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Random Points = Problem
kingu   5
N 4 hours ago by happypi31415
Source: Chinese Geometry Handout
Let $ABC$ be a triangle. Let $\omega$ be a circle passing through $B$ intersecting $AB$ at $D$ and $BC$ at $F$. Let $G$ be the intersection of $AF$ and $\omega$. Further, let $M$ and $N$ be the intersections of $FD$ and $DG$ with the tangent to $(ABC)$ at $A$. Now, let $L$ be the second intersection of $MC$ and $(ABC)$. Then, prove that $M$ , $L$ , $D$ , $E$ and $N$ are concyclic.
5 replies
kingu
Apr 27, 2024
happypi31415
4 hours ago
Combo resources
Fly_into_the_sky   1
N 5 hours ago by Fly_into_the_sky
Ok so i never did combinatorics in my life :oops: and i am willing to be able to do P1/P4 combos (or even more)
So yeah how can i start from scratch?
Remark:i don't want compuational combo resources :noo:
1 reply
Fly_into_the_sky
5 hours ago
Fly_into_the_sky
5 hours ago
Very odd geo
Royal_mhyasd   2
N 5 hours ago by Royal_mhyasd
Source: own (i think)
nevermind
2 replies
Royal_mhyasd
Yesterday at 6:10 PM
Royal_mhyasd
5 hours ago
Polynomial Application Sequences and GCDs
pieater314159   46
N 5 hours ago by cursed_tangent1434
Source: ELMO 2019 Problem 1, 2019 ELMO Shortlist N1
Let $P(x)$ be a polynomial with integer coefficients such that $P(0)=1$, and let $c > 1$ be an integer. Define $x_0=0$ and $x_{i+1} = P(x_i)$ for all integers $i \ge 0$. Show that there are infinitely many positive integers $n$ such that $\gcd (x_n, n+c)=1$.

Proposed by Milan Haiman and Carl Schildkraut
46 replies
pieater314159
Jun 19, 2019
cursed_tangent1434
5 hours ago
c^a + a = 2^b
Havu   10
N 5 hours ago by Havu
Find $a, b, c\in\mathbb{Z}^+$ such that $a,b,c$ coprime, $a + b = 2c$ and $c^a + a = 2^b$.
10 replies
Havu
May 10, 2025
Havu
5 hours ago
Own made functional equation
JARP091   0
6 hours ago
Source: Own (Maybe?)
\[
\text{Find all functions } f : \mathbb{R} \to \mathbb{R} \text{ such that:} \\
f(a^4 + a^2b^2 + b^4) = f\left((a^2 - f(ab) + b^2)(a^2 + f(ab) + b^2)\right)
\]
0 replies
JARP091
6 hours ago
0 replies
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   16
N 6 hours ago by JARP091
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
16 replies
OgnjenTesic
May 22, 2025
JARP091
6 hours ago
equal segments on radiuses
danepale   8
N Today at 3:52 PM by zuat.e
Source: Croatia TST 2016
Let $ABC$ be an acute triangle with circumcenter $O$. Points $E$ and $F$ are chosen on segments $OB$ and $OC$ such that $BE = OF$. If $M$ is the midpoint of the arc $EOA$ and $N$ is the midpoint of the arc $AOF$, prove that $\sphericalangle ENO + \sphericalangle OMF = 2 \sphericalangle BAC$.
8 replies
danepale
Apr 25, 2016
zuat.e
Today at 3:52 PM
Inequality
SunnyEvan   8
N Today at 3:37 PM by arqady
Let $a$, $b$, $c$ be non-negative real numbers, no two of which are zero. Prove that :
$$ \sum \frac{3ab-2bc+3ca}{3b^2+bc+3c^2} \geq \frac{12}{7}$$
8 replies
SunnyEvan
Apr 1, 2025
arqady
Today at 3:37 PM
Inequality conjecture
RainbowNeos   2
N Today at 3:33 PM by RainbowNeos
Show (or deny) that there exists an absolute constant $C>0$ that, for all $n$ and $n$ positive real numbers $x_i ,1\leq i \leq n$, there is
\[\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\geq C \ln n\left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}}\]
2 replies
RainbowNeos
May 29, 2025
RainbowNeos
Today at 3:33 PM
2- player game on a strip of n squares with two game pieces
parmenides51   2
N Today at 3:31 PM by Gggvds1
Source: 2023 Austrian Mathematical Olympiad, Junior Regional Competition , Problem 3
Alice and Bob play a game on a strip of $n \ge  3$ squares with two game pieces. At the beginning, Alice’s piece is on the first square while Bob’s piece is on the last square. The figure shows the starting position for a strip of $ n = 7$ squares.
IMAGE
The players alternate. In each move, they advance their own game piece by one or two squares in the direction of the opponent’s piece. The piece has to land on an empty square without jumping over the opponent’s piece. Alice makes the first move with her own piece. If a player cannot move, they lose.

For which $n$ can Bob ensure a win no matter how Alice plays?
For which $n$ can Alice ensure a win no matter how Bob plays?

(Karl Czakler)
2 replies
parmenides51
Mar 26, 2024
Gggvds1
Today at 3:31 PM
PAMO Problem 4: Perpendicular lines
DylanN   11
N Apr 23, 2025 by ATM_
Source: 2019 Pan-African Mathematics Olympiad, Problem 4
The tangents to the circumcircle of $\triangle ABC$ at $B$ and $C$ meet at $D$. The circumcircle of $\triangle BCD$ meets sides $AC$ and $AB$ again at $E$ and $F$ respectively. Let $O$ be the circumcentre of $\triangle ABC$. Show that $AO$ is perpendicular to $EF$.
11 replies
DylanN
Apr 9, 2019
ATM_
Apr 23, 2025
PAMO Problem 4: Perpendicular lines
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G H BBookmark kLocked kLocked NReply
Source: 2019 Pan-African Mathematics Olympiad, Problem 4
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DylanN
194 posts
#1 • 2 Y
Y by Adventure10, Mango247
The tangents to the circumcircle of $\triangle ABC$ at $B$ and $C$ meet at $D$. The circumcircle of $\triangle BCD$ meets sides $AC$ and $AB$ again at $E$ and $F$ respectively. Let $O$ be the circumcentre of $\triangle ABC$. Show that $AO$ is perpendicular to $EF$.
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pad
1671 posts
#3 • 2 Y
Y by Adventure10, Mango247
We know $AD$ is the $A$-symmedian of $\triangle ABC$, so it is well known that $B,C,D,O$ are concyclic. So $(BCD)=(BCO)$. Then $\angle EAO = \angle CAO=90-B$, and $\angle AEF=180-\angle CEF=180-\angle CBF=B$ since $E,C,F,B$ are concyclic. Therefore, $AO\perp EF$.
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DylanN
194 posts
#4 • 2 Y
Y by Adventure10, Mango247
It's not important that $D$ is defined as the intersection of the tangents. We can replace the circumcircle of $\triangle BCD$ with any circle that passes through $B$ and $C$. Indeed, we know that $\angle CAO = 90^\circ - \angle B$, and since $BCEF$ is cyclic, we have that $\angle B = \angle CEF$, from which the result follows.
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khanhnx
1619 posts
#5 • 1 Y
Y by Adventure10
This problem has been posted in here
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AshAuktober
1013 posts
#6
Y by
Observe that we can obtain $\Delta AEF$ by reflecting $\Delta ABC$ about its $A-$angle bisector and taking a homothety at $A$. Now since the circumcentre and orthocentre are isogonal conjugates, $O$ has to lie on the $A-$altitude of $\Delta AEF$, and therefore we are done. $\square$
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Aiden-1089
302 posts
#7
Y by
Let $H$ be the orthocenter of $ABC$.
$\measuredangle BEC = \measuredangle BDC = \measuredangle BFC \implies BCEF$ is concyclic $\implies \Delta ABC \sim \Delta AEF$.
Consider taking a suitable homothety about A then reflecting across the angle bisector of $\angle BAC$, such that $BC$ goes to $EF$. Under this transformation, line $AH$ goes to line $AO$ because they are isogonal in $\angle BAC$. Since $AH \perp BC$, we have $AO \perp EF$.
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Jishnu4414l
155 posts
#8
Y by
We know that $\measuredangle OAC=90^{\circ}-\measuredangle CBA=\measuredangle KAE$
Also, $\measuredangle CBA=\measuredangle CEF=\measuredangle AEK$
Now, $\measuredangle KAE+\measuredangle AEK+\measuredangle EKA=0$
$\implies 90^{\circ}-\measuredangle CBA+\measuredangle CBA +\measuredangle EKA=0$
$\implies \measuredangle EKA=90^{\circ}$
And we are done!
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fasttrust_12-mn
118 posts
#9
Y by
Claim: $O$ lies on the circumcircle of $\triangle BDC$
proof:
$$\measuredangle BAC =\measuredangle BCD \implies \measuredangle BDC= 180^{\circ}- 2\times \measuredangle BAC$$, we know that
$$\measuredangle BOC = 2\times \measuredangle BAC$$so
$$OBDC  \text{ is cyclic quad }$$
$$\measuredangle FBC=\measuredangle AEF $$$$\measuredangle OAC=\measuredangle OCA$$$$\measuredangle OCB=90^{\circ}$$we know that
$$\measuredangle AEG=90^{\circ}-\measuredangle BAO-\measuredangle OAE + \measuredangle BAO\implies 90^{\circ}-\measuredangle OAE$$$$\measuredangle AGE =180^{\circ}-(90^{\circ}-\measuredangle OAE+\measuredangle  OAE)=90^{\circ}$$hence we are done $\blacksquare$
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Tony_stark0094
69 posts
#10
Y by
$$\angle EBF=\angle BAF+\angle AFB=\angle A + \angle BDC=\angle A + 180 - 2\angle A = 180 - \angle A$$$$\implies FA=FB$$$$by\ symmetry\ we \ can\ get \ that\ EA=EC$$hence $F$ lies on perpendicular bisector of $AB$ and $E$ lies on perpendicular bisector of $AC$ moreover $G$ also lies on these lines
$\implies G$ is the orthocentre of $\Delta AEF$ hence $AG$ is perpendicular to $EF$
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Rayanelba
20 posts
#11 • 1 Y
Y by ATM_
To prove that $AO$ perpendicular to $EF$ , it suffies to prove that: $OAC=\frac{\pi}{2}-\measuredangle CFE$
We know that : $\measuredangle OAC=\frac{\pi}{2}-\measuredangle CBA$
And : $\measuredangle CBA=\pi-\measuredangle CBE=\measuredangle CFE$ (because CFEB is cyclic)
$\implies \measuredangle OAC=\frac{\pi}{2}-\measuredangle CFE$
Q.E.D
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ATM_
23 posts
#12
Y by
nice solution*
This post has been edited 1 time. Last edited by ATM_, Apr 23, 2025, 11:08 PM
Reason: .
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ATM_
23 posts
#13
Y by
here is a cute sketch
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