Incircle of a triangle is tangent to (ABC)

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Incircle of a triangle is tangent to (ABC)

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Source: XVII Sharygin Correspondence Round P18

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#1 h Mar 2, 2021, 3:36 AM • 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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#3 h Mar 2, 2021, 3:59 AM • 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

#4 h Mar 2, 2021, 4:01 AM • 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

#5 h Mar 2, 2021, 7:01 AM • 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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#6 h Mar 5, 2021, 1:36 PM • 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. $\blacksquare$

$[asy] size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -31.324229690975233, xmax = 26.004617634547465, ymin = -26.291270349086002, ymax = 22.529408822118842; /* image dimensions */pen zzttff = rgb(0.6,0.2,1); pen qqwuqq = rgb(0,0.39215686274509803,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); /* draw figures */draw((-14.374874070068559,18.092655468989744)--(-20.552994318817237,-1.1181418008419122), linewidth(0.8)); draw((-20.552994318817237,-1.1181418008419122)--(2.355655946615818,-2.425919079774114), linewidth(0.8)); draw((2.355655946615818,-2.425919079774114)--(-14.374874070068559,18.092655468989744), linewidth(0.8)); draw(circle((-8.67436746721582,5.660565187054879), 13.676719062236636), linewidth(0.8)); draw((-14.374874070068559,18.092655468989744)--(-10.106372746544746,-19.424216947396676), linewidth(0.8)); draw((-10.106372746544746,-19.424216947396676)--(-5.163274322663226,-1.9966888084963736), linewidth(0.8) + zzttff); draw((-20.552994318817237,-1.1181418008419122)--(-10.106372746544746,-19.424216947396676), linewidth(0.8)); draw((-10.106372746544746,-19.424216947396676)--(2.355655946615818,-2.425919079774114), linewidth(0.8)); draw((-11.436392875109853,-7.734353997228942)--(-7.047385602446938,-8.63936502684008), linewidth(0.8) + linetype("4 4") + qqwuqq); draw((-7.047385602446938,-8.63936502684008)--(-2.973860864363081,-6.771525094879985), linewidth(0.8) + linetype("4 4") + qqwuqq); draw((-8.67436746721582,5.660565187054879)--(-7.047385602446938,-8.63936502684008), linewidth(0.8) + fuqqzz); draw((-2.973860864363081,-6.771525094879985)--(-14.374874070068559,18.092655468989744), linewidth(0.8)); draw(circle((-9.390370106880283,-6.881825880170901), 12.562811527012238), linewidth(0.8) + linetype("4 4") + blue); /* dots and labels */dot((-14.374874070068559,18.092655468989744),dotstyle); label("$A$", (-13.902742517848194,19.440133480372477), NE * labelscalefactor); dot((-20.552994318817237,-1.1181418008419122),dotstyle); label("$B$", (-22.714773820534546,-2.1847939118520743), NE * labelscalefactor); dot((2.355655946615818,-2.425919079774114),dotstyle); label("$C$", (3.873251661708755,-2.1847939118520743), NE * labelscalefactor); dot((-10.106372746544746,-19.424216947396676),linewidth(4pt) + dotstyle); label("$T$", (-10.155097021303424,-20.97366525263734), NE * labelscalefactor); dot((-8.67436746721582,5.660565187054879),linewidth(4pt) + dotstyle); label("$O$", (-8.483849705276702,6.070154952158703), NE * labelscalefactor); dot((-11.436392875109853,-7.734353997228942),linewidth(4pt) + dotstyle); label("$D$", (-12.484714492128552,-8.971070892081793), NE * labelscalefactor); dot((-5.163274322663226,-1.9966888084963736),linewidth(4pt) + dotstyle); label("$E$", (-4.584272634547685,-1.2732044667465896), NE * labelscalefactor); dot((-2.973860864363081,-6.771525094879985),linewidth(4pt) + dotstyle); label("$A'$", (-2.7104498862753004,-5.932439408396844), NE * labelscalefactor); dot((-7.047385602446938,-8.63936502684008),linewidth(4pt) + dotstyle); label("$F$", (-6.407451524758654,-9.983948053310108), NE * labelscalefactor); dot((-12.905633472589205,5.179150735880404),linewidth(4pt) + dotstyle); label("$Q$", (-12.687289924374214,5.563716371544545), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$
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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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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#7 h Mar 6, 2021, 5:06 PM • 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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#8 h Mar 6, 2021, 5:53 PM • 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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#9 h Mar 9, 2021, 8:15 PM • 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

#10 h Sep 25, 2021, 9:16 PM • 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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#11 h Jan 21, 2022, 9:49 AM • 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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#12 h Jan 21, 2022, 3:21 PM • 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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#14 h Apr 22, 2025, 8:41 AM

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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