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A geometry problem involving 2 circles
Ujiandsd   1
N 9 minutes ago by Ujiandsd
Source: L
Point M is the midpoint of side BC of triangle ABC. The length of the radius of the outer circle of triangle ABM, triangle ACM
is 5 and 7 respectively find the distance between the center of their outer circles
1 reply
Ujiandsd
May 11, 2025
Ujiandsd
9 minutes ago
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Thailand MO 2025 P3
Kaimiaku   2
N 24 minutes ago by lbh_qys
Let $a,b,c,x,y,z$ be positive real numbers such that $ay+bz+cx \le az+bx+cy$. Prove that $$ \frac{xy}{ax+bx+cy}+\frac{yz}{by+cy+az}+\frac{zx}{cz+az+bx} \le \frac{x+y+z}{a+b+c}$$
2 replies
Kaimiaku
an hour ago
lbh_qys
24 minutes ago
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Burapha integer
EeEeRUT   1
N 40 minutes ago by ItzsleepyXD
Source: TMO 2025 P1
For each positive integer $m$, denote by $d(m)$ the number of positive divisors of $m$. We say that a positive integer $n$ is Burapha integer if it satisfy the following condition
[list]
[*] $d(n)$ is an odd integer.
[*] $d(k) \leqslant d(\ell)$ holds for every positive divisor $k, \ell$ of $n$, such that $k < \ell$
[/list]
Find all Burapha integer.
1 reply
EeEeRUT
an hour ago
ItzsleepyXD
40 minutes ago
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Algebra inequalities
TUAN2k8   1
N 42 minutes ago by lbh_qys
Source: Own
Is that true?
Let $a_1,a_2,...,a_n$ be real numbers such that $0 \leq a_i \leq 1$ for all $1 \leq i \leq n$.
Prove that: $\sum_{1 \leq i<j \leq n} (a_i-a_j)^2 \leq \frac{n}{2}$.
1 reply
TUAN2k8
an hour ago
lbh_qys
42 minutes ago
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Quadrilateral with Congruent Diagonals
v_Enhance   37
N an hour ago by Ilikeminecraft
Source: USA TSTST 2012, Problem 2
Let $ABCD$ be a quadrilateral with $AC = BD$. Diagonals $AC$ and $BD$ meet at $P$. Let $\omega_1$ and $O_1$ denote the circumcircle and the circumcenter of triangle $ABP$. Let $\omega_2$ and $O_2$ denote the circumcircle and circumcenter of triangle $CDP$. Segment $BC$ meets $\omega_1$ and $\omega_2$ again at $S$ and $T$ (other than $B$ and $C$), respectively. Let $M$ and $N$ be the midpoints of minor arcs $\widehat {SP}$ (not including $B$) and $\widehat {TP}$ (not including $C$). Prove that $MN \parallel O_1O_2$.
37 replies
v_Enhance
Jul 19, 2012
Ilikeminecraft
an hour ago
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Oh my god
EeEeRUT   0
an hour ago
Source: TMO 2025 P5
In a class, there are $n \geqslant 3$ students and a teacher with $M$ marbles. The teacher then play a Marble distribution according to the following rules. At the start, each student receives at least $1$ marbles from the teacher. Then, the teacher chooses a student , who has never been chosen before, such that the number of marbles that he owns in a multiple of $2(n-1)$. That chosen student then equally distribute half of his marbles to $n-1$ other students. The same goes on until the teacher is not able to choose anymore student.

Find all integer $M$, such that for some initial numbers of marbles that the students receive, the teacher can choose all the student(according to the rule above), so that each student receiving equal amount of marbles at the end.
0 replies
EeEeRUT
an hour ago
0 replies
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geometry
EeEeRUT   1
N an hour ago by ItzsleepyXD
Source: TMO 2025
Let $D,E$ and $F$ be touch points of the incenter of $\triangle ABC$ at $BC, CA$ and $AB$, respectively. Let $P,Q$ and $R$ be the circumcenter of triangles $AFE, BDF$ and $CED$, respectively. Show that $DP, EQ$ and $FR$ concurrent.
1 reply
EeEeRUT
an hour ago
ItzsleepyXD
an hour ago
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Spanish Mathematical Olympiad 2002, Problem 1
OmicronGamma   3
N an hour ago by NicoN9
Source: Spanish Mathematical Olympiad 2002
Find all the polynomials $P(t)$ of one variable that fullfill the following for all real numbers $x$ and $y$:
$P(x^2-y^2) = P(x+y)P(x-y)$.
3 replies
OmicronGamma
Jun 2, 2017
NicoN9
an hour ago
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Additive set with special property
the_universe6626   1
N 2 hours ago by jasperE3
Source: Janson MO 1 P2
Let $S$ be a nonempty set of positive integers such that:
$\bullet$ if $m,n\in S$ then $m+n\in S$.
$\bullet$ for any prime $p$, there exists $x\in S$ such that $p\nmid x$.
Prove that the set of all positive integers not in $S$ is finite.

(Proposed by cknori)
1 reply
the_universe6626
Feb 21, 2025
jasperE3
2 hours ago
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ISI UGB 2025 P4
SomeonecoolLovesMaths   8
N 2 hours ago by chakrabortyahan
Source: ISI UGB 2025 P4
Let $S^1 = \{ z \in \mathbb{C} \mid |z| =1 \}$ be the unit circle in the complex plane. Let $f \colon S^1 \longrightarrow S^2$ be the map given by $f(z) = z^2$. We define $f^{(1)} \colon = f$ and $f^{(k+1)} \colon = f \circ f^{(k)}$ for $k \geq 1$. The smallest positive integer $n$ such that $f^{(n)}(z) = z$ is called the period of $z$. Determine the total number of points in $S^1$ of period $2025$.
(Hint : $2025 = 3^4 \times 5^2$)
8 replies
SomeonecoolLovesMaths
Sunday at 11:24 AM
chakrabortyahan
2 hours ago
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So Many Terms
oVlad   7
N 3 hours ago by NuMBeRaToRiC
Source: KöMaL A. 765
Find all functions $f:\mathbb{R}\to\mathbb{R}$ which satisfy the following equality for all $x,y\in\mathbb{R}$ \[f(x)f(y)-f(x-1)-f(y+1)=f(xy)+2x-2y-4.\]Proposed by Dániel Dobák, Budapest
7 replies
oVlad
Mar 20, 2022
NuMBeRaToRiC
3 hours ago
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Incircle and circumcircle
stergiu   6
N Apr 10, 2025 by Sadigly
Source: tst- Greece 2019
Let a triangle $ABC$ inscribed in a circle $\Gamma$ with center $O$. Let $I$ the incenter of triangle $ABC$ and $D, E, F$ the contact points of the incircle with sides $BC, AC, AB$ of triangle $ABC$ respectively . Let also $S$ the foot of the perpendicular line from $D$ to the line $EF$.Prove that line $SI$ passes from the antidiametric point $N$ of $A$ in the circle $\Gamma$.( $AN$ is a diametre of the circle $\Gamma$).
6 replies
stergiu
Sep 23, 2019
Sadigly
Apr 10, 2025
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Incircle and circumcircle
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Reveal topic tags
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Source: tst- Greece 2019
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stergiu
1648 posts
stergiu
#1 h  Sep 23, 2019, 12:06 PM • 2 Y
Y by mathematicsy, Adventure10
Let a triangle $ABC$ inscribed in a circle $\Gamma$ with center $O$. Let $I$ the incenter of triangle $ABC$ and $D, E, F$ the contact points of the incircle with sides $BC, AC, AB$ of triangle $ABC$ respectively . Let also $S$ the foot of the perpendicular line from $D$ to the line $EF$.Prove that line $SI$ passes from the antidiametric point $N$ of $A$ in the circle $\Gamma$.( $AN$ is a diametre of the circle $\Gamma$).
This post has been edited 1 time. Last edited by stergiu, Sep 23, 2019, 12:08 PM
Reason: explanation
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TheDarkPrince
3042 posts
TheDarkPrince
#2 h  Sep 23, 2019, 12:09 PM • 3 Y
Y by Pluto1708, Mathasocean, Adventure10
Let $R$ be the intersection of $\odot(AI)$ and $\odot(ABC)$. Inverting about the incircle gives that $S$ maps to $R$, so $\angle ARI = 90^{\circ}$ and $R,I,S$ are collinear which completes the problem.
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stergiu
1648 posts
stergiu
#3 h  Sep 23, 2019, 3:46 PM • 1 Y
Y by Adventure10
TheDarkPrince wrote:
Let $R$ be the intersection of $\odot(AI)$ and $\odot(ABC)$. Inverting about the incircle gives that $S$ maps to $R$, so $\angle ARI = 90^{\circ}$ and $R,I,S$ are collinear which completes the problem.

Just a simple question to this nice solution:

Why $S$ goes to $R$ and not to another point $Q$ ,of the circle $(AI)$ ? Thank you !

( Ok ! The circle $(A,B,C)$ has invers the Euler circle of triangle $DEF$ and $S$ belongs to this circle.Allmost obvious, but ....)
This post has been edited 3 times. Last edited by stergiu, Sep 23, 2019, 5:41 PM
Reason: correction
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jayme
9792 posts
jayme
#5 h  Sep 24, 2019, 8:15 AM • 2 Y
Y by Adventure10, Mango247
Dear Matlinkers,

http://www.artofproblemsolving.com/community/c6h614584

Sincerely
Jean-Louis
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Euler365
143 posts
Euler365
#6 h  Sep 24, 2019, 5:04 PM • 2 Y
Y by Adventure10, Mango247
Lets complex bash
Set the incircle as the unit circle with incentre as the origin and let $d = 1$.

Then $a = \frac{2ef}{e + f} , b = \frac{2f}{1 + f} , c = \frac{2e}{1 + e}$
$\therefore \frac{a - b}{\overline{a - b}} = -f^2$.

$ nb \perp ba \implies -\frac{a - b}{\overline{a - b}} = \frac{c - n}{\overline{c - n}} = f^2$

$\therefore \overline{n} = \frac{(1+f)n-2f+2f^2}{f^2(1+f)}$

Similarly $\overline{n} = \frac{(1+e)n-2e+2e^2}{e^2(1+e)}$

$\therefore \frac{(1+e)n-2e+2e^2}{e^2(1+e)} = \frac{(1+f)n-2f+2f^2}{f^2(1+f)}$
After simplifying we obtain that
$n = (1 + e + f - ef)k$

where $k = \frac{2ef}{(e + 1)(f + 1)(e + f)}$
Now we also obtain that $\overline{k} = \frac{2ef}{(e + 1)(f + 1)(e + f)}$
So $\overline{k} = k \implies k \in\mathbb{R}$
Now also note that $s = \frac{1}{2}(1 + e + f - ef)$
So $\frac{n}{s} = 2k \in \mathbb{R}$
So $N$, $I$ and $S$ are collinear.
This post has been edited 4 times. Last edited by Euler365, Sep 25, 2019, 10:53 AM
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gambi
82 posts
gambi
#8 h  Oct 17, 2021, 2:05 PM
Y by
Let $L$ be the midpoint of arc $BAC$ in $\Gamma$, and let $M$ be the antipode of $L$ in $\Gamma$.
Let $K$ be the second intersection point of $MD$ with $\Gamma$.

Claim 1. $K$ is the Miquel point of $BFEC$.
Let $\Psi$ be the inversion around $(BIC)$. By the Incenter Lemma, we get that such inversion is centered in $M$. Clearly,
$$ \left\{\begin{array}{lll} \Psi ((ABC))=BC \\ \Psi (KM)=KM \\ \end{array} \right. \Longrightarrow \Psi (K)=D $$Therefore,
$$ MD\cdot MK=MI^2 \Longrightarrow (KID) \thickspace \text{tangent to} \thickspace IM $$Hence
$$ \measuredangle DKI=\measuredangle DIM=\measuredangle LMA=\measuredangle LKA $$But then
$$ 90^o=\measuredangle MKL=\measuredangle DKL=\measuredangle DKI+\measuredangle IKL=\measuredangle LKA+\measuredangle IKL=\measuredangle IKA $$This way $K\in (AEIF)$ and so Claim 1 is proved.

Claim 2. $K,S,I$ are collinear.
Right triangles $\triangle FSD$ and $\triangle IDC$ are similar because
$$ \measuredangle DFS=\measuredangle DFI+\measuredangle IFS=\measuredangle DBI+\measuredangle IAE=90^o-\measuredangle ICD $$Analogously, right triangles $\triangle ESD$ and $\triangle IDB$ are similar.
From these two similarities, we get
$$ \left\{\begin{array}{lll} \frac{FS}{SD}=\frac{ID}{DC} \\ \\ \frac{SE}{SD}=\frac{ID}{BD} \\ \end{array} \right. \Longrightarrow \frac{FS}{SE}=\frac{BD}{DC} \qquad (\star) $$From Claim 1, we get that $K$ is the center of the spiral similarity sending $BC$ to $FE$. But from $(\star)$, we get that such spiral similarity also sends $D$ to $S$.
Hence $\triangle KFS$ and $\triangle KBD$ are similar. Consequently,
$$ \measuredangle FKS=\measuredangle BKD=\measuredangle BKM=\measuredangle BAM=\measuredangle FAI=\measuredangle FKI $$Therefore, $K,S,I$ are collinear and so Claim 2 is proved.

Finally, from Claim 1 we have $K\in (AEFI) \Longrightarrow IK\perp KA$.
But, in light of Claim 2, this implies that lines $ISK$ and $AK$, are perpendicular, so ray $SI$ will intersect $\Gamma$ at $N$, the antipode of $A$.
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Sadigly
222 posts
Sadigly
#9 h  Apr 10, 2025, 1:46 AM
Y by
Easiest bash exercise

$(DEF)\in\mathbb{S}^1~D=d~E=e~F=f$
$A=\frac{2ef}{e+f}$
$B=\frac{2fd}{f+d}$
$C=\frac{2de}{d+e}$

$O=\frac{2def(d+e+f)}{(d+e)(e+f)(f+d)}$

$N=2O-A=\frac{2ef(d^2+de+df-ef)}{(d+e)(e+f)(f+d)}$

$S=\frac12(d+e+f-\frac{ef}{d})$

$\frac{s-i}{n-i}=\frac{(d+e+f-\frac{ef}{d})(d+e)(e+f)(f+d)}{4ef(d^2+de+df-ef)}=\frac{(d+e)(e+f)(f+d)}{4def}\in\mathbb{R}$
This post has been edited 2 times. Last edited by Sadigly, Apr 10, 2025, 1:56 AM
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