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GEOMETRY GEOMETRY GEOMETRY
Kagebaka   71
N 34 minutes ago by bin_sherlo
Source: IMO 2021/3
Let $D$ be an interior point of the acute triangle $ABC$ with $AB > AC$ so that $\angle DAB = \angle CAD.$ The point $E$ on the segment $AC$ satisfies $\angle ADE =\angle BCD,$ the point $F$ on the segment $AB$ satisfies $\angle FDA =\angle DBC,$ and the point $X$ on the line $AC$ satisfies $CX = BX.$ Let $O_1$ and $O_2$ be the circumcenters of the triangles $ADC$ and $EXD,$ respectively. Prove that the lines $BC, EF,$ and $O_1O_2$ are concurrent.
71 replies
Kagebaka
Jul 20, 2021
bin_sherlo
34 minutes ago
Equation of integers
jgnr   3
N an hour ago by KTYC
Source: Indonesia Mathematics Olympiad 2005 Day 1 Problem 2
For an arbitrary positive integer $ n$, define $ p(n)$ as the product of the digits of $ n$ (in decimal). Find all positive integers $ n$ such that $ 11p(n)=n^2-2005$.
3 replies
jgnr
Jun 2, 2008
KTYC
an hour ago
Divisibility..
Sadigly   4
N an hour ago by Solar Plexsus
Source: Azerbaijan Junior MO 2025 P2
Find all $4$ consecutive even numbers, such that the square of their product is divisible by the sum of their squares.
4 replies
Sadigly
Yesterday at 7:37 AM
Solar Plexsus
an hour ago
Surjective number theoretic functional equation
snap7822   3
N an hour ago by internationalnick123456
Source: 2025 Taiwan TST Round 3 Independent Study 2-N
Let $f:\mathbb{N} \rightarrow \mathbb{N}$ be a function satisfying the following conditions:
[list=i]
[*] For all $m, n \in \mathbb{N}$, if $m > n$ and $f(m) > f(n)$, then $f(m-n) = f(n)$;
[*] $f$ is surjective.
[/list]
Find the maximum possible value of $f(2025)$.

Proposed by snap7822
3 replies
snap7822
May 1, 2025
internationalnick123456
an hour ago
FE with devisibility
fadhool   0
an hour ago
if when i solve an fe that is defined in the set of positive integer i found m|f(m) can i set f(m) =km such that k is not constant and of course it depends on m but after some work i find k=c st c is constant is this correct
0 replies
fadhool
an hour ago
0 replies
Many Equal Sides
mathisreal   3
N an hour ago by QueenArwen
Source: Brazil EGMO TST 2023 #1
Let $ABC$ be a triangle with $BA=BC$ and $\angle ABC=90^{\circ}$. Let $D$ and $E$ be the midpoints of $CA$ and $BA$ respectively. The point $F$ is inside of $\triangle ABC$ such that $\triangle DEF$ is equilateral. Let $X=BF\cap AC$ and $Y=AF\cap DB$. Prove that $DX=YD$.
3 replies
mathisreal
Nov 10, 2022
QueenArwen
an hour ago
LOTS of recurrence!
SatisfiedMagma   4
N an hour ago by Reacheddreams
Source: Indian Statistical Institute Entrance UGB 2023/5
There is a rectangular plot of size $1 \times n$. This has to be covered by three types of tiles - red, blue and black. The red tiles are of size $1 \times 1$, the blue tiles are of size $1 \times 1$ and the black tiles are of size $1 \times 2$. Let $t_n$ denote the number of ways this can be done. For example, clearly $t_1 = 2$ because we can have either a red or a blue tile. Also $t_2 = 5$ since we could have tiled the plot as: two red tiles, two blue tiles, a red tile on the left and a blue tile on the right, a blue tile on the left and a red tile on the right, or a single black tile.

[list=a]
[*]Prove that $t_{2n+1} = t_n(t_{n-1} + t_{n+1})$ for all $n > 1$.

[*]Prove that $t_n = \sum_{d \ge 0} \binom{n-d}{d}2^{n-2d}$ for all $n >0$.
[/list]
Here,
\[ \binom{m}{r} = \begin{cases}
\dfrac{m!}{r!(m-r)!}, &\text{ if $0 \le r \le m$,} \\
0, &\text{ otherwise}
\end{cases}\]for integers $m,r$.
4 replies
SatisfiedMagma
May 14, 2023
Reacheddreams
an hour ago
combi/nt
blug   1
N an hour ago by blug
Prove that every positive integer $n$ can be written in the form
$$n=a_1+a_2+...+a_k,$$where $a_m=2^i3^j$ for some non-negative $i, j$ such that
$$a_x\nmid a_y$$for every $x, y\leq k$.
1 reply
blug
Yesterday at 3:37 PM
blug
an hour ago
Inequality, inequality, inequality...
Assassino9931   9
N an hour ago by ZeroHero
Source: Al-Khwarizmi Junior International Olympiad 2025 P6
Let $a,b,c$ be real numbers such that \[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]Find the smallest possible value of $a^2 + b^2 + c^2$.

Binh Luan and Nhan Xet, Vietnam
9 replies
Assassino9931
Today at 9:38 AM
ZeroHero
an hour ago
Vectors in a tilted square
mathwizard888   20
N an hour ago by HamstPan38825
Source: 2016 IMO Shortlist A3
Find all positive integers $n$ such that the following statement holds: Suppose real numbers $a_1$, $a_2$, $\dots$, $a_n$, $b_1$, $b_2$, $\dots$, $b_n$ satisfy $|a_k|+|b_k|=1$ for all $k=1,\dots,n$. Then there exists $\varepsilon_1$, $\varepsilon_2$, $\dots$, $\varepsilon_n$, each of which is either $-1$ or $1$, such that
\[ \left| \sum_{i=1}^n \varepsilon_i a_i \right| + \left| \sum_{i=1}^n \varepsilon_i b_i \right| \le 1. \]
20 replies
mathwizard888
Jul 19, 2017
HamstPan38825
an hour ago
Combi Geo
Adywastaken   0
2 hours ago
Source: NMTC 2024/8
A regular polygon with $100$ vertices is given. To each vertex, a natural number from the set $\{1,2,\dots,49\}$ is assigned. Prove that there are $4$ vertices $A, B, C, D$ such that if the numbers $a, b, c, d$ are assigned to them respectively, then $a+b=c+d$ and $ABCD$ is a parallelogram.
0 replies
Adywastaken
2 hours ago
0 replies
Incircle of a triangle is tangent to (ABC)
amar_04   11
N Apr 22, 2025 by Nari_Tom
Source: XVII Sharygin Correspondence Round P18
Let $ABC$ be a scalene triangle, $AM$ be the median through $A$, and $\omega$ be the incircle. Let $\omega$ touch $BC$ at point $T$ and segment $AT$ meet $\omega$ for the second time at point $S$. Let $\delta$ be the triangle formed by lines $AM$ and $BC$ and the tangent to $\omega$ at $S$. Prove that the incircle of triangle $\delta$ is tangent to the circumcircle of triangle $ABC$.
11 replies
amar_04
Mar 2, 2021
Nari_Tom
Apr 22, 2025
Incircle of a triangle is tangent to (ABC)
G H J
G H BBookmark kLocked kLocked NReply
Source: XVII Sharygin Correspondence Round P18
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amar_04
1915 posts
#1 • 4 Y
Y by A-Thought-Of-God, Bumblebee60, LoloChen, Rounak_iitr
Let $ABC$ be a scalene triangle, $AM$ be the median through $A$, and $\omega$ be the incircle. Let $\omega$ touch $BC$ at point $T$ and segment $AT$ meet $\omega$ for the second time at point $S$. Let $\delta$ be the triangle formed by lines $AM$ and $BC$ and the tangent to $\omega$ at $S$. Prove that the incircle of triangle $\delta$ is tangent to the circumcircle of triangle $ABC$.
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amar_04
1915 posts
#3 • 8 Y
Y by Aritra12, Pluto04, Dr_Vex, A-Thought-Of-God, Mathematicsislovely, Bumblebee60, PRMOisTheHardestExam, SerdarBozdag
[asy]
defaultpen(fontsize(9pt));
size(9cm);

pair A,B,C,I,T,S,M,G,X,K,N,I1,D,E,F,I2,N2,I3,D1,E1,F1,P,O,G1,G2,G3,T1,D2;

A=dir(130);
B=dir(220);
C=dir(320);
I=incenter(A,B,C);
O=circumcenter(A,B,C);
T=foot(I,B,C);
S=-T+2*foot(I,A,T);
P=-A+2*foot(O,A,I);
G=(S+T)/2;
M=(B+C)/2;
X=extension(I,G,B,C);
K=extension(X,S,A,M);
N=circumcenter(B,I,C);
I1=-I+2N;
D=foot(I1,B,C);
E=foot(I1,A,C);
F=foot(I1,A,B);
I2=incenter(X,K,M);
N2=circumcenter(K,I2,M);
I3=-I2+2N2;
D1=foot(I3,A,M);
E1=foot(I3,X,K);
F1=foot(I3,B,C);
G1=-P+2*foot(O,P,F1);
G2=foot(I2,B,C);
G3=-P+2*foot(O,P,G2);
T1=-T+2I;
D2=-D+2I1;

draw(A--B--C--cycle);
draw(circumcircle(A,B,C));
draw(incircle(A,B,C));
draw(A--T);
draw(A--M);
draw(B--X--E1);
draw(incircle(X,K,M));
draw(B--F);
draw(C--E);
draw(circumcircle(D,E,F));
draw(I1--M);
draw(A--I1,linewidth(0.3));
draw(circumcircle(D1,E1,F1));
draw(G3--P--G1,linewidth(0.4));
draw(X--I3);
draw(T--T1,linewidth(0.4));
draw(A--D--D2--T,linewidth(0.3));

dot("$A$" , A , dir(A));
dot("$B$" , B , dir(B));
dot("$C$" , C , dir(30));
dot("$S$" , S , dir(S));
dot("$T$" , T , dir(T));
dot("$M$" , M , dir(M));
dot("$X$" , X , dir(X));
dot("$I$" , I , dir(270));
dot("$I_a$" , I1 , dir(I1));
dot("$P$" , P , dir(P));
dot("$O_1$" , I2 , dir(250));
dot("$O_2$" , I3 , dir(I3));
dot("$D$" , D , dir(D));
dot("$$" , T1 , dir(T1));
dot("$D'$" , D2 , dir(D2));

[/asy]

$\textbf{LEMMA:-}$ $ABC$ be a triangle with $I$ as the incenter and let $D$ be a point on $\overline{BC}$. Let $\omega_1$ be the circle tangent to $AD,BD$ and $\odot(ABC)$ internally and $\omega_2$ be the circle tangent to $AD,CD$ and $\odot(ABC)$ internally. Let $K$ be the midpoint of $ID$ and $P=\overline{AI}\cap\odot(ABC)$. Then $\overline{PK}$ is the radical axis of $\omega_1$ and $\omega_2$.

Let $\omega_1$ and $\omega_2$ with centers $I_1,I_2$ touch $\overline{BC}$ at $X,Y$ respectively and let $M$ be the midpoint of $XY$. Let $\omega_1$ and $\omega_2$ touch $\overline{AD}$ at $P,Q$ and $\odot(ABC)$ at $R,S$ respectively . It's well known that $\overline{XR}\cap\overline{YS}=P$ and $PX\cdot PR=PY\cdot PS$. (See $\textbf{Lemma 4.33}$ of EGMO). Hence, $P$ lies on the radaical axis of $\omega_1,\omega_2$. By Sawayama-Thebault Theorem we have that $I_1,I,I_2$ are collinear and $\overline{XP}\cap\overline{YQ}=I$. Let $V$ be the reflection of $D$ over $M$. Define $\psi(\cdot)=\mathcal{P}_{\omega_1}(\cdot)-\mathcal{P}_{\omega_2}(\cdot)$ where $\mathcal{P}_{\omega}(\cdot)$ denotes the power of $\cdot$ WRT some circle $\omega$. It's well known that $\psi$ is a linear function on $\cdot$, so it suffices to show that $\psi(K)=\frac{(\psi(I)+\psi(D))}{2}=0\implies \psi(I)+\psi(D)=0$. Now, $$\psi(D)+\psi(V)=(DX^2-DY^2)+(VX^2-VY^2)=(DX^2-DY^2)+(DY^2-DX^2)=0=\psi(I)+\psi(D)\implies \psi(I)=\psi(V)\implies II_1^2-II_2^2=VI_1^2-VI_2^2$$So it further suffices to show that $\overline{IV}\perp\overline{II_1I_2}$.Notice that $\measuredangle YXI=\measuredangle YDI_2$ and $\measuredangle XIY=\measuredangle I_2YD$. Hence, $\Delta XIY\stackrel{-}{\sim}\Delta DYI_2$. Now notice that $\measuredangle VXI=\measuredangle I_2YI$ and $\frac{XI}{XV}=\frac{XI}{DY}=\frac{YI}{YI_2}$. Hnce, $\Delta IXV\stackrel{+}{\sim} \Delta IYI_2$. Thus $\overline{IV}\perp\overline{II_1I_2}$. Backtracking our series of equivalenences we get that $\overline{PK}$ is the radical axis of $\omega_1,\omega_2$. $\quad\square$
__________________________________________________________________________________________

Let $\omega_1$ with center $O_1$ be the circle tangent to $AM,BM$ and internally tangent to $\odot(ABC)$ and $\omega_2$ with center $O_2$ be the circle tangent to $AM,CM$ and internally tangent to $\odot(ABC)$. Denote $I_a$ as the $A-$ Excenter of $\Delta ABC$ , $U$ as the midpoint of $IM$ and let the $A-$ Excircle touch $\overline{BC}$ at $D$. Let $D'$ be the reflection of $D$ over $I_a$. The Homothety $\mathcal{H}$ at $A$ mapping the Incircle of $\Delta ABC$ to the $A-$ Excircle maps $T$ to $D'$, hence ,$A,T,D'$ are collinear. Combining the fact that $MD=MT$ we get that $\overline{MI_a}\parallel\overline{AT}$. Combining with $\textbf{LEMMA}$ we have that $PU\parallel\overline{I_aM}\parallel\overline{AT}\perp\overline{O_1IO_2}$. Let $\overline{O_1IO_2}\cap\overline{BC}=X$. Hence, $\overline{XS}$ is tangent to $\odot(I)$ as $AT\perp\overline{O_1IO_2}$. Hence, $\{\omega_1,\odot(I),\omega_2\}$ share the common exsimillicenter $X$ with external common tangents $\overline{BC},\overline{XS}$, thus $\omega_1,\omega_2$ are the Incircle and the $X-$ Excircle of $\Delta\{\overline{XS},\overline{AM},\overline{BC}\}$. $\quad\blacksquare$
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MP8148
888 posts
#4 • 6 Y
Y by amar_04, mijail, Pluto04, mathtiger6, hakN, PRMOisTheHardestExam
[asy]
import olympiad;
size(10cm);
defaultpen(fontsize(10pt));
defaultpen(linewidth(0.4));
dotfactor *= 1.5;

pair A = dir(135), B = dir(220), C = dir(320), I = incenter(A,B,C), T = foot(I,B,C), M = (B+C)/2, E = foot(I,C,A), F = foot(I,A,B), U = extension(E,F,B,C), S = 2*foot(T,U,I)-T, V = extension(A,M,U,S), J = incenter(M,V,U), K = foot(J,B,C), L = foot(J,A,M), T1 = 2I-T, T2 = 2M-T, G = foot(T,A,T1);

draw(A--B--C--A);
draw(incircle(A,B,C));
draw(M--U--V--M, linewidth(1));
draw(T--A--M);
draw(incircle(M,U,V)^^unitcircle, linewidth(0.75));
draw(K--L, dashed);
draw(T--T1--V);
draw(A--T2^^M--G);
draw((U+V)/2--(M+V)/2^^U--I);
dot((U+V)/2^^(M+V)/2^^L^^K);

dot("$A$", A, dir(135));
dot("$B$", B, dir(240));
dot("$C$", C, dir(320));
dot("$M$", M, dir(270));
dot("$T$", T, dir(270));
dot("$U$", U, dir(210));
dot("$V$", V, dir(80));
dot("$S$", S, dir(150));
dot("$I$", I, dir(135));
dot("$T_1$", T1, dir(80));
dot("$T_2$", T2, dir(270));
dot("$G$", G, dir(10));
[/asy]
Suppose that the tangent to $\omega$ at $S$ meets $\overline{BC}$ at $U$ and $\overline{AM}$ at $V$. By curvilinear incircle properties, the result is equivalent to proving the incenter $I$ lies on the $M$-intouch chord in $\triangle MUV$. Clearly $\overline{UI}$ bisects $\angle MUV$, so by Iran lemma it suffices to show $I$ lies on the $V$-midline.

Let $T_1$ be the reflection of $T$ over $I$, $T_2$ be the reflection of $T$ over $M$, and $G = \overline{T_1T_2} \cap \omega$. It is well-known that $A \in \overline{T_1T_2}$. Also $\overline{MG}$ is tangent to $\omega$ since $MG = MT = MT_2$ from $\angle TGT_2 = 90^\circ$. Now applying Pascal's on $TSSGT_1T_1$ and $TSGGT_1T$ implies $\overline{VT_1}$ tangent to $\omega$, and the desired result follows.
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Dr_Vex
562 posts
#5 • 2 Y
Y by amar_04, Combigeontal231
$\underline{\textbf{Solution:}}$ Let $\delta ''$ be Thebault circle of $(ABC)$ associated with $AM$ touching $\frown{AB}$ not containing $C$ We will prove that $\delta '' \equiv \delta ', $ where $\delta '$ is the incircle of $\delta$. Now, we rephrase the problem:
Rephrased Problem wrote:
In $\Delta ABC$ with $\odot (I)$ as its incircle. Let $\odot (I) \cap BC = T, AT \cap \odot (I) = S \neq T $ and $SS \cap BC = X$. Let $\delta '''$ be thebault circle of $(ABC)$ associated with $AM' (M' \in BC)$ which touches minor arc $AB$ such that $X$ is the exsimilicenter of $\delta ''$ and $\odot (I),$ then prove that $M' \equiv M$
$\newline
\underline{\textbf{Proof:}}$ First we state a short lemma,
Lemma: In $\Delta ABC$, with $I$ as its incenter and $\Delta DEF$ as its intouch triangle. Let $AI \cap (DIEC) = I \neq I_{1}.$ Let $\odot (I_{1})$ touch $AB$ and $AC$ at $K$ and $L$. Then line $\ell \parallel AB$ through $C$ is tangent to $\odot (I_{1})$
$\newline
\underline{\textbf{Proof:}}$ Let $IF \cap \ell = R, \angle CRI = 90^{\circ}$ and $\angle CEI = 90^{\circ}$ which means $R \in (DEIC).$ A homothety $(\mathcal{T})$ exists that sends $I$ to $I_{1}$ centered at, $\mathcal{T}$ sends $F$ to $K$ and $FI \cap \odot(I) = P \leftrightarrow AP \cap KI_{1} = O$ (say). Also $\angle ORI = 90^{\circ}$ as $FI \parallel KI_{1} ,$ which means that $\overline{CRO} \equiv \ell.$ Hence, the conclusion follows because $\ell \parallel AB$ and $AB$ is tangent to $\odot (I_{1}).$
$\newline
\rule{\textwidth}{0.5pt}
\newline$
Let $SS \cap AM' = D.$ Let $T-$antipode $W.R.T$ $\odot (I)$ be $T'$. Then, $DT' \parallel BC$ and is tangent to $\odot (I)$ by above lemma. Let $AS \cap \ell = W.$
$\newline$
Claim: $D$ is midpoint of $WT'$
$\newline
\underline{\textbf{Proof:}}$ Note that $$\angle DWS = \angle WTX = \angle ST'T = 90^{\circ} - \angle STT' = 90^{\circ} - \angle DST' = 180^{\circ} - (90^{\circ} + \angle DST') = \angle DSW$$Which means that $DW = DS. DT'$ and $DS$ being tangent to same circle, $DS = DT'.$ Hence, $DW = DS = DT'.$
$\newline
\newline$
It is well-known that $AT'$ is the $A-$nagel line. There exists a homothety $\mathcal{S}$ sending $\ell$ to $BC$ centered at $A$. Then $\mathcal{S}$ sends $W$ to $T$ and $T'$ to $AT' \cap BC.$ Therefore, $T$ and $AT' \cap BC$ being isotomic conjuates, $AD \cap BC = M'$ is midpoint of $BC$ or $M' \equiv M$ and $\delta ''' \equiv \delta ''$
$\newline
\rule{\textwidth}{0.5pt}
\newline$
Now, $MI$ bisects $\angle AMB$ as it is the defintion of Thebault circle. $X$ being the exsimilicenter of $\odot (I_{1})$ and $\odot (I)\Rightarrow XI$ bisects $\angle SXB.$ Therefore $\odot (I_{1})$ is incircle of $DXM$ or $\delta '' \equiv \delta '.$.
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mathaddiction
308 posts
#6 • 8 Y
Y by Pluto04, Dr_Vex, hakN, PRMOisTheHardestExam, samrocksnature, Mango247, Mango247, Mango247
The result is nice but the solution is just a combination of some (very) well-known lemmas:
We first show these three lemmas (the labelling of points in these lemma are irrelevant to the original problem).

Lemma 1. (Iran Lemma) In $\triangle ABC$, suppose $I$ is the incentre, $M_a,M_b,M_c$ are the midpoints of $BC,CA,AB$ and $T_a,T_b,T_c$ are the touching points of incircle and $BC,CA,AB$. Then $AI,T_aT_c,M_aM_b$ and the circle with diameter $CI$ are concurrent.
Proof.
Let $X$ be the projection of $C$ onto $AI$, then since $M_bX=M_bC=M_bA$ we have
$$\angle CM_bX=2\angle M_bAX=\angle BAC=\angle CM_bM_a$$and
$$\angle T_bT_aX=\angle T_bCX=90^{\circ}-\frac{\angle BAC}{2}=\angle T_bT_aT_c$$Hence the three line and the circle are concurrent at point $X$ as desired. $\blacksquare$
[asy]
       size(6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -25.113603016501955, xmax = 11.785607854595835, ymin = -13.639613371565176, ymax = 17.783389525730115;  /* image dimensions */pen qqwuqq = rgb(0,0.39215686274509803,0);  /* draw figures */draw((-15.473293189260673,11.332430976089425)--(-16.111255395965372,-2.1046480026282635), linewidth(0.8)); draw((-16.111255395965372,-2.1046480026282635)--(2.11054013303755,-2.6628649334948733), linewidth(0.8)); draw((2.11054013303755,-2.6628649334948733)--(-15.473293189260673,11.332430976089425), linewidth(0.8)); draw(circle((-11.370259667167469,2.279961860202567), 4.527724159576039), linewidth(0.8) + qqwuqq); draw((-15.473293189260673,11.332430976089425)--(-7.214272491516157,-6.889338207911946), linewidth(0.8) + linetype("4 4") + blue); draw((-15.892889384219655,2.494686119765876)--(-7.214272491516157,-6.889338207911946), linewidth(0.8) + blue); draw((-7.214272491516157,-6.889338207911946)--(-6.681376528111562,4.334783021297276), linewidth(0.8) + linetype("4 4") + blue);  /* dots and labels */dot((-15.473293189260673,11.332430976089425),dotstyle); label("$A$", (-15.171677825967656,12.209392123824001), NE * labelscalefactor); dot((-16.111255395965372,-2.1046480026282635),dotstyle); label("$B$", (-17.457852292178132,-2.7849868520637937), NE * labelscalefactor); dot((2.11054013303755,-2.6628649334948733),dotstyle); label("$C$", (3.08234875337401,-2.491618567752946), NE * labelscalefactor); dot((-11.370259667167469,2.2799618602025666),linewidth(4pt) + dotstyle); label("$I$", (-11.227504225788476,2.5282387415660117), NE * labelscalefactor); dot((-7.0003576314639115,-2.383756468061568),linewidth(4pt) + dotstyle); label("$M_{a}$", (-6.533611676814903,-1.9048819991312493), NE * labelscalefactor); dot((-6.681376528111562,4.334783021297276),linewidth(4pt) + dotstyle); label("$M_{b}$", (-6.305436344573133,4.744799111914642), NE * labelscalefactor); dot((-15.792274292613023,4.613891486730581),linewidth(4pt) + dotstyle); label("$M_{c}$", (-17.89929105579821,4.940377968121874), NE * labelscalefactor); dot((-11.508899524926496,-2.2456392109314236),linewidth(4pt) + dotstyle); label("$T_{a}$", (-12.400977363031869,-4.109516124091807), NE * labelscalefactor); dot((-8.550638429697806,5.822562902151633),linewidth(4pt) + dotstyle); label("$T_{b}$", (-8.39161081078361,6.53760529381427), NE * labelscalefactor); dot((-15.892889384219655,2.494686119765876),linewidth(4pt) + dotstyle); label("$T_{c}$", (-17.825255816143594,2.789010549842321), NE * labelscalefactor); dot((-7.214272491516157,-6.889338207911946),linewidth(4pt) + dotstyle); label("$X$", (-7.348523577678371,-8.0982124456936), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
[/asy]
{Lemma 2.} In $\triangle ABC$, $O$ is the circumcenter. Tangents at $B$ and $C$ to $(ABC)$ meets at $T$. $AT$ intersect the circumcircle again at $D\neq A$. $AO$ intersect $BC$ and $(ABC)$ at $E$ and $A'\neq A$. Then the tangents at $D$, $A'$ to $(ABC)$ and $TE$ are concurrent.
{Proof.}
Suppose the line through $O$ parallel to $AT$ intersect the line $ET$ at a point $F$. We will show that $\triangle DFO\sim\triangle DA'A$ Notice that
$$\angle DOF=\angle ADO=\angle A'AD\hspace{50pt}(1)$$hence it suffices to show $\displaystyle \frac{OF}{OD}=\frac{AA'}{DA}$. Let $R$ be the circumradius of $\triangle ABC$ and $Q$ the $A$-dumpty point of $\triangle ABC$. Then $OD\times AA'=2R^2$. Meanwhile,
$$\frac{OF}{AT}=\frac{OE}{EA}=\frac{OE}{BE}\cdot\frac{BE}{EA}=\frac{\cos A}{\sin 2C}\cdot\frac{\cos C}{\sin B}=\frac{\cos A}{2\sin B\sin C}$$$$\frac{AT}{2R\sin B}=\frac{AT}{AC}=\frac{\sin B}{\sin \angle QTC}=\frac{\sin B}{\sin \angle QBC}$$and $$AD=2R\sin\angle ACD$$Therefore, multiplying them it suffices to show
$$\sin B\cos A\sin\angle ACD=\sin C\sin\angle QBC \hspace{50pt}(2)$$Indeed we have $\cos A=\sin\angle OCB$, and that $B,Q,O,C$ are concyclic so
$$\frac{\cos A\sin\angle ACD}{\sin QBC}=\frac{\sin\angle OCB\sin\angle ACD}{\sin\angle QBC}=\frac{ AD}{2QC}=\frac{QA}{AC}=\frac{\sin\angle BAD}{\sin\angle DAC}=\frac{BD}{DC}=\frac{BA}{AC}=\frac{\sin C}{\sin B}$$as desired.
Therefore, $\triangle DFO\sim\triangle DA'A$, so $\angle ODF=\angle ADA'=90^{\circ}$ and $FD$ is tangent to $(ABC)$. Since $OF\| AD\perp DA'$ hence $FA'$ is tangent to $(ABC)$ as well. Hence the three lines mentioned in the statement of the lemma are concurrent at $F$ as desired. $\blacksquare$

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[/asy]
Lemma 3. In triangle $ABC$, Suppose the incircle touches $AB,AC,BC$ at $F,E,T$ respectively and $M$ is the midpoint of $BC$. Then $AM,TI,EF$ are concurrent.
Proof.
Let $TI\cap EF=D$, suppose the line through $D$ parallel to $BC$
meet $AB$ and $AC$ at $C_0$ and $B_0$. Then from $\angle IDC_0=\angle C_0FI=90^{\circ}$, $I,D,C_0,F$ are concyclic and similarly $I,D,B_0,E$ are concyclic.Hence
$$\angle IC_0D=\angle IFD=\angle IED=\angle IB_0D$$hence $C_0D=B_0D$, a homothety at $A$ which sends $B_0C_0$ to $CB$ will send $D$ to $M$, so $A,D,M$ are collinear. $\blacksquare$
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[/asy]
We now return to the original problem. Let $I$ be the incentre. Suppose the incircle touches $AB,AC$ at $F,E$ respectively, and $T'$ be the reflection of $T$ in $I$. Suppose the tangent to $\omega $ at $S$ intersect $AM$ and $BC$ at $L$ and $K$ respectively.
\newline Applying Lemma 3 to $\triangle TFE$ we have that $TI\cap EF$ lies on $AM$, so by Lemma $2$ $LI'$ is tangent to $\omega$. Hence
$$\text{dist}(L,BC)=\text{dist}(T',BC)=2\text{dist}(I,BC)$$which implies that $I$ lies on the line passing through the midpoint of $LM$ and $LK$. By Lemma $1$ if the incircle of $\delta$ intersects $AM$ and $BC$ at $R$ and $Q$ then $Q,I,R$ are collinear.
Let $N$ be the midpoint of minor arc of $(ABC)$ and suppose $NQ$ meet $(ABC)$ again at $P$. Then by shooting lemma we have $\triangle NIP\sim\triangle NQI$, hence letting $Z=BC\cap AN$ we have
$$\angle API=\angle APN-\angle IPN=\angle ADC-\angle QIN=\angle ZQR=\angle QRM$$so $A,P,I,R$ are concyclic. Therefore
$$\angle ARP=\angle AIP=180^{\circ}-\angle PIN=180^{\circ}-\angle IQN=\angle PQI=\angle PQR $$therefore $P$ lies on the incircle $\Omega$ of $\delta$. Since $P,Q,N$ are collinear, and the tangent at $Q$ to $\Omega$ and the tangent at $N$
to $(ABC)$ are parallel, $P$ is the homothetic center of the two circles, so they are tangent to each other at $P$ as desired.
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buratinogigle
2372 posts
#7 • 4 Y
Y by amar_04, Dr_Vex, VMF-er, LoloChen
I like this beautiful problem much. The following is a generalization of the view of the harmonic range points.

Let incircle $(I)$ of triangle $ABC$ touch $BC$ at $D$. A tangent $\ell$ of $(I)$ meets $BC$ at $S$. $P$ is a point on line $BC$ such that $\overline{SP}\cdot \overline{SD}=\overline{SB}\cdot \overline{SC}$. Line $\ell$ meets $AP$ at $T$. Prove that incircle of triangle $PST$ is tangent to circumcircle of $ABC$.

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amar_04
1915 posts
#8 • 3 Y
Y by buratinogigle, Mathematicsislovely, Bumblebee60
buratinogigle wrote:
I like this beautiful problem much. The following is a generalization of the view of the harmonic range points.

Let incircle $(I)$ of triangle $ABC$ touch $BC$ at $D$. A tangent $\ell$ of $(I)$ meets $BC$ at $S$. $P$ is a point on line $BC$ such that $\overline{SP}\cdot \overline{SD}=\overline{SB}\cdot \overline{SC}$. Line $\ell$ meets $AP$ at $T$. Prove that incircle of triangle $PST$ is tangent to circumcircle of $ABC$.

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Let $\omega_1$ and $\omega_2$ with circumcenters $O_1,O_2$ respectively be the two thebault circles of $\Delta ABC$ WRT $\overline{AP}$. Let the exsimillicenter of $\{\odot(I),\omega_1,\omega_2\}$ (which obviously exists due to Sawayama) be $S$. We show that $SD\cdot SP=SB\cdot SC$. Let $I_A$ be the $A-$ Excenter of $\Delta ABC$. From the $\textbf{LEMMA}$ in #3 we get that $\overline{PI_A}\perp\overline{O_AO_B}$. Let $\overline{PI_A}\cap\overline{IS}=K$. Clearly $K\in\odot(II_A)$. Hence, $SD\cdot SP=SI\cdot SK=SB\cdot SC$. $\blacksquare$

Remark:- Using this problem as a lemma for the case when $P$ is the $A-$ Extouch point we get the well known fact that the Thebault Circles WRT $A-$ Nagel Cevian are congruent to $\odot(I)$.
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dolly33
88 posts
#9 • 1 Y
Y by SK_pi3145
This problem clearly brings up some Thebault circles.

Generalized Lemma
Triangle $ABC$, arbitrary point $D$ on segment $BC$. Let $\omega_1, \omega_2$ circles that are tangent to $BC, (ABC), AD$, respectively from left.
If $X$ is the foot from incenter $I$ to $BC$, and $Y=\omega_1\cap (ABC), Z=\omega_2 \cap (ABC)$, $Z, Y, X, D$ are cyclic.

pf. From well-known lemma (Sawayama the bault), we easily get $O_1, I, O_2$ are collinear when $O_1, O_2$ are centers of thebault circles.
Let $P, Q=\omega_1\cap BC, Z=\omega_2 \cap BC$. Note that $\angle PIQ=90$. Let $R=(PIQ)\cap O_1O_2$.
Since $O_1P^2=O_1R\cdot O_1I$, $DR$ is the polar of $I$ wrt $\omega_1$. Thus $DR\perp O_1O_2$.
Let $S=YZ\cap BC$. $SY\cdot SZ=SP\cdot SQ=SI\cdot SR=SX\cdot SD$.

Now back to the problem. All we need to prove is that when $D$ is the midpoint of $BC$, $S$ is $EF\cap BC$ when $E, F=\omega \cap AC,AB$

From Lemma, we get $SC\cdot SB=SX\cdot SD$, since $D$ is the midpoint of $BC$, by Newton Lemma we are done.
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khina
994 posts
#10 • 1 Y
Y by PRMOisTheHardestExam
uhhhhhhhhh

solution sketch
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meysam1371
35 posts
#11 • 2 Y
Y by teomihai, PRMOisTheHardestExam
Here is another aproach:

Let tangent to $\omega$ at $S$ intersects $BC$ and $AM$ at $Q$ and $P$ respectively, and $N$ in second intersection of $IT$ with $\omega$.It is known that $PN$ is tangent to $\omega$ (its proof is simple!). Also $V$ is reflection of $P$ to $I$. We call the circle tangent to $BC$, segment $AM$ and circumcircle of triangle $\triangle ABC$, by $\gamma$. let $\gamma$ touch $AM$ and $BC$ at $U$ and $R$ respectively. We should prove that $\gamma$ is incircle of $\delta$. By Sawayama-Thebault lemma, we know that $I$ lies on $RU$.
It is enough to prove that incircle of $\delta$ touches $QM$ at $R$, or equivalently

$$ MR = \frac{1}{2}\left( MQ + MP - QP \right) \quad \Leftrightarrow \quad 2MR = MT+TQ+MP-QS-SP $$$$ \quad \Leftrightarrow \quad MT+TR+MU=MT+MU+UP-PN \quad \Leftrightarrow \quad$$$$RT+VT=UP \quad \Leftrightarrow \quad RV=WV$$where $W$ is intersection of $RU$ with the line parallel to $PU$ through $V$. Clearly triangle $\triangle VWR$ is Isosceles and $RV=WV$. We are done. $\blacksquare$

https://i.postimg.cc/7Z19pPB9/geogebra-export-2.png
This post has been edited 3 times. Last edited by meysam1371, Jan 21, 2022, 9:59 AM
Reason: some minor improvements.
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SerdarBozdag
892 posts
#12 • 3 Y
Y by teomihai, PRMOisTheHardestExam, GeoKing
Also there exists a solution by proving $ZB/ZC=(s-b)/(s-c)$ with ratio lemma and Casey's theorem. However I won't write it because I accidentally refreshed the page twice. :(
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Nari_Tom
117 posts
#14
Y by
Once you realize this condition problem will become just another challenging one, not impossible one.

Let $ABC$ be a triangle and $D$ be the point on the side $BC$. Let $\omega$ be the circle that touches $(ABC)$ internally and also touches $AD$, $CD$ at $F$ and $E$, respectively. Prove that $FE$ passes through the incenter.

Proof: Let $a, b, c$ be the side lengths. We let's use Casey's theorem on the $(ABC)$ and four other circles $A, B, C, \omega$. Then we have: $AF \cdot BC+CE \cdot AB= BE \cdot AC$ $\implies$ $AF \cdot a+ac=(b+c) \cdot BE=(b+c) \cdot (BL+LE)=(b+c)BL+(b+c)LE=ac+(b+c)BL$. $\implies$ $\frac{AF}{LE}=\frac{b+c}{a}$.
By the Menelaus theorem on $\triangle ALD$ and line I-F-E, we should prove that: $\frac{LI}{AI} \cdot \frac{AF}{FD} \cdot \frac{DE}{LE}=\frac{LI}{AI} \cdot \frac{AF}{LE}=\frac{a}{b+c} \cdot \frac{AF}{LE}=1$, which is true since we found it earlier, done.
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