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Geometry :3c
popop614   3
N an hour ago by ItzsleepyXD
Source: MINE :<
Quadrilateral $ABCD$ has an incenter $I$ Suppose $AB > BC$. Let $M$ be the midpoint of $AC$. Suppose that $MI \perp BI$. $DI$ meets $(BDM)$ again at point $T$. Let points $P$ and $Q$ be such that $T$ is the midpoint of $MP$ and $I$ is the midpoint of $MQ$. Point $S$ lies on the plane such that $AMSQ$ is a parallelogram, and suppose the angle bisectors of $MCQ$ and $MSQ$ concur on $IM$.

The angle bisectors of $\angle PAQ$ and $\angle PCQ$ meet $PQ$ at $X$ and $Y$. Prove that $PX = QY$.
3 replies
popop614
4 hours ago
ItzsleepyXD
an hour ago
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Game About Passing Pencils
WilliamSChen   0
an hour ago
A group of $n$ children sit in a circle facing inward with $n > 2$, and each child starts with an arbitrary even number of pencils. Each minute, each child simultaneously passes exactly half of all of their pencils to the child to their right. Then, all children that have an odd number of pencils receive one more pencil.
Prove that after a finite amount of time, the children will all have the same number of pencils.

I do not know the source.
0 replies
WilliamSChen
an hour ago
0 replies
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An nxn Checkboard
MithsApprentice   26
N an hour ago by NicoN9
Source: USAMO 1999 Problem 1
Some checkers placed on an $n \times n$ checkerboard satisfy the following conditions:

(a) every square that does not contain a checker shares a side with one that does;

(b) given any pair of squares that contain checkers, there is a sequence of squares containing checkers, starting and ending with the given squares, such that every two consecutive squares of the sequence share a side.

Prove that at least $(n^{2}-2)/3$ checkers have been placed on the board.
26 replies
MithsApprentice
Oct 3, 2005
NicoN9
an hour ago
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Is this FE solvable?
Mathdreams   4
N 2 hours ago by Mathdreams
Find all $f:\mathbb{R} \rightarrow \mathbb{R}$ such that \[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]for all reals $x$ and $y$.
4 replies
Mathdreams
Tuesday at 6:58 PM
Mathdreams
2 hours ago
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Coaxial circles related to Gergon point
Headhunter   0
2 hours ago
Source: I tried but can't find the source...
Hi, everyone.

In $\triangle$$ABC$, $Ge$ is the Gergon point and the incircle $(I)$ touch $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively.
Let the circumcircles of $\triangle IDGe$, $\triangle IEGe$, $\triangle IFGe$ be $O_{1}$ , $O_{2}$ , $O_{3}$ respectively.

Reflect $O_{1}$ in $ID$ and then we get the circle $O'_{1}$
Reflect $O_{2}$ in $IE$ and then the circle $O'_{2}$
Reflect $O_{3}$ in $IF$ and then the circle $O'_{3}$

Prove that $O'_{1}$ , $O'_{2}$ , $O'_{3}$ are coaxial.
0 replies
Headhunter
2 hours ago
0 replies
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Equation with powers
a_507_bc   6
N 2 hours ago by EVKV
Source: Serbia JBMO TST 2024 P1
Find all non-negative integers $x, y$ and primes $p$ such that $$3^x+p^2=7 \cdot 2^y.$$
6 replies
a_507_bc
May 25, 2024
EVKV
2 hours ago
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no numbers of the form 80...01 are squares
Marius_Avion_De_Vanatoare   2
N 2 hours ago by EVKV
Source: Moldova JTST 2024 P5
Prove that a number of the form $80\dots01$ (there is at least 1 zero) can't be a perfect square.
2 replies
1 viewing
Marius_Avion_De_Vanatoare
Jun 10, 2024
EVKV
2 hours ago
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f((x XOR f(y)) + y) = (f(x) XOR y) + y
the_universe6626   3
N 3 hours ago by jasperE3
Source: Janson MO 5 P4
Find all functions $f:\mathbb{Z}_{\ge0}\rightarrow\mathbb{Z}_{\ge0}$ such that
\[f((x\oplus f(y))+y)=(f(x)\oplus y)+y\]Note: $\oplus$ denotes the bitwise XOR operation. For example, $1001_2 \oplus 101_2 = 1100_2$.

(Proposed by ja.)
3 replies
the_universe6626
Feb 21, 2025
jasperE3
3 hours ago
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2024 8's
Marius_Avion_De_Vanatoare   3
N 3 hours ago by EVKV
Source: Moldova JTST 2024 P2
Prove that the number $ \underbrace{88\dots8}_\text{2024\; \textrm{times}}$ is divisible by 2024.
3 replies
Marius_Avion_De_Vanatoare
Jun 10, 2024
EVKV
3 hours ago
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pretty well known
dotscom26   0
3 hours ago
Let $\triangle ABC$ be a scalene triangle such that $\Omega$ is its incircle. $AB$ is tangent to $\Omega$ at $D$. A point $E$ ($E \notin \Omega$) is located on $BC$.

Let $\omega_1$, $\omega_2$, and $\omega_3$ be the incircles of the triangles $BED$, $ADE$, and $AEC$, respectively.

Show that the common tangent to $\omega_1$ and $\omega_3$ is also tangent to $\omega_2$.

0 replies
dotscom26
3 hours ago
0 replies
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Thanks u!
Ruji2018252   6
N 3 hours ago by jasperE3
Find all $f:\mathbb{R}\to\mathbb{R}$ and
\[ f(x+y)+f(x^2+f(y))=f(f(x))^2+f(x)+f(y)+y,\forall x,y\in\mathbb{R}\]
6 replies
Ruji2018252
Mar 26, 2025
jasperE3
3 hours ago
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Belgium 2004.
Virgil Nicula   6
N Aug 17, 2016 by lebathanh
PP (NMO Belgium 2004). Let $\triangle ABC$ with $c>b\ ,$ $D\in (AC$ so that $BD=CD$ and

$E\in (BC)\ ,$ $F\in AB$ so that $EF\parallel BD$ and $G\in AE\cap BD\ .$ Prove that $\widehat{BCF}\equiv\widehat{BCG}\ .$
6 replies
Virgil Nicula
Aug 15, 2016
lebathanh
Aug 17, 2016
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Belgium 2004.
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Virgil Nicula
7054 posts
Virgil Nicula
#1 h  Aug 15, 2016, 6:30 PM • 1 Y
Y by Adventure10
PP (NMO Belgium 2004). Let $\triangle ABC$ with $c>b\ ,$ $D\in (AC$ so that $BD=CD$ and

$E\in (BC)\ ,$ $F\in AB$ so that $EF\parallel BD$ and $G\in AE\cap BD\ .$ Prove that $\widehat{BCF}\equiv\widehat{BCG}\ .$
This post has been edited 4 times. Last edited by Virgil Nicula, Aug 15, 2016, 11:02 PM
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vanstraelen
8945 posts
vanstraelen
#2 h  Aug 15, 2016, 8:53 PM • 2 Y
Y by Adventure10, Mango247
$D \in [AC]$ extended?
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lebathanh
464 posts
lebathanh
#3 h  Aug 16, 2016, 9:07 AM • 1 Y
Y by Adventure10
any solution please
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Arab
612 posts
Arab
#4 h  Aug 16, 2016, 5:20 PM • 2 Y
Y by Adventure10, Mango247
Let $H$ be a point on $AD$ such that $GH\parallel BC$.

Now $FE\parallel BD\implies \frac{EF}{GB}=\frac{AE}{AG}=\frac{EC}{GH}\implies \triangle CEF\sim \triangle HGB$.

$BCHG$ is an isosceles trapezoid since $BD=CD$. Hence we have that $\angle BCF=\angle BHG=\angle BCG$.
This post has been edited 1 time. Last edited by Arab, Aug 16, 2016, 5:20 PM
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Stens
49 posts
Stens
#5 h  Aug 16, 2016, 6:07 PM • 1 Y
Y by Adventure10
Arab's solution is very nice. Below is a sketch of a not so clever solution:

Let $CF$ meet $GD$ at $F'$. By some angle chasing we observe that we are done if $DC$ is tangent to the circumcircle of $\triangle GCF'$, which is equivalent to $DB^2=DF'\cdot DG$. Finding the lengths $DF',DB$ and $DG$ in terms of $AB,BC,CA$ is not too hard, and then we just verify that we do in fact have equality above.
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Virgil Nicula
7054 posts
Virgil Nicula
#6 h  Aug 17, 2016, 2:08 AM • 2 Y
Y by Adventure10, Mango247
Thank you for your proofs. See here my proof which is more intricate (with harmonic division).
This post has been edited 2 times. Last edited by Virgil Nicula, Aug 17, 2016, 2:09 AM
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lebathanh
464 posts
lebathanh
#7 h  Aug 17, 2016, 10:36 AM • 2 Y
Y by Adventure10, Mango247
Arab wrote:
Let $H$ be a point on $AD$ such that $GH\parallel BC$.

Now $FE\parallel BD\implies \frac{EF}{GB}=\frac{AE}{AG}=\frac{EC}{GH}\implies \triangle CEF\sim \triangle HGB$.

$BCHG$ is an isosceles trapezoid since $BD=CD$. Hence we have that $\angle BCF=\angle BHG=\angle BCG$.

the proof is very nice ,thanks
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