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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
Triangle is similar to two others
gghx   3
N 2 minutes ago by LeYohan
Source: SMO junior 2024 Q2
Let $ABCD$ be a parallelogram and points $E,F$ be on its exterior. If triangles $BCF$ and $DEC$ are similar, i.e. $\triangle BCF \sim \triangle DEC$, prove that triangle $AEF$ is similar to these two triangles.
3 replies
gghx
Oct 12, 2024
LeYohan
2 minutes ago
Parallel lines lead to similar triangles
a1267ab   30
N 6 minutes ago by Ilikeminecraft
Source: USA TST for EGMO 2020, Problem 2, by Andrew Gu
Let $ABC$ be a triangle and let $P$ be a point not lying on any of the three lines $AB$, $BC$, or $CA$. Distinct points $D$, $E$, and $F$ lie on lines $BC$, $AC$, and $AB$, respectively, such that $\overline{DE}\parallel \overline{CP}$ and $\overline{DF}\parallel \overline{BP}$. Show that there exists a point $Q$ on the circumcircle of $\triangle AEF$ such that $\triangle BAQ$ is similar to $\triangle PAC$.

Andrew Gu
30 replies
a1267ab
Dec 16, 2019
Ilikeminecraft
6 minutes ago
Minimum value of a 3 variable expression
bin_sherlo   4
N 12 minutes ago by Assassino9931
Source: Türkiye 2025 JBMO TST P6
Find the minimum value of
\[\frac{x^3+1}{(y-1)(z+1)}+\frac{y^3+1}{(z-1)(x+1)}+\frac{z^3+1}{(x-1)(y+1)}\]where $x,y,z>1$ are reals.
4 replies
bin_sherlo
4 hours ago
Assassino9931
12 minutes ago
Circumcircle of ADM
v_Enhance   68
N 23 minutes ago by Giant_PT
Source: USA TSTST 2012, Problem 7
Triangle $ABC$ is inscribed in circle $\Omega$. The interior angle bisector of angle $A$ intersects side $BC$ and $\Omega$ at $D$ and $L$ (other than $A$), respectively. Let $M$ be the midpoint of side $BC$. The circumcircle of triangle $ADM$ intersects sides $AB$ and $AC$ again at $Q$ and $P$ (other than $A$), respectively. Let $N$ be the midpoint of segment $PQ$, and let $H$ be the foot of the perpendicular from $L$ to line $ND$. Prove that line $ML$ is tangent to the circumcircle of triangle $HMN$.
68 replies
v_Enhance
Jul 19, 2012
Giant_PT
23 minutes ago
Weird locus problem
Sedro   1
N Today at 4:20 PM by sami1618
Points $A$ and $B$ are in the coordinate plane such that $AB=2$. Let $\mathcal{H}$ denote the locus of all points $P$ in the coordinate plane satisfying $PA\cdot PB=2$, and let $M$ be the midpoint of $AB$. Points $X$ and $Y$ are on $\mathcal{H}$ such that $\angle XMY = 45^\circ$ and $MX\cdot MY=\sqrt{2}$. The value of $MX^4 + MY^4$ can be expressed in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
1 reply
Sedro
Today at 3:12 AM
sami1618
Today at 4:20 PM
Inequalities
sqing   4
N Today at 3:35 PM by sqing
Let $ a,b,c\geq 0 , (a+8)(b+c)=9.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{38}{23}$$Let $ a,b,c\geq 0 , (a+2)(b+c)=3.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2(2\sqrt{3}+1)}{5}$$
4 replies
sqing
Yesterday at 12:50 PM
sqing
Today at 3:35 PM
Find the range of 'f'
agirlhasnoname   1
N Today at 2:46 PM by Mathzeus1024
Consider the triangle with vertices (1,2), (-5,-1) and (3,-2). Let Δ denote the region enclosed by the above triangle. Consider the function f:Δ-->R defined by f(x,y)= |10x - 3y|. Then the range of f is in the interval:
A)[0,36]
B)[0,47]
C)[4,47]
D)36,47]
1 reply
agirlhasnoname
May 14, 2021
Mathzeus1024
Today at 2:46 PM
Function equation
hoangdinhnhatlqdqt   1
N Today at 1:52 PM by Mathzeus1024
Find all functions $f:\mathbb{R}\geq 0\rightarrow \mathbb{R}\geq 0$ satisfying:
$f(f(x)-x)=2x\forall x\geq 0$
1 reply
hoangdinhnhatlqdqt
Dec 17, 2017
Mathzeus1024
Today at 1:52 PM
Inequality with function.
vickyricky   3
N Today at 1:51 PM by SpeedCuber7
If x satisfies the inequalit$ |x - 1| + |x - 2| + |x - 3| \ge 6$, then
$(a) 0 \le x \le 4. (b) x \le 0 or x \ge 4. (c) x \le -2 or x \ge 4$. (d) None of these.
3 replies
vickyricky
May 28, 2018
SpeedCuber7
Today at 1:51 PM
Writing/Evaluating Exponential Functions
Samarthsshah   1
N Today at 1:47 PM by Mathzeus1024
Rewrite the function and determine if the function represents exponential growth or decay. Identify the percent rate of change.

y=2(9)^-x/2
1 reply
Samarthsshah
Jan 30, 2018
Mathzeus1024
Today at 1:47 PM
Functional equation
TuZo   1
N Today at 1:37 PM by Mathzeus1024
My question is, if we can determinate or not, all $f:R\to R$ continuous function with $sin(f(x+y))=sin(f(x)+f(y))$ for all real $x,y$.
Thank you!
1 reply
TuZo
Oct 23, 2018
Mathzeus1024
Today at 1:37 PM
Real parameter equation
L.Lawliet03   1
N Today at 1:12 PM by Mathzeus1024
For which values of the real parameter $a$ does only one solution to the equation $(a+1)x^{2} -(a^{2} + a + 6)x +6a = 0$ belong to the interval (0,1)?
1 reply
L.Lawliet03
Nov 3, 2019
Mathzeus1024
Today at 1:12 PM
a inequality problem
Polus425   1
N Today at 12:36 PM by Mathzeus1024
$x_1,x_2\; are\; such\; two\; different\; real\; numbers:\; $
$(x_1 ^2 -2x_1 +4ln\, x_1)+(x_2 ^2 -2x_2 +4ln\, x_2)- x_1 ^2 x_2 ^2=0$
$prove\; that:\; x_1+x_2\ge 3$
1 reply
Polus425
Dec 19, 2019
Mathzeus1024
Today at 12:36 PM
Function of Common Area [China HS Mathematics League 2021]
HamstPan38825   1
N Today at 11:35 AM by Mathzeus1024
Define the regions $M, N$ in the Cartesian Plane as follows:
\begin{align*}
M &= \{(x, y) \in \mathbb R^2 \mid 0 \leq y \leq \text{min}(2x, 3-x)\} \\
N &= \{(x, y) \in \mathbb R^2 \mid t \leq x \leq t+2 \}
\end{align*}for some real number $t$. Denote the common area of $M$ and $N$ for some $t$ be $f(t)$. Compute the algebraic form of the function $f(t)$ for $0 \leq t \leq 1$.

(Source: China National High School Mathematics League 2021, Zhejiang Province, Problem 5)
1 reply
HamstPan38825
Jun 29, 2021
Mathzeus1024
Today at 11:35 AM
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 $ABC$.
Z K Y
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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 $\Delta\{\overline{XS},\overline{AM},\overline{BC}\}$. $\quad\blacksquare$
Z K Y
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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.
Z K Y
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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
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#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]
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[/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. $\blacksquare$

[asy]
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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. $\blacksquare$
[asy]
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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.
[asy]
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[/asy]
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buratinogigle
2374 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
This post has been edited 1 time. Last edited by khina, Sep 25, 2021, 9:17 PM
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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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