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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Hard Number Theory
MuradSafarli   0
a few seconds ago
Find all odd natural numbers \( m \) such that \( m^2 - 1 \mid 3^m + 5^m \).
0 replies
MuradSafarli
a few seconds ago
0 replies
A and B take stones from a pile
WakeUp   4
N 11 minutes ago by reni_wee
Source: CentroAmerican 2003
Two players $A$ and $B$ take turns playing the following game: There is a pile of $2003$ stones. In his first turn, $A$ selects a divisor of $2003$ and removes this number of stones from the pile. $B$ then chooses a divisor of the number of remaining stones, and removes that number of stones from the new pile, and so on. The player who has to remove the last stone loses. Show that one of the two players has a winning strategy and describe the strategy.
4 replies
WakeUp
Nov 29, 2010
reni_wee
11 minutes ago
Pisano period
JARP091   3
N 17 minutes ago by JARP091
Source: At the time of writing this problem i do not know the source if any
Let $F_n$ denote the $n$th Fibonacci number, defined by the recurrence:
\[
F_0 = 0,\quad F_1 = 1,\quad \text{and} \quad F_{n+2} = F_{n+1} + F_n \quad \text{for } n \geq 0.
\]For a fixed modulus $m \in \mathbb{N}$, define the set
\[
S_m = \left\{ (i, j) \in \mathbb{N}^2 : F_i \cdot F_j \equiv 1 \pmod{m} \right\}.
\]Prove that $|S_m| = \infty$ if and only if $5 \nmid m$.

Note: Its a very beautiful problem
3 replies
JARP091
an hour ago
JARP091
17 minutes ago
FE with conditions on $x,y$
Adywastaken   1
N 26 minutes ago by jasperE3
Source: OAO
Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $\forall y>x>0$,
\[
f(x^2+f(y))=f(xf(x))+y
\]
1 reply
Adywastaken
an hour ago
jasperE3
26 minutes ago
Inequalities
sqing   3
N 5 hours ago by bogpt
Let $ a,b,c\geq 0 $ and $ab+bc+ca =1.$ Prove that
$$(a^2+b^2+c^2)(a+b+c-2)\ge 8abc(1-a-b-c) $$$$(a^2+b^2+c^2)(a+b+c-\frac{5}{2})\ge 2abc(1-a-b-c) $$
3 replies
sqing
Yesterday at 2:26 PM
bogpt
5 hours ago
Inequalities
sqing   1
N 6 hours ago by sqing
Let $ a,b,c $ be real numbers . Prove that
$$\frac{(a-1)(b-1)(c-1)}{(a^2+3)(b^2+1)(c^2+3)} \ge -\frac{1+\sqrt 2}{8}$$$$\frac{(a-1)(b-1)(c-1)}{(a^2+1)(b^2+3)(c^2+1)} \ge -\frac{3+2\sqrt{2}}{8}$$$$\frac{(a-1)(b-1)(c-1)}{(a^2+3)(b^2+2)(c^2+3)} \ge -\frac{1+\sqrt3}{16}$$$$\frac{(a-1)(b-1)(c-1)}{(a^2+2)(b^2+3)(c^2+2)} \ge -\frac{2+\sqrt{3}}{16}$$
1 reply
sqing
6 hours ago
sqing
6 hours ago
[PMO25 Qualifying II.8] A Square Can't Be A Floor
kae_3   2
N Today at 12:36 PM by tapilyoca
Determine the largest perfect square less than $1000$ that cannot be expressed as $\lfloor x\rfloor + \lfloor 2x\rfloor + \lfloor 3x\rfloor + \lfloor 6x\rfloor$ for some positive real number $x$.

Answer Confirmation
2 replies
kae_3
Feb 21, 2025
tapilyoca
Today at 12:36 PM
solve in R
zolfmark   1
N Today at 12:10 PM by Mathzeus1024
x+1/y=9/2 and y+1/z=11/4 and z+1/x=12/5
1 reply
zolfmark
Feb 16, 2016
Mathzeus1024
Today at 12:10 PM
not so hard problem (own)
BinariouslyRandom   1
N Today at 11:38 AM by BinariouslyRandom
An open rectangular box is made from exactly $900 m^2$ of cardboard, with 5 rectangular faces instead of 6 and there is no leftover cardboard. What is the largest volume this box can hold given that $100 m^2$ of cardboard is needed to cover the box?
1 reply
BinariouslyRandom
Today at 11:37 AM
BinariouslyRandom
Today at 11:38 AM
Minimize z.
Entrepreneur   3
N Today at 11:17 AM by no_room_for_error
Minimize $z = 6x + 3y.$ Subject to the constraints:
$$\begin{cases}
4x+y\ge80,\\
x+5y \ge115,\\
3x+2y\le150,\\
x,y\ge0.
\end{cases}$$
3 replies
Entrepreneur
May 21, 2025
no_room_for_error
Today at 11:17 AM
Inequalities
sqing   18
N Today at 7:48 AM by sqing
Let $ a,b,c\geq 0 ,a+b+c\leq 3. $ Prove that
$$a^2+b^2+c^2+ab +2ca+2bc +  abc \leq \frac{251}{27}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{2}{5}abc  \leq \frac{4861}{540}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{7}{20}abc  \leq \frac{2381411}{26460}$$
18 replies
sqing
May 21, 2025
sqing
Today at 7:48 AM
Inequalities
sqing   18
N Today at 7:20 AM by sqing
Let $ a,b>0   $ . Prove that
$$ \frac{a}{a^2+a +2b+1}+ \frac{b}{b^2+2a +b+1}  \leq  \frac{2}{5} $$$$ \frac{a}{a^2+2a +b+1}+ \frac{b}{b^2+a +2b+1}  \leq  \frac{2}{5} $$
18 replies
sqing
May 13, 2025
sqing
Today at 7:20 AM
helpppppppp me
stupid_boiii   1
N Today at 6:56 AM by vanstraelen
Given triangle ABC. The tangent at ? to the circumcircle(ABC) intersects line BC at point T. Points D,E satisfy AD=BD, AE=CE, and ∠CBD=∠BCE<90 ∘ . Prove that D,E,T are collinear.
1 reply
stupid_boiii
Yesterday at 4:22 AM
vanstraelen
Today at 6:56 AM
Algebra Polynomials
Foxellar   2
N Today at 5:43 AM by Foxellar
The real root of the polynomial \( p(x) = 8x^3 - 3x^2 - 3x - 1 \) can be written in the form
\[
\frac{\sqrt[3]{a} + \sqrt[3]{b} + 1}{c},
\]where \( a, b, \) and \( c \) are positive integers. Find the value of \( a + b + c \).
2 replies
Foxellar
Today at 4:52 AM
Foxellar
Today at 5:43 AM
Geo with unnecessary condition
egxa   8
N Apr 4, 2025 by ErTeeEs06
Source: Turkey Olympic Revenge 2024 P4
Let the circumcircle of a triangle $ABC$ be $\Gamma$. The tangents to $\Gamma$ at $B,C$ meet at point $E$. For a point $F$ on line $BC$ which is not on the segment $BC$, let the midpoint of $EF$ be $G$. Lines $GB,GC$ meet $\Gamma$ again at points $I,H$ respectively. Let $M$ be the midpoint of $BC$. Prove that the points $F,I,H,M$ lie on a circle.

Proposed by Mehmet Can Baştemir
8 replies
egxa
Aug 6, 2024
ErTeeEs06
Apr 4, 2025
Geo with unnecessary condition
G H J
G H BBookmark kLocked kLocked NReply
Source: Turkey Olympic Revenge 2024 P4
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egxa
211 posts
#1 • 2 Y
Y by SerdarBozdag, Rounak_iitr
Let the circumcircle of a triangle $ABC$ be $\Gamma$. The tangents to $\Gamma$ at $B,C$ meet at point $E$. For a point $F$ on line $BC$ which is not on the segment $BC$, let the midpoint of $EF$ be $G$. Lines $GB,GC$ meet $\Gamma$ again at points $I,H$ respectively. Let $M$ be the midpoint of $BC$. Prove that the points $F,I,H,M$ lie on a circle.

Proposed by Mehmet Can Baştemir
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sami1618
913 posts
#3
Y by
Consider the linear function $f(X)=\mathbb{P}(X,\Gamma)-\mathbb{P}(X,G)$. Then $$\mathbb{P}(G,\Gamma)=f(G)=\frac{f(E)+f(F)}{2}=\frac{BE^2+BF \cdot CF -EG^2-FG^2}{2}=$$$$\frac{(BM^2+BF\cdot CF+EM^2)-2EG^2}{2}=\frac{(FM^2+EM^2)-2EG^2}{2}=\frac{EF^2-2EG^2}{2}=EG$$Thus inverting about $G$ with radius $EG$ maps $F$, $I$, $H$, and $M$ all to line $BC$, finishing the problem.
Attachments:
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SerdarBozdag
892 posts
#4 • 4 Y
Y by alexanderhamilton124, bin_sherlo, egxa, ehuseyinyigit
$\angle FME = 90, GF = GE$ implies $GF = GM = GE$. Note that $G$ is on the radical axis of the point circle $(E)$ and $(ABC)$ because it lies on the $E$-midline of the triangle $EBC$. This gives
$GF^2 = GM^2 = GB \cdot GI = GH \cdot GC$ and because $B,C \in FM$; $F,I,M,H,G$ are cyclic.
This post has been edited 2 times. Last edited by SerdarBozdag, Aug 6, 2024, 2:23 PM
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hukilau17
288 posts
#5 • 1 Y
Y by ehuseyinyigit
We complex bash. Let $\Gamma$ be the unit circle (and let $j$, rather than $i$, denote the coordinate of $I$), so that
$$|a|=|b|=|c|=1$$$$e=\frac{2bc}{b+c}$$$$m=\frac{b+c}2$$$$\overline{f} = \frac{b+c-f}{bc}$$$$g = \frac{e+f}2 = \frac{2bc+bf+cf}{2(b+c)}$$$$\overline{g} = \frac{\frac2{bc}+\frac{b+c-f}{b^2c}+\frac{b+c-f}{bc^2}}{2\left(\frac1b+\frac1c\right)} = \frac{b^2+4bc+c^2-bf-cf}{2bc(b+c)}$$$$j = \frac{b-g}{b\overline{g}-1} = \frac{\frac{2b^2-bf-cf}{2(b+c)}}{\frac{b^2+2bc-c^2-bf-cf}{2c(b+c)}} = \frac{c(2b^2-bf-cf)}{b^2+2bc-c^2-bf-cf}$$$$h = \frac{c-g}{c\overline{g}-1} = \frac{\frac{2c^2-bf-cf}{2(b+c)}}{\frac{-b^2+2bc+c^2-bf-cf}{2b(b+c)}} = \frac{b(2c^2-bf-cf)}{-b^2+2bc+c^2-bf-cf}$$$$j-f = \frac{bf^2+cf^2-b^2f-3bcf+2b^2c}{b^2+2bc-c^2-bf-cf} = \frac{(b-f)(2bc-bf-cf)}{b^2+2bc-c^2-bf-cf}$$$$h-f = \frac{bf^2+cf^2-c^2f-3bcf+2bc^2}{-b^2+2bc+c^2-bf-cf} = \frac{(c-f)(2bc-bf-cf)}{-b^2+2bc+c^2-bf-cf}$$$$j-m = \frac{-b^3+b^2c-bc^2+c^3+b^2f-c^2f}{2(b^2+2bc-c^2-bf-cf)} = \frac{(c-b)(b^2+c^2-bf-cf)}{2(b^2+2bc-c^2-bf-cf)}$$$$h-m = \frac{b^3-b^2c+bc^2-c^3-b^2f+c^2f}{2(-b^2+2bc+c^2-bf-cf)} = \frac{(b-c)(b^2+c^2-bf-cf)}{2(-b^2+2bc+c^2-bf-cf)}$$Thus
$$\frac{(j-f)(h-m)}{(h-f)(j-m)} = -\frac{b-f}{c-f}$$which is real since $F$ lies on line $BC$. $\blacksquare$
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ohhh
48 posts
#6
Y by
Fix triangle $ABC$ and move the point $F$ along the line $BC$ (Yes, ignore the "which is not on the segment" condition).
$degF = 1 \implies degG  = 1 \implies degH = degI =1$
Notice that the condition $(IMGF)$ cyclic has degree at most $degGI + degMI + degMF +deg GF= 1 + 1 + 0 + 1= 3$.
So $4$ points are enough!
$1: F = B$
$2: F = M$
$3: F$ such that $G = M_{bc}$
$4: F$ such that $G = M_a$
Okay! Similarly, we can prove $(HMGF)$ and consequently $(IMHGF)$, as desired.
This post has been edited 3 times. Last edited by ohhh, Aug 7, 2024, 12:41 AM
Reason: Typos
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Anancibedih
18 posts
#7 • 2 Y
Y by ehuseyinyigit, Ege_Saribass
Let $T$ be the point that satisfies $FBET$ parallelogram. $\angle{TFB}=\angle{EBC}=\angle{BCE},FC||TE\Rightarrow \hspace{1mm}\text{FTEC is cyclic}$ $\angle{TEC}+\angle{FCE}=180^\circ=\angle{TEC}+\angle{TIC}\Rightarrow \text{I lies on FTEC circle}$. So $FB\cdot BC=2FB\cdot BM=IB\cdot BT=2BG\cdot IB\Rightarrow \text{FGMI is cyclic}$. We know that $FG=GM$ so $\angle{FIG}=\angle {GIM}=\angle{FMG}$ so $\triangle {GMB}\sim \triangle {GIM}\Rightarrow GM^2=GB\cdot GI=GH\cdot GC$ so $\angle{GMF}=\angle{MHC}$ and we get $\angle{GIM}+\angle{GHM}=180^\circ$ so $H$ lies on $FGMI$ circle. We're done.
This post has been edited 3 times. Last edited by Anancibedih, Aug 21, 2024, 2:01 PM
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khanhnx
1618 posts
#8 • 1 Y
Y by MS_asdfgzxcvb
We have $BC$ is polar of $E$ WRT $\Gamma$ so $\bigodot(EF)$ is orthogonal to $\Gamma$. From this, we have $\mathcal{I}^{k = GM^2}_G: \Gamma \longleftrightarrow \Gamma$. Hence $\mathcal{I}^{k = GM^2}_G: F \longleftrightarrow F, M \longleftrightarrow M, B \longleftrightarrow I, C \longleftrightarrow H$. But $B, C, M, F$ are collinear then $G, I, H, F, M$ lie on a circle
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ehuseyinyigit
839 posts
#10
Y by
The most natural way is constructing parallelograms. Let $C'$ and $B'$ be reflection of the points $C$ and $B$ wrt to $G$ respectively. Then $\angle BHC=\angle BAC=\angle BCE=\angle BFC'$ implies $BHC'F$ is cyclic and thus $FHMG$ (Since $BC.CF=2MC.CF=C'C.CH=2GC.CH$). On the other hand, $BE=EC$ implies $CEB'F$ is cyclic-isosceles trapezoid. Furthermore $\angle BHC=\angle CIB'=\angle CEB'$ gives $B'EICF$ is cyclic. Then PoP gives $BC.BF=2BM.BF=BI.BB'=2BI.BG$ implies $FGMI$ is cyclic. The result follows from $(FGMI)$ and $(FHMG)$ as desired.
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ErTeeEs06
67 posts
#11
Y by
$G$ is on the radax of $(ABC)$ and pointcircle $E$, since that radax is the midparallel in triangle $EBC$. So inversion with radius $GE$ around $G$ sends $H$ to $C$ and $I$ to $B$. Now since $G$ is midpoint of $EF$ and $\angle EMF=90^\circ$ we see that $GE=GM=GF$ so $M$ and $F$ remain fixed. So the images of $F, I, H, M$ are collinear so $F, I, H, M$ are concyclic.
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