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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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0 replies
jlacosta
May 1, 2025
0 replies
IMO Shortlist 2008, Geometry problem 2
April   43
N 10 minutes ago by ezpotd
Source: IMO Shortlist 2008, Geometry problem 2, German TST 2, P1, 2009
Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE = \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$.

Proposed by Charles Leytem, Luxembourg
43 replies
April
Jul 9, 2009
ezpotd
10 minutes ago
In Cyclic Quadrilateral ABCD, find AB^2+BC^2-CD^2-AD^2
Darealzolt   0
an hour ago
Source: KTOM April 2025 P8
Given Cyclic Quadrilateral \(ABCD\) with an area of \(2025\), with \(\angle ABC = 45^{\circ}\). If \( 2AC^2 = AB^2+BC^2+CD^2+DA^2\), Hence find the value of \(AB^2+BC^2-CD^2-DA^2\).
0 replies
Darealzolt
an hour ago
0 replies
Plz give me the solution
Madunglecha   1
N an hour ago by top1vien
For given M
h(n) is defined as the number of which is relatively prime with M, and 1 or more and n or less.
As B is h(M)/M, prove that there are at least M/3 or more N such that satisfying the below inequality
|h(N)-BN| is under 1+sqrt(B×2^((the number of prime factor of M)-3))
1 reply
Madunglecha
4 hours ago
top1vien
an hour ago
King's Constrained Walk
Hellowings   1
N 2 hours ago by Hellowings
Source: Own
Given an n x n chessboard, with a king starting at any square, the king's task is to visit each square in the board exactly once (essentially an open path); this king moves how a king in chess would.
However, we are allowed to place k numbers on the board of any value such that for each number A we placed on the board, the king must be in the position of that number A on its Ath square in its journey, with the starting square as its 1st square.
Suppose after we placed k numbers, there is one and only one way to complete the king's task (this includes placing the king in a starting square), find the minimum value of k set by n.

Didn't know I could post it here xd; I'm unsure how hard this question could be.
1 reply
Hellowings
4 hours ago
Hellowings
2 hours ago
Vieta Substitution
Darealzolt   0
2 hours ago
Let \(\alpha,\beta,\gamma,\delta\) be the roots of the equation \(x^4-3x^3+6x^2+5x-25\). Hence, find the value of \(Z\) if
\[
Z=\frac{\alpha+\beta+\gamma+\delta}{\alpha(\beta+\gamma+\delta)+\beta(\gamma+\delta)}+\alpha\beta\gamma\delta[\alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta)]
\]
0 replies
Darealzolt
2 hours ago
0 replies
Floor Function Series
Darealzolt   2
N 2 hours ago by pieMax2713
Let \( \lfloor x \rfloor \) denote the greatest integer less than or equal to \(x\). Hence find the value of \(M\), if
\[
M = \left\lfloor \frac{1^2}{3} \right\rfloor + \left\lfloor \frac{2^2}{3}  \right\rfloor + \left\lfloor \frac{3^2}{3}  \right\rfloor + \dots + \left\lfloor \frac{39^2}{3}  \right\rfloor + \left\lfloor \frac{40^2}{3}  \right\rfloor
\]
2 replies
Darealzolt
2 hours ago
pieMax2713
2 hours ago
Vincentian Numbers
Darealzolt   0
3 hours ago
A number is called \(Vincentian\) if within that number exists a digit \(k \in \{1,2,3,4,5,6,7\}\) that appears exactly \(k^2\) times in that number, hence find the number of \(Vincentian\) that consist of 4 digits (Numbers may contain a 0)
0 replies
Darealzolt
3 hours ago
0 replies
9 Isogonal and isotomic conjugates
V0305   13
N 3 hours ago by ohiorizzler1434
1. Do you think isogonal conjugates should be renamed to angular conjugates?
2. Do you think isotomic conjugates should be renamed to cevian conjugates?

Please answer truthfully :)

Credit to Stead for this renaming idea
13 replies
V0305
May 26, 2025
ohiorizzler1434
3 hours ago
Prove atleast one from a,b,c is 2
Darealzolt   2
N 4 hours ago by sqing
Let \(a,b,c\) be real numbers, such that
\[
a^2+b^2+c^2+abc=5
\]\[
a+b+c=3
\]Prove that atleast one of the numbers \(a,b,c\) is equal to \( 2\).
2 replies
Darealzolt
Yesterday at 11:31 AM
sqing
4 hours ago
Interesting Geometry
captainmath99   4
N Yesterday at 8:01 PM by captainmath99
Let ABC be a right triangle such that $\angle{C}=90^\circ, CA=6, CB=4$. A circle O with center C has a radius of 2. Let P be a point on the circle O.

a)What is the minimum value of $(AP+\dfrac{1}{2}BP)$?
Answer Check

b) What is the minimum value of $(\dfrac{1}{3}AP+BP)$?
Answer Check
4 replies
1 viewing
captainmath99
May 25, 2025
captainmath99
Yesterday at 8:01 PM
Looking for even one person to study math.
abduqahhor_math   2
N Yesterday at 6:25 PM by EaZ_Shadow
Hi guys,I am looking for a person to study math topics related to olympiad.I have just finished 10th grade
2 replies
abduqahhor_math
Yesterday at 5:22 PM
EaZ_Shadow
Yesterday at 6:25 PM
Inequalities
lgx57   0
Yesterday at 3:55 PM
Let $a,b,c,d,e \ge 0$,$\sum \dfrac{1}{a+4}=1$.Prove that:
$$\sum \dfrac{a}{a^2+4} \le 1$$
Let $x,y,z>0$.Prove that:
$$\sum (y+z)\sqrt{\dfrac{yz}{(z+x)(y+x)}} \ge x+y+z$$
0 replies
lgx57
Yesterday at 3:55 PM
0 replies
Find the sum of all the products a_i a_j
Darealzolt   1
N Yesterday at 2:13 PM by alexheinis
Among the 100 constants \( \{ a_1,a_2,a_3,\dots,a_{100} \} \),there are \(39\) equal to \( -1\), and \(61\) equal to \(1\). Find the sum of all the products \(a_i a_j\) , where \(a \leq i < j \leq 100\).
1 reply
Darealzolt
Yesterday at 11:24 AM
alexheinis
Yesterday at 2:13 PM
IOQM P26 2024
SomeonecoolLovesMaths   5
N Yesterday at 1:11 PM by SomeonecoolLovesMaths
The sum of $\lfloor x \rfloor$ for all real numbers $x$ satisfying the equation $16 + 15x + 15x^2 = \lfloor x \rfloor ^3$ is:
5 replies
SomeonecoolLovesMaths
Sep 8, 2024
SomeonecoolLovesMaths
Yesterday at 1:11 PM
Factor of P(x)
Brut3Forc3   20
N Apr 23, 2025 by IceyCold
Source: 1976 USAMO Problem 5
If $ P(x),Q(x),R(x)$, and $ S(x)$ are all polynomials such that \[ P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x),\] prove that $ x-1$ is a factor of $ P(x)$.
20 replies
Brut3Forc3
Apr 4, 2010
IceyCold
Apr 23, 2025
Factor of P(x)
G H J
G H BBookmark kLocked kLocked NReply
Source: 1976 USAMO Problem 5
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Brut3Forc3
1948 posts
#1 • 8 Y
Y by samrocksnature, Amir Hossein, Rounak_iitr, Adventure10, Mango247, and 3 other users
If $ P(x),Q(x),R(x)$, and $ S(x)$ are all polynomials such that \[ P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x),\] prove that $ x-1$ is a factor of $ P(x)$.
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xpmath
2735 posts
#2 • 3 Y
Y by samrocksnature, Adventure10, Mango247
Solution
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math154
4302 posts
#3 • 3 Y
Y by samrocksnature, Adventure10, Mango247
Unfortunately,
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xpmath
2735 posts
#4 • 2 Y
Y by samrocksnature, Adventure10
Blah yeah I knew I'd make some mistakes last night anyway.
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Kunihiko_Chikaya
14514 posts
#5 • 3 Y
Y by samrocksnature, Adventure10, Mango247
Can we generalize the problem?
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JSGandora
4216 posts
#6 • 3 Y
Y by samrocksnature, Adventure10, Mango247
A Complete Solution
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Kunihiko_Chikaya
14514 posts
#7 • 4 Y
Y by samrocksnature, Adventure10, Mango247, and 1 other user
!976 USA MO, Problem 5.
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Kolumbus
45 posts
#8 • 4 Y
Y by samrocksnature, geniusofart, Adventure10, Mango247
Sorry, but why can't I just
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AkshajK
4820 posts
#9 • 2 Y
Y by samrocksnature, Adventure10
bump; also curious about above question
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va2010
1276 posts
#10 • 11 Y
Y by mathtastic, Math1331Math, maXplanK, targo___, khan.academy, llplp, samrocksnature, Adventure10, Mango247, Ritwin, clarkculus
Here's a faster solution. Let $\omega$ be a primitive $5$th root of unity, and observe that for $x = 1, 2, 3, 4$, $P(1)+\omega^xQ(1)+\omega^{2x}R(1) = 0$. Hence $\omega, \omega^2, \omega^3$, and $\omega^4$ are roots of the quadratic equation $P(1)+xQ(1)+x^2R(1)$. A quadratic null at 3 different places must be null everywhere, implying $R(1)=Q(1)=P(1)=0$, so we're done.
This post has been edited 1 time. Last edited by va2010, Sep 28, 2015, 11:28 PM
Reason: latex darn
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OlympusHero
17020 posts
#11 • 1 Y
Y by samrocksnature
My Start

Is there a way to conclude without using roots of unity or something like that? Not too familiar with such things.
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lrjr24
967 posts
#12 • 1 Y
Y by samrocksnature
OlympusHero wrote:
My Start

Is there a way to conclude without using roots of unity or something like that? Not too familiar with such things.

I don’t think there is a way without roots of unity.
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jasperE3
11388 posts
#13
Y by
Let $K(x)$ denote the given assertion, and let $\omega$ be a primitive $5$th root of unity.

$$\begin{cases}K(1)&\Rightarrow P(1) P(1)+Q(1)+R(1)=5S(1)\\
K(\omega)&\Rightarrow P(1)+\omega Q(1)+\omega^2R(1)=0\\
K\left(\omega^2\right)&\Rightarrow P(1)+\omega^2Q(1)+\omega^4R(1)=0\\
K\left(\omega^3\right)&\Rightarrow P(1)+\omega^3Q(1)+\omega R(1)=0\\
K\left(\omega^4\right)&\Rightarrow P(1)+\omega^4Q(1)+\omega^3R(1)=0\end{cases}$$Adding up all of these yields $P(1)=S(1)$. Hence:
$$\begin{cases}Q(1)+R(1)=4S(1)\\
\omega Q(1)+\omega^2R(1)+S(1)=0\\
\omega^2Q(1)+\omega^4R(1)+S(1)=0\\
\omega^3Q(1)+\omega R(1)+S(1)=0\\
\omega^4Q(1)+\omega^3 R(1)+S(1)=0\end{cases}$$which is more than enough. Solving the system, we easily get $P(1)=0$, which completes the proof. $\square$
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Math_Is_Fun_101
159 posts
#14 • 2 Y
Y by jasperE3, clarkculus
lrjr24 wrote:
OlympusHero wrote:
My Start

Is there a way to conclude without using roots of unity or something like that? Not too familiar with such things.

I don’t think there is a way without roots of unity.
There is a way.

Let $T\colon\mathbb R[x]\to\mathbb R^5$ be the function given by
\[ \sum a_kx^k\mapsto\left(\sum_{5\mid k}a_k,\sum_{5\mid k-1}a_k,\ldots,\sum_{5\mid k-4}a_k\right). \]It is clear that
\[ T(P(x^5)+xQ(x^5)+x^2R(x^5))=(P(1),Q(1),R(1),0,0). \]It is also easy to see that
\[ T((x^4+x^3+x^2+x+1)S(x))=(S(1),S(1),S(1),S(1),S(1)). \]Since $P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$, these two must be equal, giving
\[ P(1)=Q(1)=R(1)=S(1)=0. \]Hence, $x-1\mid P(x)$. $\blacksquare$
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huashiliao2020
1292 posts
#15
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sketch
This post has been edited 5 times. Last edited by huashiliao2020, Apr 14, 2023, 5:30 AM
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primemystic
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#16
Y by
Are there any easy solution without roots of unity?
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Rohit-2006
245 posts
#17
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Finding this problem from one week....finally....
Attachments:
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IceyCold
211 posts
#19
Y by
primemystic wrote:
Are there any easy solution without roots of unity?

There actually is.
Use the fact that $a-b$ divides $P(a)-P(b)$
And by that,I mean you can always factor out $a-b$ from $P(a)-P(b)$,of course assuming that $a-b$ is not $0$.
Applying the lemma three times should get you something that's somewhat obvious.If it isn't already,then I'll post the solution later.
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xytunghoanh
39 posts
#20
Y by
IceyCold wrote:
primemystic wrote:
Are there any easy solution without roots of unity?

There actually is.
Use the fact that $a-b$ divides $P(a)-P(b)$
And by that,I mean you can always factor out $a-b$ from $P(a)-P(b)$,of course assuming that $a-b$ is not $0$.
Applying the lemma three times should get you something that's somewhat obvious.If it isn't already,then I'll post the solution later.

I have solved it.
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xytunghoanh
39 posts
#21
Y by
By Bezout, we have
\[x^4+x^3+x^2+x+1 \mid x^5-1 \mid \left[P(x^5)+xQ(x^5)+x^2R(x^5)\right]-\left[P(1)+xQ(1)+x^2R(1)\right]\]Then
\[x^4+x^3+x^2+x+1 \mid P(1)+xQ(1)+x^2R(1)\]But $deg \ (P(1)+xQ(1)+x^2R(1))<deg \ (x^4+x^3+x^2+x+1)$ then
\[P(1)+xQ(1)+x^2R(1) \equiv 0 \ \forall \ x\]Hence, $P(1)=Q(1)=R(1)=0 \ \blacksquare$
This post has been edited 1 time. Last edited by xytunghoanh, Apr 22, 2025, 4:36 PM
Reason: fix
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IceyCold
211 posts
#22
Y by
xytunghoanh wrote:
By Bezout, we have
\[x^4+x^3+x^2+x+1 \mid x^5-1 \mid \left[P(x^5)+xQ(x^5)+x^2R(x^5)\right]-\left[P(1)+xQ(1)+x^2R(1)\right]\]Then
\[x^4+x^3+x^2+x+1 \mid P(1)+xQ(1)+x^2R(1)\]But $deg \ (P(1)+xQ(1)+x^2R(1))<deg \ (x^4+x^3+x^2+x+1)$ then
\[P(1)+xQ(1)+x^2R(1) \equiv 0 \ \forall \ x\]Hence, $P(1)=Q(1)=R(1)=0 \ \blacksquare$

Yep,that's the solution I was aiming for,thank you!
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