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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.

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[*]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
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
HCSSiM results
SurvivingInEnglish   54
N 7 minutes ago by NoSignOfTheta
Anyone already got results for HCSSiM? Are there any point in sending additional work if I applied on March 19?
54 replies
+3 w
SurvivingInEnglish
Apr 5, 2024
NoSignOfTheta
7 minutes ago
9 ARML Location
deduck   25
N 13 minutes ago by wc31415
UNR -> Nevada
St Anselm -> New Hampshire
WCU -> North Carolina


Put your USERNAME in the list ONLY IF YOU WANT TO!!!! !!!!!

I'm going to UNR if anyone wants to meetup!!! :D

Current List:
Iowa
UNR
PSU
St Anselm
WCU
25 replies
deduck
Yesterday at 4:19 PM
wc31415
13 minutes ago
Prove that 4p-3 is a square - Iran NMO 2005 - Problem1
sororak   24
N 34 minutes ago by EmersonSoriano
Let $n,p>1$ be positive integers and $p$ be prime. We know that $n|p-1$ and $p|n^3-1$. Prove that $4p-3$ is a perfect square.
24 replies
sororak
Sep 21, 2010
EmersonSoriano
34 minutes ago
Mathcounts state
happymoose666   38
N 37 minutes ago by tikachaudhuri
Hi everyone,
I just have a question. I live in PA and I sadly didn't make it to nationals this year. Is PA a competitive state? I'm new into mathcounts and not sure
38 replies
happymoose666
Mar 24, 2025
tikachaudhuri
37 minutes ago
IMO ShortList 1999, combinatorics problem 4
orl   27
N 41 minutes ago by cursed_tangent1434
Source: IMO ShortList 1999, combinatorics problem 4
Let $A$ be a set of $N$ residues $\pmod{N^{2}}$. Prove that there exists a set $B$ of of $N$ residues $\pmod{N^{2}}$ such that $A + B = \{a+b|a \in A, b \in B\}$ contains at least half of all the residues $\pmod{N^{2}}$.
27 replies
orl
Nov 14, 2004
cursed_tangent1434
41 minutes ago
purple comet discussion
ConfidentKoala4   65
N an hour ago by abbominable_sn0wman
when can we discuss purple comet
65 replies
+1 w
ConfidentKoala4
May 2, 2025
abbominable_sn0wman
an hour ago
Geometry
Lukariman   2
N an hour ago by Lukariman
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that <HDM = 2∠AMP.
2 replies
Lukariman
Yesterday at 12:43 PM
Lukariman
an hour ago
IMO 2010 Problem 1
canada   119
N an hour ago by lpieleanu
Find all function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all $x,y\in\mathbb{R}$ the following equality holds \[
f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor \] where $\left\lfloor a\right\rfloor $ is greatest integer not greater than $a.$

Proposed by Pierre Bornsztein, France
119 replies
canada
Jul 7, 2010
lpieleanu
an hour ago
f(n) <= f(a(G)) + f(b(G))
dangerousliri   5
N an hour ago by awesomeming327.
Source: FEOO, Problem 2, Shortlist C2
Given a group $G$ of people, we define $a(G)$ to be the least number of tables needed such that each person from the group sits in one of them with no two friends sitting on the same table, and $b(G)$ to be the least number of tables needed such that each person from the group sits in one of them with no two enemies sitting on the same table.
Consider all functions $f:\mathbb{N} \to \mathbb{N}$ such that for each group $G$ of $n$ people we have
\[ f(n) \leqslant f(a(G)) + f(b(G)) \quad \text{and} \quad f(2) = 1.\]Find all possible values of $f(2020)$.

Note: We assume that for every pair of people in a group, they are either both friends with each other or both enemies with each other. Also, $\mathbb{N}$ is the set of all positive integers.

Proposed by Demetres Christofides, Cyprus
5 replies
dangerousliri
May 30, 2020
awesomeming327.
an hour ago
f(w^2+x^2+y^2+z^2)=f(w^2+x^2)+f(y^2+z^2)
dangerousliri   23
N 2 hours ago by awesomeming327.
Source: FEOO, Problem 1, Shortlist N1
Find all functions $f:\mathbb{N}_0\rightarrow\mathbb{R}$ such that,
$$f(w^2+x^2+y^2+z^2)=f(w^2+x^2)+f(y^2+z^2)$$for all non-negative integers $w,x,y$ and $z$.

Note: $\mathbb{N}_0$ is the set of all non-negative integers and $\mathbb{R}$ is the set of all real numbers.

Proposed by Dorlir Ahmeti, Kosovo
23 replies
dangerousliri
May 30, 2020
awesomeming327.
2 hours ago
Darboux cubic
srirampanchapakesan   0
2 hours ago
Source: Own
Let P be a point on the Darboux cubic (or the McCay Cubic ) of triangle ABC.

P1P2P3 is the circumcevian or pedal triangle of P wrt ABC.

Prove that P also lie on the Darboux cubic ( or the McCay Cubic) of P1P2P3 .
0 replies
srirampanchapakesan
2 hours ago
0 replies
primitive polyominoes
N.T.TUAN   29
N 2 hours ago by Disjunction
Source: USAMO 2007
An animal with $n$ cells is a connected figure consisting of $n$ equal-sized cells[1].

A dinosaur is an animal with at least $2007$ cells. It is said to be primitive it its cells cannot be partitioned into two or more dinosaurs. Find with proof the maximum number of cells in a primitive dinosaur.

(1) Animals are also called polyominoes. They can be defined inductively. Two cells are adjacent if they share a complete edge. A single cell is an animal, and given an animal with $n$ cells, one with $n+1$ cells is obtained by adjoining a new cell by making it adjacent to one or more existing cells.
29 replies
N.T.TUAN
Apr 26, 2007
Disjunction
2 hours ago
Only consecutive terms are coprime
socrates   37
N 2 hours ago by blueprimes
Source: 7th RMM 2015, Problem 1
Does there exist an infinite sequence of positive integers $a_1, a_2, a_3, . . .$ such that $a_m$ and $a_n$ are coprime if and only if $|m - n| = 1$?
37 replies
socrates
Feb 28, 2015
blueprimes
2 hours ago
Cyclic roots are not real, they can't hurt you
anantmudgal09   21
N 2 hours ago by bjump
Source: INMO 2023 P2
Suppose $a_0,\ldots, a_{100}$ are positive reals. Consider the following polynomial for each $k$ in $\{0,1,\ldots, 100\}$:
$$a_{100+k}x^{100}+100a_{99+k}x^{99}+a_{98+k}x^{98}+a_{97+k}x^{97}+\dots+a_{2+k}x^2+a_{1+k}x+a_k,$$where indices are taken modulo $101$, i.e., $a_{100+i}=a_{i-1}$ for any $i$ in $\{1,2,\dots, 100\}$. Show that it is impossible that each of these $101$ polynomials has all its roots real.

Proposed by Prithwijit De
21 replies
anantmudgal09
Jan 15, 2023
bjump
2 hours ago
Polynomial Roots
gauss1181   51
N Apr 25, 2025 by Ilikeminecraft
Source: 1984 USAMO #1
The product of two of the four roots of the quartic equation $x^4 - 18x^3 + kx^2+200x-1984=0$ is $-32$. Determine the value of $k$.
51 replies
gauss1181
Aug 11, 2009
Ilikeminecraft
Apr 25, 2025
Polynomial Roots
G H J
G H BBookmark kLocked kLocked NReply
Source: 1984 USAMO #1
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Tafi_ak
309 posts
#40 • 4 Y
Y by Mango247, Mango247, Mango247, cubres
The answer is $86$. By Vieta we have
\begin{align*}
    \sum r_i&=18\tag{$1$} \\
    \sum r_ir_j&=k\tag{$2$} \\
    \sum r_ir_jr_k&=-200\tag{$3$} \\
    r_1r_2r_3r_4&=-1984\tag{$4$}
\end{align*}Let WLOG $r_1r_2=-32$. Using $(2)$ we have $r_3r_4=-1984/-32=62$. Using $r_1r_2$, $r_3r_4$ and equation $(1)$ in equation $(3)$ we have \[ 62(r_1+r_2)-32(r_3+r_4)=-200\tag{$5$} \]Rewriting the equation $(2)$ we have \[ (r_1+r_2)(r_3+r_4)+30=k\tag{$6$} \]Solving the equation $(1)$ and $(5)$ in terms of variable $r_1+r_2$ and $r_3+r_4$ we get $4, 14$. Putting these value in the equation $(6)$ we get out answer $k=86$.
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HamstPan38825
8860 posts
#41 • 2 Y
Y by centslordm, cubres
Let $$x^4-18x^3+kx^2+200x-1984 = (x^2+ax-32)(x^2+bx+32)$$for some $a, b$. Equating the $x^3$ and $x$ coefficients yields
\begin{align*}
a+b &= -18 \\
62a-32b &= 200
\end{align*}so $a = -4, b = -14$, and $k = 56-32+62=\boxed{86}$.
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cinnamon_e
703 posts
#42 • 1 Y
Y by cubres
solution
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mathmax12
6051 posts
#43 • 1 Y
Y by cubres
By Veita the product of the other 2 roots are 62, let the 4 roots be $a,b,c,d$ by vieta's we have $a+b+c+d=18,ab+ac+Ad+bc+bd+cd$, and $abc+acd+abd+bcd=-200$

And then we can do some calculator from here to get that $k=\boxed{86}$
This post has been edited 1 time. Last edited by mathmax12, Apr 21, 2023, 1:08 PM
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Rounak_iitr
456 posts
#44 • 1 Y
Y by cubres
Assume the roots of the equation $x^4-18x^3+kx^2+200x-1984=0$ are $\alpha  ,\beta  ,\gamma  ,\delta$ Given product of any two roots of this equation is -32. By Vietas Relation we get $$\alpha+\beta+\gamma+\delta=18$$$$\alpha\beta+\beta\gamma+\gamma\alpha+\alpha\delta+\beta\delta+\gamma\delta=k$$$$\alpha\beta\gamma+\beta\gamma\delta+\alpha\beta\delta+\alpha\gamma\delta=-200$$$$\alpha\beta\gamma\delta=-1984$$Lets assume $\alpha\beta=-32$ from product of roots we get $\alpha\beta\gamma\delta=-1984\implies\gamma\delta=62$ $$\alpha\beta\gamma+\beta\gamma\delta+\alpha\gamma\delta+\alpha\beta\delta=-200\implies62(\alpha+\beta)-32(\gamma+\delta)\implies62(\alpha+\beta)-32(18-(\alpha+\beta))=-200\implies94(\alpha+\beta)=376\implies\boxed{\alpha+\beta=4}$$From these we also get $\boxed{\gamma+\delta=14}$ Therefore, $$\alpha\beta+\beta\gamma+\gamma\alpha+\delta\alpha+\beta\delta+\gamma\delta=k\implies(\alpha+\beta)(\gamma+\delta)+\alpha\beta+\gamma\delta=14.4+62-32=56+30\implies86.$$Is the correct answer...
Vietas Relation :love:
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pqr.
174 posts
#45 • 2 Y
Y by Rounak_iitr, cubres
Let the roots be $r_1, r_2, r_3, r_4$. By Vieta, if $r_1r_2=-32$, then $r_3r_4=62$. Additionally, Vieta gives $$r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4=-200 \implies -32(r_3+r_4)+62(r_1+r_2).$$But notice that $r_3+r_4=18-(r_1+r_2)$ again by Vieta. Then, solving we get that $r_1+r_2=4$ and $r_3+r_4=14$. Finally, by Vieta, we see have $$k=r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4=(r_1r_2+r_3r_4)+(r_1r_3+r_1r_4+r_2r_3+r_2r_4)=-32+62+(r_1+r_2)(r_3+r_4)=30+56=86,$$and we're done.
This post has been edited 2 times. Last edited by pqr., Aug 1, 2023, 2:59 AM
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exp-ipi-1
1074 posts
#46 • 2 Y
Y by Rounak_iitr, cubres
easiest usamo ever
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joshualiu315
2534 posts
#47 • 1 Y
Y by cubres
solution
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peppapig_
281 posts
#48 • 1 Y
Y by cubres
Let the roots of the quartic be $a$, $b$, $c$ and $d$. WLOG, let $ab=-32$, and by Vieta's, we have that $abcd=-1984$, meaning that $cd=-62$. Let $x=a+b$ and $y=c+d$. By Vieta's, we have that
\[18=a+b+c+d=(a+b)+(c+d)=x+y,\]and
\[-200=abc+abd+acd+bcd=cd(a+b)+ab(c+d)=62x-32y.\]Solving the system of equations gives us that $x=4$ and $y=14$. Again using Vieta's, we have that
\[k=ac+ad+bc+bd+ab+cd=(a+b)(c+d)+ab+cd=xy-32+62,\]or $xy+30$, giving us that $k$ is $86$, finishing the problem.
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Markas
105 posts
#49 • 1 Y
Y by cubres
By Vieta's we have that 1) $x_1 + x_2 + x_3 + x_4 = 18$, 2) $x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 = k$, 3) $x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4 + x_2x_3x_4 = -200$, 4) $x_1x_2x_3x_4 = -1984$. Now we can assume $x_1x_2$ = -32 $\Rightarrow$ $x_3x_4$ = 62. We now plug this into 3) and we get $-32x_3 - 32x_4 + 62x_1 + 62x_2 = 62(x_1 + x_2) - 32(x_3 + x_4)$ and after using 1) we get $62(x_1 + x_2) - 32(18 - x_1 - x_2) = -200$. Let $x_1 + x_2 = a$, we get $62a - 32(18 - a) + 200 = 94a - 576 + 200 = 94a - 376 = 0$ $\Rightarrow$ a = 4 $\Rightarrow$ $x_1 + x_2 = 4$ and $x_3 + x_4 = 14$. Now we plug in everything we know in 2) so $x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 = - 32 + x_1(x_3 + x_4) + x_2(x_3 + x_4) + 62 = 30 + (x_1 + x_2)(x_3 + x_4) = 30 + 4.14 = 30 + 56 = 86 = k$ $\Rightarrow$ k = 86 $\Rightarrow$ the answer is 86.
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combowomborhombo
11 posts
#50 • 1 Y
Y by cubres
The product of all the roots in this polynomial is $-1984$. Since we know the product of one pair of roots, we can find the product of the other pair of roots as $62$. We can split the original polynomial into two quadratic polynomials:

\[
x^4 - 18x^3 + kx^2 + 200x - 1984 = (x^2 + ax - 32)(x^2 + bx + 62)
\]
By matching coefficients, we find the following relationships:

\begin{align*}
a + b &= -18, \\
ab + 30 &= k, \\
62a - 32b &= 200.
\end{align*}
Using the first and last relationships, we solve for $a$ and $b$:

\[
(a, b) = (-4, -14).
\]
Substituting into the equation $ab + 30 = k$, we find:

\[
k = \boxed{86}.
\]
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P162008
182 posts
#51 • 1 Y
Y by cubres
Let's assume that the biquadratic equation x⁴ - 18x³ + kx² + 200x - 1984 = 0 has roots a,b,c and d respectively such that ab = -32.

By Vieta's Relation we've

abcd = -1984 —————(1)

abc + bcd + cda + dab = -200

ab(c + d) + cd(a + b) = -200 —————(2)

ab + ac + ad + bc + bd + cd = k

ab + cd + (a + b)c + (a + b)d = k

ab + cd + (a + b)(c + d) = k —————(3)

and, (a + b) + (c + d) = 18

or, (a + b) = 18 - (c + d) —————(4)

Substituting ab = -32 in equation (1), we get

(-32)cd = -1984

Therefore, cd = 62

On substituting ab = -32, cd = 62 and (a + b) = 18 - (c + d) in equation (2), we get

-32(c + d) + 62[18 - (c + d)] = -200

1116 - 62(c + d) - 32(c + d) = -200

94(c + d) = 1316

Therefore, (c + d) = 14

Similarly, (a + b) = 18 - (c + d) = 4

Therefore, k = -32 + 62 + 4(14) = 86
This post has been edited 5 times. Last edited by P162008, Jan 28, 2025, 11:57 PM
Reason: Typo
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cubres
119 posts
#52
Y by
Wordless solution
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jb2015007
1941 posts
#53 • 1 Y
Y by cubres
trivial by vietas
i think everyone who knows that vietas exists would be able to apply and at least know how to do around half of it
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Ilikeminecraft
616 posts
#54
Y by
Let the four roots be $a, b, c, d$ with $ab = -32.$ By Vieta's, we get the following:
\begin{align*}
    a + b + c + d & = 18 \\
    abcd & = -1984 \implies cd = 62 \\
    abc + abd + acd + bcd & = -200 \implies -32 c - 32 d + 62a + 62b = -200 
\end{align*}The last equation implies that $a + b = \frac{376}{94} = 4,$ and $c + d = 14.$ Thus,
\begin{align*}
    ab + ac + ad + bc + bd + cd & = -32 + 62 + (a + b)(c + d) \\
    & = 30 + 4 \cdot 14 = \boxed{86}
\end{align*}
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