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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.

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[*]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
9 Fun Proof Endings
elasticwealth   32
N 19 minutes ago by Unicode_Master03B8
It seems like AOPS is going through a stressful phase right now.

Let's lighten the mood by voting on the best proof endings!
32 replies
elasticwealth
Today at 12:26 AM
Unicode_Master03B8
19 minutes ago
2025 USAMO Rubric
plang2008   24
N 29 minutes ago by Mathdreams
1. Let $k$ and $d$ be positive integers. Prove that there exists a positive integer $N$ such that for every odd integer $n>N$, the digits in the base-$2n$ representation of $n^k$ are all greater than $d$.

Rubric for Problem 1

2. Let $n$ and $k$ be positive integers with $k<n$. Let $P(x)$ be a polynomial of degree $n$ with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers $a_0,\,a_1,\,\ldots,\,a_k$ such that the polynomial $a_kx^k+\cdots+a_1x+a_0$ divides $P(x)$, the product $a_0a_1\cdots a_k$ is zero. Prove that $P(x)$ has a nonreal root.

Rubric for Problem 2

3. Alice the architect and Bob the builder play a game. First, Alice chooses two points $P$ and $Q$ in the plane and a subset $\mathcal{S}$ of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair $A,\,B$ of cities, they are connected with a road along the line segment $AB$ if and only if the following condition holds:
[center]For every city $C$ distinct from $A$ and $B$, there exists $R\in\mathcal{S}$ such[/center]
[center]that $\triangle PQR$ is directly similar to either $\triangle ABC$ or $\triangle BAC$.[/center]
Alice wins the game if (i) the resulting roads allow for travel between any pair of cities via a finite sequence of roads and (ii) no two roads cross. Otherwise, Bob wins. Determine, with proof, which player has a winning strategy.

Note: $\triangle UVW$ is directly similar to $\triangle XYZ$ if there exists a sequence of rotations, translations, and dilations sending $U$ to $X$, $V$ to $Y$, and $W$ to $Z$.

Rubric for Problem 3

4. 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 $XY$.

Rubric for Problem 4

5. Determine, with proof, all positive integers $k$ such that \[\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k\]is an integer for every positive integer $n$.

Rubric for Problem 5

6. Let $m$ and $n$ be positive integers with $m\geq n$. There are $m$ cupcakes of different flavors arranged around a circle and $n$ people who like cupcakes. Each person assigns a nonnegative real number score to each cupcake, depending on how much they like the cupcake. Suppose that for each person $P$, it is possible to partition the circle of $m$ cupcakes into $n$ groups of consecutive cupcakes so that the sum of $P$'s scores of the cupcakes in each group is at least $1$. Prove that it is possible to distribute the $m$ cupcakes to the $n$ people so that each person $P$ receives cupcakes of total score at least $1$ with respect to $P$.

Rubric for Problem 6
24 replies
plang2008
Apr 2, 2025
Mathdreams
29 minutes ago
MOP Emails Out! (not clickbait)
Mathandski   77
N an hour ago by jlcong
What an emotional roller coaster the past 34 days have been.

Congrats to all that qualified!
77 replies
Mathandski
Tuesday at 8:25 PM
jlcong
an hour ago
USA(J)MO Statistics Out
BS2012   33
N an hour ago by jlcong
Source: MAA edvistas page
https://maa.edvistas.com/eduview/report.aspx?view=1561&mode=6
who were the 2 usamo perfects
33 replies
+2 w
BS2012
Yesterday at 10:07 PM
jlcong
an hour ago
No more topics!
Catch those negatives
cappucher   44
N Apr 6, 2025 by Apple_maths60
Source: 2024 AMC 10A P11
How many ordered pairs of integers $(m, n)$ satisfy $\sqrt{n^2 - 49} = m$?

$
\textbf{(A) }1 \qquad
\textbf{(B) }2 \qquad
\textbf{(C) }3 \qquad
\textbf{(D) }4 \qquad
\textbf{(E) } \text{Infinitely many} \qquad
$
44 replies
cappucher
Nov 7, 2024
Apple_maths60
Apr 6, 2025
Catch those negatives
G H J
G H BBookmark kLocked kLocked NReply
Source: 2024 AMC 10A P11
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cappucher
93 posts
#1 • 1 Y
Y by IanJung
How many ordered pairs of integers $(m, n)$ satisfy $\sqrt{n^2 - 49} = m$?

$
\textbf{(A) }1 \qquad
\textbf{(B) }2 \qquad
\textbf{(C) }3 \qquad
\textbf{(D) }4 \qquad
\textbf{(E) } \text{Infinitely many} \qquad
$
This post has been edited 3 times. Last edited by cappucher, Nov 8, 2024, 4:46 AM
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mithu542
1570 posts
#2
Y by
I got $(D)$. Just squared and used difference of squares :)

I almost did 2 until I realized that negative solutions are allowed lol
This post has been edited 1 time. Last edited by mithu542, Nov 7, 2024, 5:10 PM
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Vivaandax
79 posts
#3 • 3 Y
Y by IanJung, wangzrpi, Aaronjudgeisgoat
Maa really loved to troll people with the negatives on this test
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xiaoya.peng
120 posts
#4
Y by
I got 2 because m can't be negative and the absolute value of n can't be less than 7 so that got rid of some solutions
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mithu542
1570 posts
#5 • 1 Y
Y by sadhase25
oh oops i forgot about that
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MathRook7817
662 posts
#6
Y by
mithu542 wrote:
I got $(D)$. Just squared and used difference of squares :)

I almost did 2 until I realized that negative solutions are allowed lol

bro same
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LostDreams
144 posts
#7
Y by
Square it and apply differences of squares and you notice that there are 4 possibilities when you bash which is (24, 25), (24, -25), (0, 7), and (0,-7)

Giving us the answer D
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Siddharthmaybe
106 posts
#9
Y by
Lol people really thought they tricked the question but the question tricked them, m can't be -ve cuz its square root function
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Alex-131
5361 posts
#10
Y by
I have no idea whether I did C or D, rip. I got the negative solutions, but I don't remember whether I put (0,-7) in. kms
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Siddharthmaybe
106 posts
#11
Y by
Alex-131 wrote:
I have no idea whether I did C or D, rip. I got the negative solutions, but I don't remember whether I put (0,-7) in. kms

ya i just realized that too i don't remember that :(
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stjwyl
1262 posts
#12
Y by
wait so wats the answer
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ostriches88
1528 posts
#13
Y by
stjwyl wrote:
wait so wats the answer

D
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ethanzhang1001
1060 posts
#14
Y by
Vivaandax wrote:
Maa really loved to troll people with the negatives on this test

they also should've put 8 as an answer choice in E instead because what if you accidentally set m negative :P
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ElaineGu
388 posts
#15
Y by
7 24 25 makes it trivial lmao
25, -25, 7, -7 for n
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wangzrpi
159 posts
#16
Y by
Factor by difference of squares and you see that there are 8
There are 4 extraneous
So I picked the largest one (4)
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pingpongmerrily
3569 posts
#17
Y by
i put 3 bc i thought that square roots couldn't be negative or smtg

:noo:
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golden_star_123
206 posts
#18
Y by
Nobody has put a full solution so here it is:

$\sqrt{n^2-49} = m \implies n^2 - m^2 = 49$. We have a difference of squares, so $(n-m)(n+m) = 49$.

If $(n-m)(n+m) = 7\cdot 7$, we have $n=7, n=-7$ and $m=0$. This gives two possibilities.

If $(n-m)(n+m) = 49\cdot 1$, we have $n=25, m=-24$. This doesn't work, however, as a square root is always positive.

If $(n-m)(n+m) = 1\cdot 49$, we have $n=25, m=24$. One possibility.

If $(n-m)(n+m) = -49\cdot -1$, we have $n=-25, m=24$. One possibility.

This adds up to a total of $\boxed{4}$ possibilities.
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cyberhacker
401 posts
#19
Y by
bro i put 4 and changed to 2?? i was fenting
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williamxiao
2510 posts
#20
Y by
Why did I put 3 it’s literally a square problem so it has to be even :(
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ElaineGu
388 posts
#21
Y by
williamxiao wrote:
Why did I put 3 it’s literally a square problem so it has to be even :(

:noo:
which case did u forget
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pingpongmerrily
3569 posts
#22
Y by
williamxiao wrote:
Why did I put 3 it’s literally a square problem so it has to be even :(

sameee :sob:
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ChaitraliKA
1004 posts
#23
Y by
williamxiao wrote:
Why did I put 3 it’s literally a square problem so it has to be even :(

Unless if 0 had worked

Anyways, I got this question right in the wrong way :rotfl: who cares honestly
This post has been edited 1 time. Last edited by ChaitraliKA, Nov 7, 2024, 9:34 PM
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awzhang10
72 posts
#24
Y by
i had brainrot during the test and put infinitely many LOL
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Akang11
36 posts
#25
Y by
Sillied on the negative roots, please send help :sob:
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SkatingKitty
223 posts
#26
Y by
I crashed out and I left it blank
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goldenuni678
237 posts
#27
Y by
I got D, from squaring and then using difference of squares
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SkatingKitty
223 posts
#28
Y by
I forgot the formula tears
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andrewcheng
525 posts
#29
Y by
I got C because I forgot -7,0
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gicyuraok2
1059 posts
#30
Y by
dang i probably would have sillied this one but anyway $0,\pm7$ and $\pm24,25$ fun stuff
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smarty101
322 posts
#31
Y by
forgot about 7-24-25 triangles :wallbash:
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BrighterFrog11
1116 posts
#32
Y by
aww man. I got this one wrong. I didn't think about 0 and +/- 7
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lpieleanu
2913 posts
#34
Y by
Solution
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jb2015007
1917 posts
#35
Y by
lol I actually got this right in like 2 mins cuz I was being cautious
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NS0004
186 posts
#36
Y by
How can m be negative because of the square root?
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pingpongmerrily
3569 posts
#37
Y by
NS0004 wrote:
How can m be negative because of the square root?

this is what i thought during the test but think a little harder, there are still 4 cases without m being negative
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Serengeti22
1135 posts
#38
Y by
I but 2 , I didn’t realize the negatives. The answer is 4
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Challengees24
1098 posts
#39
Y by
Did anyone else make a graph lol

n^2+m^2=49, circle with 4 roots

(0,7),(7,0),(-7,0),(0,-7)
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golden_star_123
206 posts
#40
Y by
Your approach doesn't work because m can't be negative.
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Challengees24
1098 posts
#41
Y by
oml lmho i did it wrong way and still got the answer write

daym
This post has been edited 1 time. Last edited by Challengees24, Nov 9, 2024, 5:33 PM
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Challengees24
1098 posts
#42 • 1 Y
Y by PikaPika999
I just realized it should have been $n^2-m^2=49$

heyyy maybe this test wasnt so bad lol(joke kinda sorta, cuz it wasnt that bad)
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Apple_maths60
24 posts
#43 • 1 Y
Y by PikaPika999
I don't think (7,0)or (-7,0) would be a solution because if we put zero in the main equation we get √(-49)=m×m .
But √(-49) is not possible
So the answer must be 2
(25,-24)&(25,24)
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NS0004
186 posts
#44 • 1 Y
Y by PikaPika999
Bruh amc said the answer was 4....
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AbhayAttarde01
1484 posts
#45 • 1 Y
Y by PikaPika999
Apple_maths60 wrote:
I don't think (7,0)or (-7,0) would be a solution because if we put zero in the main equation we get √(-49)=m×m .
But √(-49) is not possible
So the answer must be 2
(25,-24)&(25,24)

swap the numbers
think you accidently swapper the numbers for m and n
it shoult be (0, -7), (0, 7), (24, -25), (24, 25)
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sadas123
1232 posts
#46 • 1 Y
Y by PikaPika999
I got (4) possible values
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Apple_maths60
24 posts
#47 • 1 Y
Y by PikaPika999
AbhayAttarde01 wrote:
Apple_maths60 wrote:
I don't think (7,0)or (-7,0) would be a solution because if we put zero in the main equation we get √(-49)=m×m .
But √(-49) is not possible
So the answer must be 2
(25,-24)&(25,24)

swap the numbers
think you accidently swapper the numbers for m and n
it shoult be (0, -7), (0, 7), (24, -25), (24, 25)


Oh, ok got it. I actually did that mistake
Thank you!
This post has been edited 1 time. Last edited by Apple_maths60, Apr 6, 2025, 7:06 PM
Reason: .
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