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find positive n so that exists prime p with p^n-(p-1)^n$ a power of 3
parmenides51   14
N an hour ago by MathLuis
Source: JBMO Shortlist 2017 NT5
Find all positive integers $n$ such that there exists a prime number $p$, such that $p^n-(p-1)^n$ is a power of $3$.

Note. A power of $3$ is a number of the form $3^a$ where $a$ is a positive integer.
14 replies
parmenides51
Jul 25, 2018
MathLuis
an hour ago
GCD of terms in a sequence
BBNoDollar   1
N an hour ago by mashumaro
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
1 reply
BBNoDollar
5 hours ago
mashumaro
an hour ago
Diodes and usamons
v_Enhance   46
N 2 hours ago by HamstPan38825
Source: USA December TST for the 56th IMO, by Linus Hamilton
A physicist encounters $2015$ atoms called usamons. Each usamon either has one electron or zero electrons, and the physicist can't tell the difference. The physicist's only tool is a diode. The physicist may connect the diode from any usamon $A$ to any other usamon $B$. (This connection is directed.) When she does so, if usamon $A$ has an electron and usamon $B$ does not, then the electron jumps from $A$ to $B$. In any other case, nothing happens. In addition, the physicist cannot tell whether an electron jumps during any given step. The physicist's goal is to isolate two usamons that she is sure are currently in the same state. Is there any series of diode usage that makes this possible?

Proposed by Linus Hamilton
46 replies
v_Enhance
Dec 17, 2014
HamstPan38825
2 hours ago
China South East Mathematical Olympiad 2013 problem 2
s372102   3
N 2 hours ago by AGCN
$\triangle ABC$, $AB>AC$. the incircle $I$ of $\triangle ABC$ meet $BC$ at point $D$, $AD$ meet $I$ again at $E$. $EP$ is a tangent of $I$, and $EP$ meet the extension line of $BC$ at $P$. $CF\parallel PE$, $CF\cap AD=F$. the line $BF$ meet $I$ at $M,N$, point $M$ is on the line segment $BF$, the line segment $PM$ meet $I$ again at $Q$. Show that $\angle ENP=\angle ENQ$
3 replies
s372102
Aug 10, 2013
AGCN
2 hours ago
f(x+y) = max(f(x), y) + min(f(y), x)
Zhero   50
N 2 hours ago by lpieleanu
Source: ELMO Shortlist 2010, A3
Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x+y) = \max(f(x),y) + \min(f(y),x)$.

George Xing.
50 replies
Zhero
Jul 5, 2012
lpieleanu
2 hours ago
Similar to iran 1996
GreekIdiot   1
N 2 hours ago by Lufin
Let $f: \mathbb R \to \mathbb R$ be a function such that $f(f(x)+y)=f(f(x)-y)+4f(x)y \: \forall x,y \: \in \: \mathbb R$. Find all such $f$.
1 reply
GreekIdiot
Apr 26, 2025
Lufin
2 hours ago
Prove that 4p-3 is a square - Iran NMO 2005 - Problem1
sororak   24
N 2 hours 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
2 hours ago
IMO ShortList 1999, combinatorics problem 4
orl   27
N 2 hours 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
2 hours ago
Geometry
Lukariman   2
N 2 hours 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
2 hours ago
IMO 2010 Problem 1
canada   119
N 2 hours 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
2 hours ago
Question 2
Valentin Vornicu   88
N Apr 20, 2025 by Nari_Tom
Consider five points $ A$, $ B$, $ C$, $ D$ and $ E$ such that $ ABCD$ is a parallelogram and $ BCED$ is a cyclic quadrilateral. Let $ \ell$ be a line passing through $ A$. Suppose that $ \ell$ intersects the interior of the segment $ DC$ at $ F$ and intersects line $ BC$ at $ G$. Suppose also that $ EF = EG = EC$. Prove that $ \ell$ is the bisector of angle $ DAB$.

Author: Charles Leytem, Luxembourg
88 replies
Valentin Vornicu
Jul 25, 2007
Nari_Tom
Apr 20, 2025
Question 2
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TechnoLenzer
55 posts
#75
Y by
Let $\Gamma = (BCED)$, $P = DA \cap \Gamma$, $Q = BA \cap \Gamma$. We have $F = \lambda \cap CD$, and define $X = QF \cap \Gamma$.

By Pascal on $PXQBCD$, we have $A = BQ \cap DP$, $F = XQ \cap CD$, and $BC \cap XP$ collinear. Hence they are collinear on $\lambda$, and since $G = \lambda \cap BC$ then $G$ lies on $XP$.

Claim 1: $X$ is the $C$-dumpty point of $\triangle CGF$.
Proof: $\angle XGC = \angle PGC = \angle GPD = \angle XPD = \angle XCD = \angle XCF$. Similarly, $\angle GCX = \angle BPX = \angle BQX = \angle CFX$. Hence, by AA we have $\triangle XCG \sim \triangle XFC$, and $X$ is the $C$-dumpty point. $\square$

It's well-known that $CX \perp EX$. Hence, since $C, X, E$ lie on $\Gamma$, either $X = E$ or $E$ is the $C$-antipode in $\Gamma$.

Case 1: $X = E \Rightarrow \angle ECG = \angle EFC = \angle FCE \Rightarrow CE$ is the angle bisector of $\angle FCG$. Hence $\triangle CGF$ is isosceles. Since $\triangle FDA \sim \triangle FCG$ by AA we have $\angle FAB = \angle AFD = \angle DAF$, and $\lambda$ is the bisector of $\angle DAB$.

Case 2: $E$ is the $C$-antipode. Then $(CGF)$ is the image of $\Gamma$ under homothety scale factor 2 at $C$, and so no other point on $(CGF)$ lies inside $\Gamma$. But $F \in (CGF)$ lies on segment $CD$ within $\Gamma$, contradiction.
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JAnatolGT_00
559 posts
#76
Y by
Note that the Simson line of $E$ wrt $\triangle BCD$ is homothetic to $\ell$ wrt $C$ with factor $1/2,$ and therefore passes through the center of $ABCD$. This yields $|EB|=|ED|\implies \angle FCE=\angle DBE=\angle BDE=\angle GCE,$ so it's trivial that $|CF|=|CG|$ and hence $\angle BAF=\angle CFG=\angle CGF=\angle DAF,$ as desired.
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HamstPan38825
8860 posts
#77
Y by
The key is to construct the Simson Line of $E$ with respect to triangle $BCD$. By taking a homothety at $C$ with ratio $2$, it follows that this line is parallel to $\ell$.

On the other hand, it is well-known that the the Simson Line is parallel to $\overline{CJ}$, where $J$ is the point at which the altitude at $E$ to $\overline{BD}$ meets $(BCD)$. Thus, it suffices to show $J$ is the arc midpoint.

But the aforementioned homothety implies that the Simson Line bisects $\overline{AC}$, so it bisects $\overline{BD}$ too. This common midpoint $O$ is thus both the midpoint and foot from $E$ in triangle $EAB$, implying that it is isosceles. Thus, $\overline{EJ}$ bisects $\widehat{BD}$, which is enough.
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awesomeming327.
1714 posts
#78
Y by
Let $M,N,O$ be midpoints of $CG$, $CF$, and $BD$, respectively. Note that $EN\perp CD$ and $EM\perp BC$. Note that the perpendicular from $E$ onto $BD$ must lie on $MN$. Since $O,M,N$ are midpoints of $AC$, $FC$, and $GC$ respectively, $O$ lies on $MN$. Therefore, $EO\perp BD$.

Now, $EB=ED$, $\angle EMB=\angle END$ and $\angle NDE=\angle EBM$ so $\triangle BME\cong \triangle DNE$. Therefore, $EM=EN$ which implies $CG=CF$. Thus, \[\angle FAD=\angle CGF=\angle CFG=\angle BAF\]as desired.
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Marinchoo
407 posts
#79
Y by
Let $H_{b}$ and $H_{d}$ be the midpoints of $CG$ and $CF$, respectively. Notice that $\triangle ADF \sim \triangle GCF$ and $\triangle EH_{d}D \sim \triangle EH_{b}B$. The first follows from $ABCD$ being a parallelogram and the second follows from $\angle EDC = \angle EBC$ combined with $\overline{EH_{d}} \perp \overline{FC}$ and $\overline{EH_{b}} \perp \overline{CG}$. These give:
\[\frac{\frac{1}{2}CF\cdot\tan\angle EFC}{\frac{1}{2}CG\cdot\tan\angle ECG} = \frac{EH_{d}}{EH_{b}} = \frac{DH_{d}}{BH_{b}} = \frac{DF + FH_{d}}{BC + CH_{b}} = \frac{DF + \frac{1}{2}CF}{BC + \frac{1}{2}CG} = \frac{CF}{CG}\cdot \frac{\frac{DF}{CF}+\frac{1}{2}}{\frac{AD}{CG}+\frac{1}{2}} = \frac{CF}{CG}\]where the last equality follows from $\triangle ADF \sim \triangle GCF$. Now $\tan\angle EFC = \tan\angle ECG$ and the angles are acute, hence $\angle EFC = \angle ECG$. This implies that $\triangle EFC \cong \triangle EGC$, so $CF = CG$ and therefore $AD = DF$, implying that $\ell$ is the angle bisector of $\angle BAC$ as desired.
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huashiliao2020
1292 posts
#80
Y by
Got a hint to construct midpoints, but I think the finish is pretty unique!
Let J,K,L be the midpoints of GC,FC,BD. Notice that by a homothety of 1/2 ratio at C we get AFG is a line implies LKJ is a line; Since EJ perp. BC, EK perp. CD, by converse of Simson line we get that EL perp. BD. Now note by pop we get $$\frac{DF}{FC}=\frac{AD}{CG}=\frac{BC}{CG}\implies\frac{DF}{BC}=\frac{FC}{CG}=\frac{AB}{BG}=\frac{CD}{BG}\implies BC\cdot CD=DF\cdot BG\implies^{BC\cdot BG=DF\cdot CD}CD=BG,BC=DF\implies BAG=BGA=GAD,$$as desired. $\blacksquare$
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IAmTheHazard
5001 posts
#81 • 2 Y
Y by centslordm, ike.chen
Let $M$ be the midpoint of $\overline{BD}$, so by a homothety at $C$, $M$ is collinear with the feet of the altitudes from $E$ to $\overline{CB}$ and $\overline{CD}$. On the other hand, by Simson lines, these two feet are also collinear with the foot from $E$ to $\overline{BD}$, hence this foot is just $M$, so $E$ is the midpoint of arc $BCD$.

Then $\angle FCE=\angle GCE$, so $CF=CG$ and thus $\angle GFC=\angle FGC \implies \angle GAB=\angle FAD$, which finishes the problem. $\blacksquare$

EDIT: bro ive done this problem before too whatttttttttttt
This post has been edited 1 time. Last edited by IAmTheHazard, Aug 30, 2023, 8:57 PM
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asdf334
7585 posts
#82
Y by
had to get hint ;-; my brain

Just draw Simson line of $E$ to $BCD$, which passes through midpoint of $BD$. So $E$ is arc midpoint, finish since $\triangle ECF\sim\triangle EBD\sim \triangle EGC$
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EpicBird08
1751 posts
#83 • 1 Y
Y by GeoKing
I have absolutely no clue how I even conjured this solution.

We start with a key claim:

Claim: Let $XYZ$ be a triangle, and let $P$ and $Q$ be on lines $XY$ and $XZ$ such that $PQ \parallel YZ.$ Let $R$ be on $YZ$ such that $YPQR$ is a parallelogram. Then regardless of the position of points $P$ and $Q$, the circumcircle of $\triangle YPR$ will pass through a fixed point other than $Y.$

Proof: It suffices to show that all circles of this type share a radical axis. To do so, we will use barycentric coordinates with $\triangle XYZ$ as the reference triangle. Let $X = (1,0,0), Y = (0,1,0),$ and $Z = (0,0,1).$ Also let $P = (1 - \lambda, \lambda, 0)$ and $Q = (1 - \lambda, 0, \lambda),$ where $\lambda$ is an arbitrary real number, so that $P - Q = \lambda(0,1,-1) = \lambda \cdot (Y-Z)$ and therefore $PQ \parallel YZ.$ Then $R = Q + Y - P = (0,1-\lambda, \lambda).$

Now we find the circumcircle of $\triangle YPR.$ A point $(x,y,z)$ on this circle must satisfy an equation of the form
\[
a^2 yz + b^2 zx + c^2 xy = ux + vy + wz,
\]where $a = YZ, b = ZX,$ and $c = XY.$ Plugging in $Y = (0,1,0)$ gives $v = 0,$ so
\[
a^2 yz + b^2 zx + c^2 xy = ux + wz.
\]Plugging in $P = (1 - \lambda, \lambda, 0)$ gives $u(1-\lambda) = c^2 \lambda(1-\lambda),$ so $u = c^2 \lambda.$ Similarly, plugging in $R = (0, 1-\lambda, \lambda)$ gives $w\lambda = a^2 \lambda(1-\lambda),$ so $w = a^2 (1-\lambda).$

Therefore, our circumcircle is
\[
\boxed{a^2 yz + b^2 zx + c^2 xy = c^2 \lambda x + a^2 (1-\lambda) z}.
\]If we let $\lambda_1$ and $\lambda_2$ be two distinct real values of $\lambda,$ then the radical axis of the two corresponding circles is given by
\[
(c^2 \lambda_1 - c^2 \lambda_2) x + (a^2 (1 - \lambda_1) - a^2 (1-\lambda_2))z = 0.
\]Dividing both sides by $\lambda_1 - \lambda_2$ gives $c^2 x - a^2 z = 0,$ which is clearly fixed. This concludes our (lengthy) proof of the claim.

Now apply the claim with $(X,Y,Z,P,Q,R) = (G,C,F,B,A,D).$ Since $E$ is the circumcenter of $GCF,$ it has no relevance to $B,A,$ or $D,$ so $E$ is the fixed point as shown to exist by the claim. Taking $\lambda$ to be arbitrarily close to $0,$ we discover that $(FEC)$ is tangent to $CG.$ Hence $\measuredangle GCF = \measuredangle CEF = 2 \measuredangle CGF.$ However, by sum of angles in a triangle, $\measuredangle GCF = \measuredangle CGF + \measuredangle GFC,$ so $\measuredangle CGF = \measuredangle GFC.$ Since $CF \parallel AB$ and $DA \parallel CG,$ we get $\measuredangle CGF = \measuredangle DAF$ and $\measuredangle GFC = \measuredangle FAB.$ Thus $\measuredangle DAF = \measuredangle FAB,$ and we conclude.

Ok reading the solutions above there is a Simson line??????
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Z4ADies
64 posts
#84
Y by
Different and short solution :D
Assume that,the angle bisector of $\angle DAB \cap CD=G$ and $\angle DAB \cap BC=F$ call triangle $CGF$'s center $E$ we will prove $BCED$ cyclic.Let,$\angle DAG=\angle GAB=\angle AGD=\angle CGF=\angle CFG=\alpha$ from Ptolemy's sine lemma that is equivalent to show $CE\sin(2\alpha)+CB\sin(90-\alpha)=CD\sin(90+\alpha)$.We know $AD=DG=BC=x$ and $GC=CF=y$ ,we also know $AB=BF$.Sine Thrm to $\triangle CEG$ $\implies$ $CE\sin(2\alpha)=y\sin(90-\alpha)$.So, $\sin(90-\alpha)(x+y)=CD\sin(90+\alpha)=(x+y)\sin(90+\alpha)$ which is indeed true.
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cursed_tangent1434
622 posts
#86
Y by
Magical solution, needed a pointer to look at Simson Lines. Let $M$ and $N$ denote the midpoints of $GC$ and $FC$. Now, let $P = \overline{AC} \cap \overline{MN}$. By the Midpoint Theorem, it follows that $P$ is the midpoint of $AC$. Since diagonals of a parallelogram bisect each other, it also follows that $P$ is the midpoint of $BD$. But now, consider the Simson Line of $E$ with respect to $\triangle PCD$. Clearly the feet of the perpendicular from $E$ to $GC$ and $FC$ and $M$ and $N$ and thus, $EP\perp BD$. But note that this means $\overline{EP}$ is in fact the perpendicular bisector of segment $BD$, which implies that $EB=ED$. Thus,
\[\measuredangle BAD = \measuredangle DCB = \measuredangle DEB = 2\measuredangle DBE =  2 \measuredangle FCE = \measuredangle FEC = 2 \measuredangle FGC = 2\measuredangle FAD \]which indeed implies that $\ell$ is the internal $\angle BAD-$bisector.
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AngeloChu
470 posts
#87
Y by
denote the perpendiculars from $E$ to $BC$, $CD$, and $BD$ as $P_1$, $P_2$, and $P_3$ respectively, and they are collinear by simson line
it is easy to tell that $P_1$ is the midpoint of $CG$ and $P_2$ is the midpoint of $CF$, so $P_1P_2P_3$ is parallel to $AFG$ so $P_3$ is the midpoint of $BD$ and $AC$
$DBE=DCE$ since $BCDE$ is cyclic, and $GCE=BDE$ also since $BCDE$ is cyclic
thus $FCE=GCE$ and we can prove $CE$ is perpendicular to $FG$ and $P_1P_2$
thus, $CP_1=CP_2$, and angle chasing gives $BAG=DAG=CP_1P_2$
thus, $ \ell$ is the bisector of angle $DAB$
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ihatemath123
3446 posts
#88
Y by
Let $\ell'$ be the image of $\ell$ under a homothety of factor $\tfrac{1}{2}$ centered at $C$. Simson's theorem on $E$ WRT $(BCD)$ tells us that the foot from $E$ to $\overline{BD}$ lies on $\ell'$. But the intersection of $\ell'$ with $\overline{BD}$ is always the midpoint of $\overline{BD}$, so $E$ is one of the arc midpoints. This determines $\ell$, as desired.
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ItsBesi
146 posts
#89
Y by
My idea is the same but I am posting it for storage, solved it on 10.10.2024
Here is my long solution:

Let $K$,$M$ and $N$ be the midpoints of the sides $CG$,$CF$ and $AC$ respectively.

Claim: Points $\overline{N-M-K}$ are collinear
Proof:


Claim: $ED=EB$
Proof:

Claim: $\ell$ is the angle bisector of $\angle DAB$
Proof:
Attachments:
This post has been edited 1 time. Last edited by ItsBesi, Mar 30, 2025, 7:21 PM
Reason: typo
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Nari_Tom
117 posts
#90
Y by
My approach was also nice.

Of course we have $\frac{BC}{CG}=\frac{DF}{FC}$, and $\angle CDE= \angle GBE$. Since we have $E$ lies on perpendicular bisectors of $CG$ and $CE$, we get $EBCG \sim EDFG$ which concludes the result instantly.
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