)\).
Answer: Click to reveal hidden text
Solution. Subtracting the two given equations, we get
\begin{align*}
        a^4+b^4+c^4=38.
    \end{align*}Taking the square root of the second equation, we get

\begin{align*}
        a^2+b^2+c^2 = 10.
    \end{align*}
Then,
\begin{align*}
        a^4+b^4+c^4-(a^2+b^2+c^2) &= a^4-a^2+b^4-b^2+c^4-c^2 \\
        &= a^2(a^2-1)+b^2(b^2-1)+c^2(c^2-1) = \boxed{28}.
    \end{align*}
0 replies
qrxz17
Yesterday at 1:32 PM
0 replies
17th PMO Area Stage #5
yes45   0
Yesterday at 1:31 PM
Triangle \(ABC\) has a right angle at \(B\), with \(AB = 3\) and \(BC = 4\). If \(D\) and \(E\) are points
on \(AC\) and \(BC\), respectively, such that \(CD = DE = \frac{5}{3}\), find the perimeter of quadrilateral
\(ABED\).
Answer Confirmation
Solution
0 replies
yes45
Yesterday at 1:31 PM
0 replies
Sipnayan 2019 SHS Orals Semifinal Round B Difficult \#3
qrxz17   0
Yesterday at 1:30 PM
Problem. Suppose that \(a\), \(b\), \(c\) are positive integers such that \((a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 63\). Find all possible values of \(a+b+c\).
Answer. Click to reveal hidden text
Solution. Expanding and cancelling "like" terms, we get

\begin{align*}
        (a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) &= (a^4 +b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2) - 2(a^4+b^4+c^4) \\
        &= 2a^2b^2+2a^2c^2+2b^2c^2 - a^4 -b^4-c^4
    \end{align*}
We multiply the entire equation by \(-1\) to rearrange the terms and place the higher-degree terms at the front of the expression in order to have a more recognizable form.

\begin{align*}
        a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2 = -63.
    \end{align*}
Simplifying the left-hand expression, we get

\begin{align*}
        a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2 &=a^4-2a^2(b^2+c^2)+b^4-2b^2c^2+c^4 \\
        &= a^4-2a^2(b^2+c^2)+(b^2+c^2)^2 - 4b^2c^2 \\
        &= [a^2 - (b^2+c^2)]^2-(2bc)^2 \\
        &= (a^2-b^2-c^2+2bc)(a^2-b^2-c^2-2bc)\\
        &=[a^2-(b-c)^2][a^2-(b+c)^2] \\
        &= (a+b-c)(a-b+c)(a+b+c)(a-b-c) = -63.
    \end{align*}
If \(a=b=c\), we get
\begin{align*}
     -3a^4 &=-63 \\
     a^4 &= 21
    \end{align*}

in which there are no real solutions for \(a\).

If we either have \(a=b\), \(a=c\), or \(b=c\), we get something of the form

\[
    a^2(a+2b)(a-2b)=-63.
    \]
Since \(1\) and \(9\) are the only square factors of \(63\), we have \(a = \{1, 3\}\).

In this case, the solutions for \((a, b, c)\) are \((1, 4, 4)\) and \((3, 2, 2)\), including all their permutations.

If \(a, b, c\) are all distinct positive integers, there are no solutions.

Therefore, all possible values of \(a+b+c\) are \(\boxed{7 \text{ and }9}\).
0 replies
qrxz17
Yesterday at 1:30 PM
0 replies
MATHirang MATHibay 2011 Orals Survival Round Average \#1
qrxz17   0
Yesterday at 1:27 PM
Problem. Solve for all possible values of y: $\sqrt{y^2 + 4y + 8} + \sqrt{y^2+4y+4}=\sqrt{2(y^2+4y+6)}$
Answer. Click to reveal hidden text
Solution. Let \( N = y^2+4y+6\). We have
\begin{align*}
        \sqrt{N+2} + \sqrt{N-2} = \sqrt{2N}
    \end{align*}
Taking the square of the equation, we get

\begin{align*}
        (N+2)+(N-2)+2\sqrt{N^2-4}&=2N \\
        \sqrt{N^2-4} &=0
    \end{align*}
Taking the square of the equation again, we get

\begin{align*}
        N^2-4&=0 \\
        N^2&=4
    \end{align*}
If N is -2, then \(\sqrt{2N}\) is an imaginary number. Thus, N must be equal to +2. We then have

\begin{align*}
        y^2 +4y+6&=2 \\
        y^2+4y+4&=0 \\
        (y+2)^2&=0 \\
        y+2&=0 \\
        y &= \boxed{-2}.
    \end{align*}
0 replies
qrxz17
Yesterday at 1:27 PM
0 replies
Original Problem (qrxz17)
qrxz17   0
Yesterday at 1:12 PM
Problem. Given that \( 2^{\log_{16} 71} = a \) and \( x^{\log_{256} 71} = 71a \), find the value of \(x\).
Answer. Click to reveal hidden text
Solution. Let \(k=\log_{16}71\). We have
\begin{align*}
    2^k &= a \\   
    16^k &= 71
    \end{align*}
So,
\begin{align*}
        2^k\cdot16^k&=a\cdot71 \\
        32^k &= 71a = x^{\log_{256} 71}
    \end{align*}
Substituting the value of k, we get

\begin{align*}
       32^{\log_{16}71} &= x^{\log_{256} 71}
   \end{align*}
By using the change-of-base formula, we get

\begin{align*}
       32^{\frac{\log_{256}71}{\log_{256}16}}=32^{\frac{\log_{256}71}{\frac{1}{2}}} = 32^{2 \cdot \log_{256}71} = 1024^{\log_{256}71}
       \end{align*}
Therefore, \(x=\boxed{1024}\).
0 replies
qrxz17
Yesterday at 1:12 PM
0 replies
MathILy 2025 Decisions Thread
mysterynotfound   47
N May 20, 2025 by quavante
Discuss your decisions here!
also share any relevant details about your decisions if you want
47 replies
mysterynotfound
Apr 21, 2025
quavante
May 20, 2025
MathILy 2025 Decisions Thread
G H J
G H BBookmark kLocked kLocked NReply
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mysterynotfound
16 posts
#1
Y by
Discuss your decisions here!
also share any relevant details about your decisions if you want
This post has been edited 2 times. Last edited by mysterynotfound, Apr 21, 2025, 4:28 PM
Z K Y
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dragonborn56
1536 posts
#2 • 1 Y
Y by trk08
rejected
Z K Y
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Turtle09
1817 posts
#3
Y by
rejected lol

edit: freshman 1st time applicant
This post has been edited 1 time. Last edited by Turtle09, Apr 21, 2025, 8:34 PM
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KevinChen_Yay
242 posts
#4
Y by
yo wut im on school computer rn but am i cooked if i submitted yesterday night
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Alex-131
5398 posts
#5
Y by
rejected as sophomore, 2nd time applicant. Nearly fullsolve pset(except for the last problem, which I fake solved). Decent-ish essays about how I got into math, cooperation in sports, and hobbies like chess and physics.
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ConfidentKoala4
611 posts
#6
Y by
rejected as sophomore 2nd time applicant
almost same as @above
sad
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BossLu99
1399 posts
#7
Y by
rejected as 8th grade 1st time applicant
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clarkculus
249 posts
#8 • 1 Y
Y by centslordm
i got a "non-decision" email delaying my decision for a few weeks after a lot of complications regarding my letter of rec ... did anyone else get this? is this a de facto reject?

for reference i submitted my app mid-march but i got this email 4/12 when they finally got all my app stuff together
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CrunchyCucumber
66 posts
#9
Y by
@above The same thing happened to me, but I don’t know if it was because they lost parts of my application.
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cowstalker
297 posts
#10
Y by
KevinChen_Yay wrote:
yo wut im on school computer rn but am i cooked if i submitted yesterday night

isnt it due the 29th....
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rbcubed13
105 posts
#11
Y by
are decisions on a rolling basis
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Turtle09
1817 posts
#12
Y by
rbcubed13 wrote:
are decisions on a rolling basis

yeah
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GP102
855 posts
#13
Y by
to the people that got results: when did u submit?
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Alex-131
5398 posts
#14
Y by
GP102 wrote:
to the people that got results: when did u submit?

I submitted app on Thursday, recommender sent on Friday, and rejection on Thursday.
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Turtle09
1817 posts
#15
Y by
GP102 wrote:
to the people that got results: when did u submit?

early march, got mine back in a week
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