Let ABCD be a cyclic convex quadrilateral and Γ 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 Γ not containing A and D. Prove that the excenter of triangle BCE opposite E lies on the line LM.
Problem
Source: 2021 APMO P3
Tags: geometry, circumcircle, APMO
09.06.2021 09:48
09.06.2021 10:25
dame dame
09.06.2021 10:41
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.)
Attachments:
APMO 2021 P3 makeshift diagram.pdf (492kb)
09.06.2021 10:47
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.
Attachments:

09.06.2021 11:09
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.
09.06.2021 11:18
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
09.06.2021 11:49
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 .
09.06.2021 11:51
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.
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09.06.2021 12:02
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
09.06.2021 12:37
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
09.06.2021 14:50
Adding onto post 3, see here... this problem is not very new it seems
09.06.2021 14:51
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.
09.06.2021 17:23
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.
09.06.2021 18:23
Also pretty much the exact same problem as this one from the 2018 Romania TST
09.06.2021 20:04
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.
09.06.2021 21:19
11.06.2021 22:58
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
12.06.2021 05:25
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.
12.06.2021 21:49
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 BQCas required. \square
13.06.2021 05:16
Attachments:

30.06.2021 00:08
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_1Now, 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
07.02.2022 10:24
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.
10.03.2022 03:16
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.
16.03.2022 08:22
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.
22.02.2023 10:27
mofumofu wrote:
How do you use this problem when solving this problem?
05.03.2023 19:43
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.
26.06.2023 15:29
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.
14.08.2023 16:34
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.
11.03.2024 00:43
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
25.06.2024 08:22
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
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27.07.2024 16:39
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.
18.08.2024 14:02
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.