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Probably a good lemma
Zavyk09   0
12 minutes ago
Source: found when solving exercises
Let $ABC$ be a triangle with circumcircle $\omega$. Arbitrary points $E, F$ on $AC, AB$ respectively. Circumcircle $\Omega$ of triangle $AEF$ intersects $\omega$ at $P \ne A$. $BE$ intersects $CF$ at $I$. $PI$ cuts $\Omega$ and $\omega$ at $K, L$ respectively. Construct parallelogram $QFRE$. Prove that $A, R, P$ are collinear.
0 replies
Zavyk09
12 minutes ago
0 replies
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Gergonne point Harmonic quadrilateral
niwobin   2
N 13 minutes ago by Lil_flip38
Triangle ABC has incircle touching the sides at D, E, F as shown.
AD, BE, CF concurrent at Gergonne point G.
BG and CG cuts the incircle at X and Y, respectively.
AG cuts the incircle at K.
Prove: K, X, D, Y form a harmonic quadrilateral. (KX/KY = DX/DY)
2 replies
niwobin
Yesterday at 8:17 PM
Lil_flip38
13 minutes ago
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Inequality on non-nagative numbers
TUAN2k8   0
20 minutes ago
Source: My book
Let $a,b,c$ be non-nagative real numbers such that $a+b+c=3$.
Prove that $ab+bc+ca-abc \leq \frac{9}{4}$.
0 replies
1 viewing
TUAN2k8
20 minutes ago
0 replies
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Incircle in an isoscoles triangle
Sadigly   2
N 35 minutes ago by Sadigly
Source: own
Let $ABC$ be an isosceles triangle with $AB=AC$, and let $I$ be its incenter. Incircle touches sides $BC,CA,AB$ at $D,E,F$, respectively. Foot of altitudes from $E,F$ to $BC$ are $X,Y$ , respectively. Rays $XI,YI$ intersect $(ABC)$ at $P,Q$, respectively. Prove that $(PQD)$ touches incircle at $D$.
2 replies
Sadigly
Friday at 9:21 PM
Sadigly
35 minutes ago
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Prove that the triangle is isosceles.
TUAN2k8   7
N an hour ago by TUAN2k8
Source: My book
Given acute triangle $ABC$ with two altitudes $CF$ and $BE$.Let $D$ be the point on the line $CF$ such that $DB \perp BC$.The lines $AD$ and $EF$ intersect at point $X$, and $Y$ is the point on segment $BX$ such that $CY \perp BY$.Suppose that $CF$ bisects $BE$.Prove that triangle $ACY$ is isosceles.
7 replies
1 viewing
TUAN2k8
May 16, 2025
TUAN2k8
an hour ago
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Locus of Mobile points on Circle and Square
Kunihiko_Chikaya   1
N an hour ago by Mathzeus1024
Source: 2012 Hitotsubashi University entrance exam, problem 4
In the $xyz$-plane given points $P,\ Q$ on the planes $z=2,\ z=1$ respectively. Let $R$ be the intersection point of the line $PQ$ and the $xy$-plane.

(1) Let $P(0,\ 0,\ 2)$. When the point $Q$ moves on the perimeter of the circle with center $(0,\ 0,\ 1)$ , radius 1 on the plane $z=1$,
find the equation of the locus of the point $R$.

(2) Take 4 points $A(1,\ 1,\ 1) , B(1,-1,\ 1), C(-1,-1,\ 1)$ and $D(-1,\ 1,\ 1)$ on the plane $z=2$. When the point $P$ moves on the perimeter of the circle with center $(0,\ 0,\ 2)$ , radius 1 on the plane $z=2$ and the point $Q$ moves on the perimeter of the square $ABCD$, draw the domain swept by the point $R$ on the $xy$-plane, then find the area.
1 reply
Kunihiko_Chikaya
Feb 28, 2012
Mathzeus1024
an hour ago
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Circle is tangent to circumcircle and incircle
ABCDE   73
N 2 hours ago by AR17296174
Source: 2016 ELMO Problem 6
Elmo is now learning olympiad geometry. In triangle $ABC$ with $AB\neq AC$, let its incircle be tangent to sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. The internal angle bisector of $\angle BAC$ intersects lines $DE$ and $DF$ at $X$ and $Y$, respectively. Let $S$ and $T$ be distinct points on side $BC$ such that $\angle XSY=\angle XTY=90^\circ$. Finally, let $\gamma$ be the circumcircle of $\triangle AST$.

(a) Help Elmo show that $\gamma$ is tangent to the circumcircle of $\triangle ABC$.

(b) Help Elmo show that $\gamma$ is tangent to the incircle of $\triangle ABC$.

James Lin
73 replies
ABCDE
Jun 24, 2016
AR17296174
2 hours ago
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Mathematical Olympiad Finals 2013
parkjungmin   0
2 hours ago
Mathematical Olympiad Finals 2013
0 replies
parkjungmin
2 hours ago
0 replies
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n^k + mn^l + 1 divides n^(k+1) - 1
cjquines0   37
N 2 hours ago by alexanderhamilton124
Source: 2016 IMO Shortlist N4
Let $n, m, k$ and $l$ be positive integers with $n \neq 1$ such that $n^k + mn^l + 1$ divides $n^{k+l} - 1$. Prove that
[list]
[*]$m = 1$ and $l = 2k$; or
[*]$l|k$ and $m = \frac{n^{k-l}-1}{n^l-1}$.
[/list]
37 replies
cjquines0
Jul 19, 2017
alexanderhamilton124
2 hours ago
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A very beautiful geo problem
TheMathBob   4
N 2 hours ago by ravengsd
Source: Polish MO Finals P2 2023
Given an acute triangle $ABC$ with their incenter $I$. Point $X$ lies on $BC$ on the same side as $B$ wrt $AI$. Point $Y$ lies on the shorter arc $AB$ of the circumcircle $ABC$. It is given that $$\angle AIX = \angle XYA = 120^\circ.$$Prove that $YI$ is the angle bisector of $XYA$.
4 replies
TheMathBob
Mar 29, 2023
ravengsd
2 hours ago
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Difficult combinatorics problem
shactal   0
2 hours ago
Can someone help me with this problem? Let $n\in \mathbb N^*$. We call a distribution the act of distributing the integers from $1$
to $n^2$ represented by tokens to players $A_1$ to $A_n$ so that they all have the same number of tokens in their urns.
We say that $A_i$ beats $A_j$ when, when $A_i$ and $A_j$ each draw a token from their urn, $A_i$ has a strictly greater chance of drawing a larger number than $A_j$. We then denote $A_i>A_j$. A distribution is said to be chicken-fox-viper when $A_1>A_2>\ldots>A_n>A_1$ What is $R(n)$
, the number of chicken-fox-viper distributions?
0 replies
shactal
2 hours ago
0 replies
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Cubic and Quadratic
mathisreal   3
N 2 hours ago by macves
Source: CIIM 2020 P2
Find all triples of positive integers $(a,b,c)$ such that the following equations are both true:
I- $a^2+b^2=c^2$
II- $a^3+b^3+1=(c-1)^3$
3 replies
mathisreal
Oct 26, 2020
macves
2 hours ago
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Functional Geometry
GreekIdiot   2
N Apr 30, 2025 by Double07
Source: BMO 2024 SL G7
Let $f: \pi \to \mathbb R$ be a function from the Euclidean plane to the real numbers such that $f(A)+f(B)+f(C)=f(O)+f(G)+f(H)$ for any acute triangle $\Delta ABC$ with circumcenter $O$, centroid $G$ and orthocenter $H$. Prove that $f$ is constant.
2 replies
GreekIdiot
Apr 27, 2025
Double07
Apr 30, 2025
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Source: BMO 2024 SL G7
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GreekIdiot
245 posts
GreekIdiot
#1 h  Apr 27, 2025, 1:08 PM
Y by
Let $f: \pi \to \mathbb R$ be a function from the Euclidean plane to the real numbers such that $f(A)+f(B)+f(C)=f(O)+f(G)+f(H)$ for any acute triangle $\Delta ABC$ with circumcenter $O$, centroid $G$ and orthocenter $H$. Prove that $f$ is constant.
This post has been edited 1 time. Last edited by GreekIdiot, Apr 27, 2025, 1:08 PM
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ItzsleepyXD
147 posts
ItzsleepyXD
#2 h  Apr 30, 2025, 3:50 AM
Y by
Consider $\triangle ABC$ with circumcenter $O$, centroid $G$ and orthocenter $H$.
$D,E,F,M_{OG},M_{OH},M_{HG}$ is midpoint of $BC,CA,AB,OG,OH,HG$
Define $O_A,G_A,H_A$ be circumcenter, centroid and orthocenter of $\triangle AEF$.
and $H_B,O_B,G_B,H_C,O_C,G_C$ analogously
by $\triangle ABC, \triangle AFE , \triangle BFD , \triangle CDE , \triangle DEF$ and some calculation
known that $f(O)+f(G)=f(M_{OG})+f(M_{HG})$
Let $M'$ be reflection of $M_{HG}$ wrt point $M_{OG}$ notice that $f(M_{HG}=f(M')$
so f is constant. done $\square$
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Double07
91 posts
Double07
#3 h  Apr 30, 2025, 8:29 PM
Y by
Use complex numbers:

For a point $A(a)$ we use $f(a)=f(A)$.

WLOG suppose $f(0)=0$.

Let $\omega=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$.

We have $f(a)+f(\omega a)+f(\omega^2 a)=3\cdot f(0)=0, \forall a\in \mathbb{C}$.

We also have $f(x)+f(\omega a)+f(\omega^2 a)=0+f\left(\dfrac{x-a}{3}\right)+f(x-a), \forall a,x\in \mathbb{C}$ with $|a|=|x|>|a-x|$ (so that the triangle is acute).

So $f(x)-f(a)=f\left(\dfrac{x-a}{3}\right)+f(x-a)$.

Because $f(x)-f(y)$ depends on just $x-y$ (for $x,y$ complex with $|x|=|y|$), we can consider function $g$ such that $g(x-y)=f(x)-f(y)$.

So $g(x-a)=f\left(\dfrac{x-a}{3}\right)+f(x-a)$.

Notice that $g$ is additive: $g(x-y)+g(y-z)=f(x)-f(y)+f(y)-f(z)=f(x)-f(z)=g(x-z)$.

So $g(nx)=ng(x)$, $\forall n\in \mathbb{Z}$ and $x\in\mathbb{C}$.

So $g(x)+g(-x)=0\implies g(a-x)+g((-a)-(-x))=0\implies f(a)-f(x)+f(-a)-f(-x)=0\implies$
$\implies f(x)+f(-x)=f(a)+f(-a)$, so $f(x)+f(-x)$ is a constant $c$.

But $f(1)+f(\omega)+f(\omega^2)=0=-(f(-1)+f(-\omega)+f(-\omega^2))\implies 3c=0\implies c=0\implies f$ is odd.

$f(x)-f(y)=g(x-y)\implies f(x)+f(-y)=g(x-y)\implies f(x)+f(y)=g(x+y)$, where $|x|=|y|$.

Choose $x=\omega a$ and $y=\omega^2 a$ in the above to get $-f(a)=f(\omega a)+f(\omega^2 a)=g(-a)=-g(a)\implies f\equiv g$.

But we know that $g(x)=f(x)+f\left(\frac{x}{3}\right)\implies f\left(\frac{x}{3}\right)=0, \forall x\in \mathbb{C}$, so we're done.
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