dame dame

A not-so-hard problem; given that I got spoiled on using some trigonometry and $``\text{to look at the direction of} \ LM"$ and ignore $\triangle BCE$. While that does tell me to not add anything unnecessary, but the hint didn't spoil any of the core ideas --- so that counts as a half-solve in my book. $\color{green} \rule{8.8cm}{2pt}$ $\color{green} \diamondsuit$ $\boxed{\textbf{Angles and Angles and Lemma with Angles.}}$ $\color{green} \diamondsuit$ $\color{green} \rule{8.8cm}{2pt}$ The main way the problem can solve (a.k.a. destroy) itself is to look at quadrilateral $LBI_EC$ with $I_E$ the excenter of $\triangle BCE$. Let $P = \overline{AB} \cap \overline{DC}$, and let $\angle P = x, \angle B = y, \angle C = z$. Furthermore, to capture all four degrees of freedom, let $\angle ABD = \theta$. By crude angle-chasing we can obtain that: $\angle EBC = y-\theta$ and so $\angle I_EBC = 90^{\circ} - \dfrac{y-\theta}{2}$ which is equal to $\dfrac{x+z+\theta}{2}$, $\angle MBC = \dfrac{\angle BAC}{2} = \dfrac{x+\theta}{2} = \angle MCB$, $\angle LBC = \dfrac{y}{2}$ and similarly, $\angle LCB = \dfrac{y}{2}$, finally, $\angle MBI_E = \dfrac{x+z+\theta}{2} - \angle MBC$, which is equal to $\dfrac{z}{2}$ or $\angle LCB$. Similarly, $\angle MCI_E$ must be equal to $\angle LCB$. For simplicity of writeup, let $\angle \dfrac{y}{2} = \beta$ (half of what $\angle B$ should be) and $\angle \dfrac{z}{2} = \gamma$. $\color{green} \rule{4.5cm}{0.4pt}$ $\color{green} \clubsuit$ $\boxed{\text{Conjuring the Lemma.}}$ $\color{green} \clubsuit$ $\color{green} \rule{4.5cm}{0.4pt}$ The existence of $M$ as almost an isogonal conjugate might surprise the reader, when the angles from $``\text{the opposite sides}"$ are actually equal. Keeping that in mind, we will guess that for every $LBI_EC$ with $\angle LBI_E, \angle LCI_E$ equal, there exists a point $M$ that satisfies the four properties (the third directly follows from the $\textit{first two}$ conditions plus the fact that the $\textit{big angles are equal}$.) $\angle MBI_E = \angle BCL$, $\angle MCI_E = \angle CBL$, $MB = MC$, and $L,M,I_E$ collinear. $\blacksquare$ $\blacksquare$ $\color{red} \rule{4cm}{2pt}$ $\color{red} \spadesuit$ $\color{red} \boxed{\textbf{The Final Push.}}$ $\color{red} \spadesuit$ $\color{red} \rule{4cm}{2pt}$ We prove the Lemma in two (almost) equivalent ways. (See Attachment for diagrams: I'm a bit too lazy to find reasons to manually post two pictures from GeoGebra, when they are really simple and virtually equal to each other.)

Let $I_1$ and $I_2$ be the incircle of $\triangle ABC$ and $\triangle DBC$(Note that $L$ must be the intersection of line $BI_1$ and $CI_2$), $F$ the excircle of $EBC$, and $X = AI_1 \cap CF$, $Y = DI_2 \cap BF$. Claim: $I_1I_2 // XY$ Proof. By incenter - excenter lemma, $AI_1$ and $DI_2$ insterect at $\Gamma$ at $M$. Moreover, since $BF$ and $CF$ are both external bisectors of $\angle DBC$ and $\angle ACB$, $MX = MI_2 = MI_1 = MB = MC = MY$ which means that $I_1 I_2 C X Y B$ is cylic with $M$ as its center. It is easy to see $I_1 I_2 // XY$ because $I_1X$ and $I_2Y$ are both diameters of $(I_1 I_2 C X Y B)$. $\blacksquare$ Now, we let $U = BI_1 \cap CF$ and $V = CI_2 \cap BF$. Claim: $BCUV$ is cyclic Proof. Easy angle chasing gives $\angle UBV = \angle UCV$. $\blacksquare$ Claim: $\triangle FXY$ and $\triangle LI_1I_2$ are perspective Proof. Since, $BCXY$ and $BCUV$ are cylic. $\angle FXY = \angle FBC = \angle FUV$. So, $UV // XY // I_1I_2$. Notice that $U = FX \cap LI_1$, $V = FY \cap LI_2$, and $P_{\infty}$ (on line $XY$) $= XY \cap I_1I_2$ are collinear. So, $\triangle FXY$ and $\triangle LI_1I_2$ are perspective. $\blacksquare$ By the third claim, we get that $LF$, $XI_1$, and $YI_2$ are concurrent at $M$ (since $M$ = $XI_1 \cap YI_2$). Hence, $L, M$, and $F$ are collinear as needed.

Simple angle-chasing solution. Let $K$ be the excenter of $\triangle EBC$, and let $L'$ be the isogonal conjugate of $L$ w.r.t. $\triangle BMC$. Notice that $$\begin{align*} \angle L'BC = \angle LBM &= \angle LBC + \angle CBM \\ &= \frac{\angle ABC}{2} + \frac{\angle BAC}{2} \\ &= 90^{\circ} - \frac{\angle BCA}{2} \\ &= \angle KCB, \end{align*}$$ so $L'KBC$ is an isosceles trapezoid. By symmetry, this means that $(MK, ML')$ is isogonal with respect to $\angle BMC$, which gives the desired result.

Let $I_B$, $I_C$ denote the incenters of $\triangle ABC$, $\triangle DBC$, and let $J_B$, $J_C$ denote the excenters of $\triangle ABC$, $\triangle DBC$ opposite $A$, $D$. The desired collinearity follows from Pascal theorem on $BI_AJ_ACI_DJ_D$. $\square$

Firstly note that $L$ is the incentre of $\triangle TAC ; T= AB \cap CD$ Let $I_a ,I_d$ be the incentres of $\triangle BAC ,\triangle DBC$ and let $E_a ,E_d$ be the excentres of $\triangle ABC ,\triangle DBC$ and $P$ be the excentre of $\triangle EBC$. Notice $\angle BI_aC= 90+\dfrac{\angle BAC}{2}=90+\dfrac{\angle BDC}{2}=\angle BI_dC$ So $B,I_a,I_d,C$ is cyclic then by incentre excentre lemma $B,I_a,I_d,C,E_a,E_d$ lies on the same circle . Also by incentre excentre lemma, $D,I_d,M,E_d$ and $A,I_a,M,E_a$ are collinear . Also $L=CI_a \cap BI_d$ and $CE_d$ is the external angle bisector of $\angle ACB=\angle ECB$ which implies $C,E_d,K$ are collinear .similarly $B,E_a,K$ are collinear .therefore by Pascal's theorem on hexagon$I_dE_dCI_aE_aB \Longrightarrow I_dE_d \cap I_aE_a, E_dC \cap E_aB ,I_aC \cap BI_d =M,K,L$ are collinear .

Let $I$ be the excenter of $\triangle BCE$ opposite $E$. $BI \cap \Gamma= X ,CI \cap \Gamma= Y , BL \cap \Gamma= Z ,CL \cap \Gamma= T$ We have: $B(CIML)=B(CXMZ)=\dfrac{\overline{MC}}{\overline{MX}}:\dfrac{\overline{ZC}}{\overline{ZX}}$ $C(BIML)=C(BYMT)=\dfrac{\overline{MB}}{\overline{MY}}:\dfrac{\overline{TB}}{\overline{TY}}$ We easily have $X,Y$ is the midpoint of the major arc $\stackrel\frown{CD},\stackrel\frown{AB}$, $Z,T$ is the midpoint of arc $\stackrel\frown{CDA},\stackrel\frown{DAB}$ We have: $\stackrel\frown{XT}=\stackrel\frown{XD}-\stackrel\frown{TD}=\dfrac{1}{2}(\stackrel\frown{DXC}-\stackrel\frown{DB})=\stackrel\frown{BC}=MB$ Simlarity, we have: $\stackrel\frown{ZY}=\stackrel\frown{MC} \Rightarrow \stackrel\frown{XT}=\stackrel\frown{ZY}=\stackrel\frown{MB}=\stackrel\frown{MC}$ Thus, we have: $\overline{ZX}=\overline{TY}, \overline{ZC}=\overline{MY};\overline{MX}=\overline{TB}$ Thus, $B(CIML)=C(BIML) \Rightarrow I,M,L$ are collinear.

Let $F$ be the E-excenter in $\triangle BCE$, Let $X$ be the incenter of $\triangle ABC$ and let $Y$ be the incenter of $\triangle BCD$. Let $I$ be the incenter of $\triangle BCE$. Finally, let $Z$ be the miquel point of the self intersecting quadrilateral $XCBY$ By Fact 5, we know that $MB = MC = MX = MY$ and so $BCXY$ is cyclic with center $M$. Obviously, by definition, $Z$ lies on $(BIC), (XIY), (XBM)$ and $(YCM)$ (because $M$ is the center of $BCXY$). Let $P = XY \cap BC$. By radical axis theorem on $(BCXY), (BICZ), (XIYZ)$, we see that $Z,I,P$ are also collinear. Since $Z$ is the miquel point, we must have $MZ \perp PI$ and so $\angle MZI = 90^\circ$. By radical axis on $(BXZM), (YZMC), (BCXY)$, we see that $L$ lies on $MZ$ So, it suffices to show that $M,Z,F$ are collinear or that $\angle FZI = 90^\circ$. But $F \in (BIC)$ since its the excenter and $\angle FZI = \angle FCI = 90^\circ$ and so we are done. $\blacksquare$

adj0109 wrote: Let $I_1$ and $I_2$ be the incircle of $\triangle ABC$ and $\triangle DBC$(Note that $L$ must be the intersection of line $BI_1$ and $CI_2$), $F$ the excircle of $EBC$, and $X = AI_1 \cap CF$, $Y = DI_2 \cap BF$. Claim: $I_1I_2 // XY$ Proof. By incenter - excenter lemma, $AI_1$ and $DI_2$ insterect at $\Gamma$ at $M$. Moreover, since $BF$ and $CF$ are both external bisectors of $\angle DBC$ and $\angle ACB$, $MX = MI_2 = MI_1 = MB = MC = MY$ which means that $I_1 I_2 C X Y B$ is cylic with $M$ as its center. It is easy to see $I_1 I_2 // XY$ because $I_1X$ and $I_2Y$ are both diameters of $(I_1 I_2 C X Y B)$. $\blacksquare$ Now, we let $U = BI_1 \cap CF$ and $V = CI_2 \cap BF$. Claim: $BCUV$ is cyclic Proof. Easy angle chasing gives $\angle UBV = \angle UCV$. $\blacksquare$ Claim: $\triangle FXY$ and $\triangle LI_1I_2$ are perspective Proof. Since, $BCXY$ and $BCUV$ are cylic. $\angle FXY = \angle FBC = \angle FUV$. So, $UV // XY // I_1I_2$. Notice that $U = FX \cap LI_1$, $V = FY \cap LI_2$, and $P_{\infty}$ (on line $XY$) $= XY \cap I_1I_2$ are collinear. So, $\triangle FXY$ and $\triangle LI_1I_2$ are perspective. $\blacksquare$ By the third claim, we get that $LF$, $XI_1$, and $YI_2$ are concurrent at $M$ (since $M$ = $XI_1 \cap YI_2$). Hence, $L, M$, and $F$ are collinear as needed. I had a similar idea the last part...using perspectivity

Adding onto post 3, see here... this problem is not very new it seems

Let $I_1,I_2,I$ be the incenters of triangles $ABC, BDC,BEC$, respectively. It's trivial that $I_1,I,C$ and $I_2,I,B$ are collinear. In addition, since the circle with center $L$ is tangent to $AB,BC,CD$, we infer that $L$ lies on the bisectors of angles $\angle ABC$ and $\angle BCD$. Hence, $B,I_1,L$ and $C,I_2,L$ are also collinear. To end, note that $A,I_1,M$ and $D,I_2,M$ are collinear as well, since $M$ is the midpoint of the small arc $BC$. Therefore, by Fact 5, $MB=MI_1=MI_2=MC$ i.e., $BI_1I_2C$ is cyclic with its center being point $M$. Let $P \equiv I_1I_2 \cap BC$. Then, by Brokard's theorem, $ML \perp PI$. Hence, in order to prove that $L,M,I_E$ are collinear, where $I_E$ is the $E-$excenter of triangle $BEC$, we need to prove that $PI \perp MI_E$. We make the following Claim: Claim: $\angle I_1IP=\angle MI_EC$ and $\angle BIP=\angle BI_EM$. Proof: Firstly, note that $$\angle I_1IP+\angle BIP=\angle I_1IB=\angle BI_EC=\angle MI_EC+\angle BI_EM$$ Hence, if we let $\angle I_1IP=x, \angle BIP=y, \angle MI_EC=z, \angle BI_EM=w$, we obtain that $x+y=z+w$. In addition, by LoS we obtain: $$\frac{PI_1}{\sin x}=\frac{PI}{\sin \angle PI_1I}=\frac{PI}{\sin \angle PBI}=\frac{PB}{\sin y} \Rightarrow \frac{\sin x}{\sin y}=\frac{PI_1}{PB}=\frac{\sin \angle I_1BC}{\sin \angle PCI_2}=\frac{\sin \angle ABC/2}{\sin \angle BCD/2}$$ and $$\frac{\sin z}{\sin w}=\frac{\frac{BM}{\sin w}}{\frac{CM}{\sin z}}=\frac{\frac{MI_E}{\sin \angle MBI_E}}{\frac{MI_E}{\sin \angle MCI_E}}=\frac{\sin \angle MCI_E}{ \sin \angle MBI_E}=\frac{\sin \angle ABC/2}{\sin \angle BCD/2},$$ therefore $\frac{\sin x}{\sin y}=\frac{\sin z}{\sin w}$. Let $x+y=z+w=T$, then $\frac{\sin x}{\sin (T-x)}=\frac{\sin z}{\sin (T-z)}, \,\, (*)$. Note that $T=\angle BI_EC=180^\circ-\angle BIC=90^\circ-\frac{\angle BEC}{2}<90^\circ$, therefore $T$, $x$ and $z$ are acute. Hence, the function $f(t)=\frac{\sin t}{\sin (T-t)}$ where $t \in (0, \frac{\pi}{2})$ is strictly increasing, hence injective. Therefore, $(*)$ implies that $x=z$, which proves the Claim $\blacksquare$ To the problem, by the Claim we have that $$\angle I_2PI=\angle PIB-\angle PI_2B=\angle BI_EM-\angle ICB=\angle BI_EM-\angle II_EB=\angle II_EM,$$ hence $\angle I_2PI=\angle II_EM$. To conclude, note that $I_1I_2 \perp EI_E$ by simple angle-chasing: $$\angle I_1IE+\angle I_2I_1I=\angle 180^\circ-\angle EIC+\angle I_2BC=90^\circ-\frac{\angle EBC}{2}+\frac{\angle EBC}{2}=90^\circ,$$ therefore we infer that $PI \perp MI_E$, since the pairs of lines $(PI,PI_2)$ and $(I_EE,I_EM)$ form equal angles, and two lines, one from each pair, are perpendicular. Hence, the problem is solved.

Let $M_1$ be the midpoint of $\widehat{ADC}$, let $M_2$ be the midpoint of $\widehat{DAB},$ and let $X$ be the $E$-excenter of $\triangle EBC.$ Note that $$2\angle ACB = \widehat{AB}=\widehat{BM_{1}}-\widehat{AM_1}=\widehat{BM_1}-\widehat{CM_1}=2(\angle BCM_{1}-\angle M_{1}BC).$$ Therefore, $$\begin{align*}\angle XCB = \frac{180^\circ-\angle ACB}{2}&=\frac{180^\circ-\angle BCM_1+\angle M_{1}BC}{2}\\&=\frac{\angle CM_{1}B+2\angle LBC}{2}=\frac{180^\circ-\angle BMC}{2}+\angle LBC.\end{align*}$$ We can similarly show that $$\angle CBX=\frac{180^\circ-\angle BMC}{2}+\angle BCL.$$ Now construct $M'$ such that $\overline{MC}\parallel\overline{BL}$ and $\overline{MB}\parallel\overline{CL}.$ Remark that $$\angle XCM'=\angle XCB-\angle LBC=\frac{180^\circ-\angle BMC}{2}=\angle MCB,$$ $$\angle M'BX=\angle CBX-\angle BCL=\frac{180^\circ-\angle BMC}{2}=\angle CBM.$$ Hence, $M'$ is the isogonal conjugate of $M$ with respect to $\triangle XBC.$ Moreover, by the second isogonality lemma, $\overline{XL}$ is isogonal to $\overline{XM'}.$ Thus, $X,M,L$ collinear, as desired.

Also pretty much the exact same problem as this one from the 2018 Romania TST

Let $Q$ and $P$ be the incenters of $ABC$ and $DBC$ respectively. From the incenter-excenter lemma $M$ is the center of $(CPQB)$. Let $O$ be the excenter of $BEC$. $BO$ and $CO$ intersect $(CPQB)$ at $R$ and $S$ respectively. Pascal Theorem on $BQSCPR$ finishes the proof.

We begin with the following lemma. Lemma. wrote: Let circumcircle of $ADE$ be $\omega$. Let $B,C$ lie on $\omega$ such that $AB=AC$. Let $F$ be the intersection of line parallel to $AD$ through $B$ and line parallel to $AE$ through $C$. Prove that $AF,CD,BE$ are concurrent. Proof. We use trig Ceva. We need $$\frac{\sin{\angle ADC}}{\sin{\angle CDE}}\cdot \frac{\sin{\angle CFA}}{\sin{\angle AFB}}\cdot \frac{\sin{\angle DEB}}{\sin{\angle BEA}}=1.$$ But, we have $\angle ADC=\angle BEA$, $\angle CDE=\angle ACF$ and $\angle DEB=\angle FBA$, thus we need $$\frac{\sin{\angle FBA}}{\sin{\angle ACF}}\cdot \frac{\sin{\angle CFA}}{\sin{\angle AFB}}=1,$$ but this is true by sine law on $\triangle ABF$ and $\triangle ACF$. $\square$ Call the excenter of triangle $BCE$ opposite to $E$ as $I$ and call the midpoint of arc $BC$ as $M\equiv M_{BC}$, for other midpoints define similarly. Since $IB\perp IM_{CD}\perp M_{BC}M_{BD}$ and $IC\perp IM_{AB}\perp M_{BC}M_{AC}$, we use the lemma and we are done. $\blacksquare$

Let $O$ be the circumcenter of $ABCD$. First, note point $E$ is irrelevant: instead, phrase the problem as showing that lines $LM$, the external angle bisector of $\angle ACB$, and the external angle bisector of $\angle DBC$ concur. Now, define the points $N$ the midpoint of arc $\widehat{DABC}$, $P$ the midpoint of arc $\widehat{ADCB}$, $Q$ the midpoint of arc $\widehat{BAD}$, and $R$ the midpoint of arc $\widehat{ADC}$. Then the problem is equivalent to showing $BR\cap CQ, M, BN\cap PC$ are collinear. The problem is immediate by complex at this stage, but we choose to goof around instead. By immediate angle chase, we have $\angle POR = \angle BAC$. Similarly, $\angle QON = \angle BAC$. Then we can disregard $A$ and $D$ now and just focus on the information $\angle POQ = \angle NOR$ and $\angle POR = \angle QON = \angle BAC$. Now the problem is even easier to complex, but we continue to goof around. In particular, we just nuke the problem with moving points after noting $CPRB$ and $CQNB$ convex. We need to check the maps $Q\mapsto QC\mapsto QC\cap BR$ and $Q\mapsto N\mapsto NB\mapsto NB\cap PC=K\mapsto MK\cap BR$ are the same noting $Q\mapsto N$ is a projective map. At $Q=P$, we have $L = BR\cap CP$ but also $K = BN\cap PC = BR\cap CP$, so collinearity certainly holds. At $Q=M$, we get $L = BR\cap CQ = BR\cap CM$ and $K = BN \cap PC = BC\cap PC = C$ so collinearity certainly holds. At $Q=B$, we get $L = BR\cap CQ = BR\cap CB = B$ and $K=BN\cap PC = BM\cap PC$ so collinearity certainly holds. gg.

Solution. Let $I_A$ and $I_D$ be the incenters of $\bigtriangleup BAC$ and $\bigtriangleup BDC$, respectively, and $K$ the $E$-excenter of $\bigtriangleup BCE$. It's clear that $L = \overline{CI_D}\cap \overline{BI_A}$ and $M = \overline{DI_D}\cap \overline{AI_A}$. Since $M$ is the circumcenter of $\bigtriangleup CI_DB$ and $\bigtriangleup CI_AB$, we conclude that $BI_AI_DC$ is a cyclic quadrilateral. Consider $Q$ as the intersection point of lines $BI_D$ and $CI_A$ (i.e. the incenter of $\bigtriangleup BCE$) and construct $P = \overline{I_AI_D}\cap \overline{BC}$, $R = \overline{LM}\cap \overline{PQ}$. By La Hire's theorem, we know that $\angle QRM = 90^\circ$. Because $QBKC$ is a cyclic quadrilateral with diameter $\overline{QK}$, it suffices to show that $R$ lies on the circumcircle of $\bigtriangleup BQC$. Recall that $L$ and $R$ are inverses of each other with respect to $(BI_AI_DC)$, hence $MB^2=MR\cdot ML = MC^2$, which implies that $MB$ and $MC$ are tangent to the circumcircles of $\bigtriangleup BLR$ and $\bigtriangleup CLR$, respectively. Therefore $$\angle BRC = \angle BRM + \angle MRC = \angle MBL + \angle MCL = \angle BAC + \dfrac{1}{2}\left(\angle ABC + \angle BCD\right) = 90^\circ + \dfrac{\angle BEC}{2} = \angle BQC$$ as required. $\square$

Quite a nice problem - posting for storage and for some reason I didn't notice that $X,Y$ are also excenters Let $I_1,I_2,I_3$ be the incenters of $\triangle DBC, \triangle ABC, \triangle EBC$, respectively. Then notice that $M$ lies on $AI_1$ and $DI_2$ and $I = BI_2 \cap CI_1$, moreover, $L$ is the intersection of $BI_1$ and $CI_2$. Notice that $BI_2I_1C$ is cyclic as $$\angle I_2BI_1 = \frac{\angle ABD}{2} = \frac{\angle ACD}{2} = \angle I_2CI_1$$ Now, let $AI_1$ and $DI_2$ intersect the circle through $B,I_2,I_1,C$ at points $X$ and $Y$, respectively. Notice that $$\angle BI_1X = \angle BI_1M = \angle I_1BM = 90^{\circ} - \frac{\angle ADB}{2}$$ meaning that $X$ lies on the external angle bisector of $\angle BCE$ and $Y$ lies on the external angle bisector of $\angle EBC$ and consequently, $CX \cap BY = T$ is the $E$ excenter of $\triangle EBC$. Now, using Pascal's Theorem on $YI_1CXI_1B$ we get that $L,M,T$ are collinear and consequently, line $LM$ passes through the $E$ excenter of $\triangle EBC$. $\blacksquare$

Let S be out excenter. ∠MBS = 90 - ∠DBC/2 - ∠CBM = 90 - ∠DAB/2 = ∠DCB/2 = ∠LCB so if MB meets BLC at J then ∠LCB = ∠LJB = ∠JBS so LJ || BS. Let CM meet BLC at I. with same approach we can prove LI || CS. M is midpoint of arc BC so BC || IJ. So by homothety LS, CI and BJ meet at single point which is M. so S lies on LM as wanted. we're Done.

Let $I_E$ be the $E-$excenter of $\triangle BCE$. Fix $\triangle LBC$, and consider the below diagram.

If we let $\theta(X)$ and $\theta'(X)$ be the corresponding angles as shown in the figure for the point $X$ below $BC$, we want to show that $\theta(I_E) = \theta(M)$. Since $\theta(X)+\theta'(X)=\angle BLC$ is constant, it is equivalent to show $f(I_E)=f(M)$, where $f(X) := \frac{\sin \theta'(X)}{\sin \theta(X)}$. (Similar idea used in Russia 2019/11.6.) Let us calculate $f(X)$ in terms of $\alpha_2$ and $\beta_2$ as denoted in the diagram. As in the diagram, let $\alpha=\angle LBC=\tfrac12\angle ABC$ and let $\beta=\angle LCB=\tfrac12 \angle DCB$. By the Law of Sines, $$\frac{CX}{\sin \theta}=\frac{LX}{\sin (\beta+\beta_2)}, \quad \frac{BX}{\sin \theta'}=\frac{LX}{\sin(\alpha+\alpha_2)} \implies f(X) = \frac{BX}{CX}\cdot \frac{\sin(\alpha+\alpha_2)}{\sin(\beta+\beta_2)}.$$ We now verify $f(L)=f(M)$, by plugging in the corresponding appropriate values of $\alpha_2$ and $\beta_2$. For $X=M$: Then $\alpha_2=\beta_2=\tfrac12\angle BAC$. So $$\begin{align*} f(M) &= \frac{BM}{CM}\cdot \frac{\sin(\alpha+\alpha_2)}{\sin(\beta+\beta_2)} \\ &= 1\cdot \frac{\sin(\tfrac12 \angle ABC+ \tfrac12 \angle BAC)}{\sin(\tfrac12\angle DCB+\tfrac12\angle BAC)} \\ &= \frac{\sin(90^\circ - \tfrac12 \angle ACB)}{\sin(90^\circ - \tfrac12 \angle DBC)}= \frac{\cos(\tfrac12 \angle ACB)}{\cos(\tfrac12 \angle DBC)}. \end{align*}$$ For $X=I_E$: Then $\overline{BI_E}$ and $\overline{CI_E}$ bisect the angles external to $\angle CBD$ and $\angle BCA$ respectively. So $\alpha_2 = 90^\circ - \tfrac12 \angle DBC$ and $\beta_2=90^\circ-\tfrac12\angle ACB$. So $$\begin{align*} f(I_E) &= \frac{BI_E}{CI_E} \cdot \frac{\sin(\alpha + \alpha_2)}{\sin(\beta+\beta_2)} \\ &= \frac{BI_E}{CI_E} \cdot \frac{\sin(\tfrac12\angle ABC+90^\circ-\tfrac12\angle DBC)}{\sin(\tfrac12\angle DCB+90^\circ-\tfrac12\angle ACB)} \\ &= \frac{BI_E}{CI_E}\cdot \frac{\sin(90^\circ+\tfrac12\angle ABD)}{\sin(90^\circ+\tfrac12 \angle DCA)} \\ &= \frac{BI_E}{CI_E} \cdot 1 \end{align*}$$ since $\angle ABD=\angle DCA$. Now, we want to find the above ratio, and by Law of Sines, $$\frac{BI_E}{CI_E} = \frac{\sin \angle BCI_E}{\sin \angle CBI_E} = \frac{\sin(90^\circ-\tfrac12\angle ACB)}{\sin(90^\circ-\tfrac12\angle DBC)} = \frac{\cos(\tfrac12 \angle ACB)}{\cos(\tfrac12 \angle DBC)}.$$ The two ratios are equal, so we are done.

What's Pascal Let $K$ be the $E$-excenter of $\triangle BCE$. Let lines $BM$ and $CM$ intersect the circumcircle of $\triangle BCL$ again at $P$ and $Q$, respectively. By angle chasing, triangles $LPQ$ and $KBC$ are homothetic, so their center of homothety is $M$. Thus, $L$, $M$, and $K$ are collinear.

mofumofu wrote:

How do you use this problem when solving this problem?

Quite nice. Let $E_x$ be the $E$-excenter mentioned in the problem. Let $A' = BE_x \cap CL$ and $D' = CE_x \cap BL$. Notice by angle chasing that $\angle BA'C = \angle BD'C = \frac{1}{2} \angle BAC$. So let $\Omega = (BA'D'C)$, then it follows that $BM = BB, CM = CC$ on $\Omega$. Now by Brocard's theorem on $\Omega$, $E_xL$ is the polar of $A'D' \cap BC$, and $M$ is the pole of $BC$, so by La Hire's theorem, $E_xLM$ is concurrent.

I didn't expect this to work at all but oh well. Let $F$ be the intersection of lines $AB$ and $CD$. Let $M$, $N$, $P$ be the midpoints of arc $BC$ in circles $(ABCD)$, $(EBC)$ and $(FBC)$, and let $I_E$ be the $E$-excenter in $\triangle BCE$. A quick angle chase shows that $LP \parallel NI_E$, so we just need to show that $$\frac{MP}{MN} = \frac{PL}{NI_E} = \frac{PB}{NB},$$ where $PL = PB$ and $NI_E = NB$ by the incenter-excenter lemma. Now note that $$\angle NBM = \angle NBE - \angle MBA + \angle ABE = \angle EBC + \frac{1}{2} \angle BEC - \angle ABC - \frac{1}{2} \angle A = \frac{1}{2}\angle ABD.$$ Similarly, we can show that $\angle MBP = \frac{1}{2}\angle ABD$, so we are done by the angle bisector theorem.

Too easy for problem #3, because it is just trig bashing. Let $T$ be the excenter of $(B C E)$. Define point T such that T is intersection of external bisectors of angles $\angle EBC$ and $\angle ECB$. Let $ML\cap BC$=$K$, $BM\cap CT$=$X$ and $CM \cap BT$=$Y$ To prove that $M$, $K$, $T$ are collinear, we will prove the reverse of cheva theorem in triangle $BCT$, which we can easily prove by trigonometry. Firstly do the angles chasing, then use ratio lemma in triangle $BCT$ respect to lines $BM$ and $CM$ to find $CX$/$XT$ and $TY$/$YB$, then use ratio lemma in triangle $BLC$ to find $BK$/$KC$, then use sine theorem in triangles $CLM$ and $BLM$ to find $CL$/$BL$, using these we can find that reverse of cheva in triangle $BCT$ with respect to lines $BM$, $CM$ to respect point $K$, which finishes the problem.

won't post full sol cuz someone did already but i just really like how in the complex bash, usually i think intersection formula is scary but like the entire six term numerator cancels out if you do it right which is just so spicy i had to post about it

Miku3D wrote: Let $ABCD$ be a cyclic convex quadrilateral and $\Gamma$ be its circumcircle. Let $E$ be the intersection of the diagonals of $AC$ and $BD$. Let $L$ be the center of the circle tangent to sides $AB$, $BC$, and $CD$, and let $M$ be the midpoint of the arc $BC$ of $\Gamma$ not containing $A$ and $D$. Prove that the excenter of triangle $BCE$ opposite $E$ lies on the line $LM$. Let $F$ be the intersection of lines $AB$ and $CD$; $J$ be the excenter of $\triangle BCE$ opposite $E$ $M'$, $M''$ be the intersection of lines $BM$, $CM$ with line $LJ$ We see that the intersection of segment $LJ$ and circle $(ABCD)$ alway lies on arc $BMC$, so $M'$, $M''$ all lie on segment $LJ$ Since $BL$ is the internal bisector of $\angle ABC$, $CJ$ is the exterior bisector of $\angle ACB$, we have : $\, \,$$\angle LBM = \angle LBC + \angle CBM = \frac12 \angle ABC + \frac12 \angle CAB = \frac12 (180^\circ - \angle ACB) = \angle BCJ$ Similar, we have : $\angle CBJ = \angle LCM$, $\angle LBC = \angle JCM$, $\angle JCB = \angle JBM$\ Therefore : $\frac{M'J}{M'L} = \frac{BJ. \frac{\sin \angle MBJ}{\sin \angle BM'J}}{BL. \frac{\sin \angle MBL}{\sin \angle BM'L}} = \frac{BJ. \sin \angle LCB}{BL. \sin \angle JCB} = \frac{CJ. \sin \angle LBC}{CL. \sin \angle JBC} = \frac{M''J}{M''L}$ Combine with $M'$, $M''$ all lie on segment $LJ$, we see that $M \equiv M' \equiv M''$. Which means $J$ lie on line $ML$, done

Let $J$ be the $E$-excenter of $BCE$. Let $CE$ and $BE$ meet $\Gamma$ at $P$ and $Q$ respectively, and note that $PB = PA$ and $QC = QD$. On the other hand, let $BL$ and $CL$ meet $\Gamma$ at $X$ and $Y$ respectively, and note that $AX = XC$ an $BY = YD$. Note that $QY = MC = MB = PX$, hence $QP \parallel YX$, $PM \parallel XL$ and $MQ \parallel LY$. Hence triangles $QPM$ and $YXL$ are homothetic. Hence $ML$ goes through $YQ \cap XP$. Finally, Pascal on $PCYQBX$ yields $J$, $L$ and $YQ \cap XP$ are collinear. Thus we conclude all four points $J$, $L$, $YQ \cap XP$ and $M$ are collinear.

Let $X$ and $Y$ be the incenters of $\triangle ABC$ and $\triangle DBC$, respectively. Then $\{B, X, L\}$, $\{C, Y, L\}$, $\{A, X, M\}$, and $\{D, Y, M\}$ are all collinear triples. By the incenter-excenter lemma, $B$, $X$, $Y$, and $C$ all lie on the same circle $\omega$ centered at $M$. Let $K$ be the excenter of $\triangle BCE$ opposite $E$. Let $DY$ intersect $BK$ at $P$ and let $AX$ intersect $CK$ at $Q$. We now claim that $P, Q \in \omega$. To prove that we will show $P \in (BYC) = \omega$; the proof that $Q \in \omega$ is similar. We have $$\begin{align*} \angle BPY &= 180^\circ - \angle BDP - \angle DBP \\ &= 180^\circ - \frac12 \angle BDC - \angle DBC - \frac12 (180^\circ - \angle DBC) \\ &= 90^\circ - \frac12 \angle BDC - \frac12 \angle DBC \\ &= \frac12 \angle BCD \\ &= \angle BCY, \end{align*}$$ proving the claim. To finish, Pascal's theorem on $BPYCQX$ gives $L, M, K$ collinear, as desired.