First, note that $\angle EBA = \angle ECA = \angle SCR = \angle SBR = \angle ABR$, so it follows that $AB$ is the angle bisector of $\angle EBR$. Similarly $AC$ is the angle bisector of $\angle SCD$. It follows that since $\triangle EBR$ is isosceles, we have $ES = SR$ and $\angle SER = \angle SRE$. Now, note that \begin{eqnarray*} \angle BRC &=& \angle BSC \\ &=& 180 - \angle ESB \\ &=& \angle EBS + \angle SEB \\ &=& \angle EBS + \angle SRB \\ &=& \angle EBS + \angle SCB \\ &=& \frac{\text{arc } AB}{2} \\ &=& \angle RCB \end{eqnarray*} Thus $\triangle BRC$ is isosceles and therefore $BC = BR = EB$. EDIT : Oops I did not finish my copy and paste. Here is the rest. Now, note that $\angle DEC = \angle RBC = \angle RSC$, so it follows $DE \| RS$. But then $\angle DER = \angle ERS = \angle REL$, implying $KE$ bisects $\angle DEL$. Since $DK$ is an angle bisector, it follows $K$ is the incenter of $\triangle DEL$ so $2 \angle ELK = \angle ELD$ so it suffices to show $\angle ELD = \angle BCD$. Note that $\angle ELD = \frac{\text{arc ED} + \text{arc BC}}{2}$ and $\angle BCD = \angle ECD + \angle ECB = \frac{\text{arc ED}}{2} + \angle BEC = \frac{\text{arc ED}}{2} + \frac{\text{arc BC}}{2}$ so $\angle ELD = \angle BCD$ and therefore we are done. EDIT2 : Fixed a latex error.

Nice. For reference, here's a diagram. [asy][asy]/* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(7cm); real lsf=0.9001; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair B = (2.0, -3.0); pair C = (10.0, -3.0); pair S = (4.76, 2.38); pair R = 2*foot(C,relpoint(Line(circumcenter(B,S,C),B,lisf),0.5-10/lisf),relpoint(Line(circumcenter(B,S,C),B,lisf),0.5+10/lisf))-C; pair A = IntersectionPoint(Line(B,S,lisf),Line(C,R,lisf)); path w = circumcircle(A,B,C); pair D = IntersectionPoint(w,Line(B,R,lisf),0); pair E = IntersectionPoint(w,Line(C,S,lisf),0); path CE = C--E; path BD = B--D; pair L = IntersectionPoint(BD,CE); path ER = E--R; pair K = IntersectionPoint(ER,Line(L,bisectorpoint(E,L,D),lisf)); /* Draw objects */ dot(B); dot(C); dot(S); dot(R); dot(A); draw(A--B, rgb(0.6,0.2,0.0)); draw(B--C, rgb(0.6,0.2,0.0)); draw(C--A, rgb(0.6,0.2,0.0)); draw(circumcircle(B,C,S), rgb(0.6,0.6,0.0) + linewidth(1.0) + linetype("4 4")); draw(w, rgb(0.0,0.4,0.4) + linewidth(1.2)); dot(D); dot(E); draw(CE, rgb(0.0,0.6,0.2)); draw(BD, rgb(0.0,0.6,0.2)); dot(L); draw(ER, rgb(0.8,0.2,0.0)); dot(K); draw(E--D, rgb(0.4,0.4,0.4) + linewidth(1.0) + linetype("4 4")); draw(D--K, rgb(0.4,0.4,0.4) + dotted); /* Label points */ label("$B$", B, lsf*dir(225)); label("$C$", C, lsf*dir(315)); label("$S$", S, 2*lsf*dir(95)); label("$R$", R, 2*lsf*dir(85)); label("$A$", A, lsf*dir(75)); label("$D$", D, lsf*dir(45)); label("$E$", E, lsf*dir(135)); label("$L$", L, lsf*dir(85)); label("$K$", K, lsf*dir(85)); [/asy][/asy] The "main configuration" in the problem is that $BE = BR = BC$, as dinoboy showed above. You can see this as follows: by angle chasing it is not hard to show that $\angle EBR = 2\angle ECR$. So we see that $C$ lies on the circle at $B$ with radius $BE = BR$. Note that $R$ is the incenter of $\triangle ECD$, but more-over, $B$ serves as the infamous center in what I've often colloquially called Fact 5.

From cyclic quadrilaterals $EACB$ and $RSBC$ we have $\angle EBA=\angle RCS=\angle RBS$. there fore $BS$ is the perpendicular bisector of $ER$ and $\angle ESR=\angle SRE$. Since $\angle BCE=\angle BDE=\angle LRS$ by cyclic quads, $SR || ED$. Then $\angle LER=\angle SRE=\angle RED$, and we have $K$ the incenter of $DLE$. Additionally, $\angle LCB=\angle SRL=\angle CEB=\angle BDC$, therefore $BE=BC$ and $\angle BDE=\angle BDC$, implying $\triangle ELD\sim\triangle BCD$ and $\angle ELD=\angle BCD$. This finishes as $\angle ELK$ is half of $\angle ELD$.

Although sjaelee's is the most elegant solution, here's another way to end: $BE=BR\implies \angle BAE=\angle BAC\iff BE=BC$, i.e. inversion of pole $B$ and power $BC^2$ sends the circle $\odot(ABC)$ to line $CE$, i.e. $ASLD$ is cyclic, or $\angle SLD=\angle BCD$ (both have the same supplement, $\angle SAD$). With the previously shown $K$ incenter of $\triangle DEL$, we are done. Best regards, sunken rock

when will ELMO shortlist be out

easily get $AE=AR=AD$ and $R$ is incenter of the triangle $ECD.$ So, $K$ is incenter of the triangle $ELD$ and \[ \angle ELK = \frac{ELD}{2} =\frac{\overarc{BD}}{2}= \frac{\angle BCD}{2} .\]

Short solution: $\angle EBR=\angle EBA+\angle SBR=\angle ECR+\angle ECR=2\angle ECR$ and $BE=BR$, so $B$ is the circumcentre of $\triangle ERC$. So $\angle BEC=\angle BCE \implies \angle LDC=\angle LDE$. Also $\angle DCR=\angle ECR$ (both equal $\angle SBR$) so $R$ is the incentre of $\triangle EDC$. Hence $ER$ bisects $\angle DEL$ and since we are given that $DK$ bisects $\angle EDL$, $K$ is the incentre of $\triangle DEL$. So $\angle ELK=\frac{1}{2} \angle ELD$. Now from earlier observations $BE=BC$ so $\angle EDL=\angle BDC$ and we also have $\angle DEL=\angle DBC$ so $\triangle DEL ~ \triangle DBC$. Thus $\angle ELD=\angle BCD$. So $\angle ELK = \frac{1}{2} \angle BCD$.

Note that $\widehat{EBA}=\widehat{ECA}=\widehat{EBD}$, hence $BA$ is the bisector of $\widehat{EBD}$. As $BE=BR$, we get that $A$ is on the perpendicular bisector of $ER$. But $A$ is also on the perpendicular bisector of $DE$ due to $BA$ being bisector, hence $AR=AD$. Therefore $\widehat{BRC}=\widehat{ARD}=\widehat{ADR}=\widehat{BCR}$ i.e. $BC=BR$. A simple angle chasing yields now that $ER$ is the bisector of $\widehat{DEL}$ so $K$ is the incenter of $\triangle{DEL}$, whence $\widehat{ELK}=\dfrac{1}{2}\widehat{ELD}=\dfrac{1}{2}\widehat{BCD}$.

$B$ is center of spiral send $S->E$ and $R->A$ so $\frac{BE}{BS}=\frac{BA}{BR}$ then $ {BR}^2=BA.BS ->\widehat{ASR}=\widehat{BRC}->\widehat{BCR}=\widehat{BRC}->BE=BR=CR $. A simple angle chasing to prove $K$ is the incenter of $\triangle{DEL}$

Could someone explain why it happens to be that $K$ lies on the angle bisector of $\angle LDE$? I seem to understand everything else from the solutions except that, and I don't see any solution explaining it. (I see how $K$ lies on the bisector of $\angle LED$ though).

@above it’s given in the question

Here's an angle chase that's really easy to find but quite long. Claim: $K$ is the incenter of $\triangle ELD.$ Proof: Note that $\angle BDK = \angle EDK$ is given. Now we will show that $\angle LEK = \frac{1}{2} \angle LED.$ But $\angle LED = \angle CED = \angle CBD = \angle CBR = \angle CSR,$ so it suffices to show that $ES=SR.$ But now $\angle EBA = \angle ECA = \angle SCR = \angle SBR$ and $EB=BR,$ which implies the desired result. Claim: $R$ is the incenter of $\triangle CDE.$ Proof: Note that $\angle CER = \angle DER,$ so it suffices to show that $\angle RCE = \angle DCR.$ But $\angle ECA = \angle SCR = \angle SBR = \angle ABD = \angle ACD.$ This implies that $B$ is the arc midpoint of $\widehat{EC},$ so $\angle BDC = \angle ECB,$ so $\triangle LBC = \triangle CBD.$ Now notice that $\angle ELK = \frac{1}{2} \angle ELD = \frac{1}{2} \angle BLC = \frac{1}{2} \angle BCD,$ as desired.

Note that $$\angle ABD = \angle SBR = \angle SCR + \angle ECA$$so $AE=AD$ and because $BE=BR$ then $SE=SR$. Now, $$\angle BRC = \angle BSC = \angle ASE = \angle RSA = \angle BCR$$which means that $BR=BC=BE$. We see that now because $\angle DCR = \angle RCE$ and $\angle CDR = \angle RDE$ then $R$ is the incenter of $\triangle CDE$ which means that $\angle DEK = \angle KEL$ and so similarly, $K$ is the incenter of $\triangle DEL$. As a result, $2\angle ELK = \angle ELD$ and now since $\angle EDL = \angle EDB = \angle BDC$ and $\angle DEL = \angle DEC = \angle DBC$, then we see that $\triangle DEL \sim \triangle DBC$ meaning that $\angle DLE = \angle DCB$ so we are done.

Observe that $\angle EBA = \angle ECA = \angle SCR = \angle SBR = \angle ABD$. This means $\angle EBR = 2 \angle ECR$ and so $B$ is circumcenter of $(ERC)$. Since $BE = BC$ and $AD = AE$, by Fact $5$, we see that $R$ is incenter of $\triangle CDE$. So since $KE$ is angle bisector of $\angle CED$, we see that $K$ is the incenter of $\triangle ELD$. Since now, $2 \angle ELK = \angle ELD$, it suffices to show $\angle ELD = \angle BCD$. But $\angle BCD = \angle BCA + \angle ACD = \angle BCR + \angle ACE = \angle BRC + \angle ACE = \angle ELD$ and we are done. $\blacksquare$

Notice $$\angle EBA = \angle ECA = \angle SCR = \angle SBR = \angle ABD = \angle ACD$$so $A$ is the midpoint of arc $DE$. Because $BE = BR$ and $BA$ bisects $\angle EBR$, we know $BA \perp ER$ so $AEBR$ is a kite. Thus, $$\angle ECB = \angle EAB = \angle RAB = \angle CAB = \angle CEB$$implying $BC = BE = BR$. Claim: $R$ is the incenter of $CDE$. Proof. Observe $R \in CA$, which bisects $\angle DCE$. In addition, $$\angle CDR = \angle CDB = \angle CEB = \angle ECB = \angle EDB = \angle EDR$$implying the desired conclusion. $\square$ Claim: $\angle ELD = \angle BCD$. Proof. Notice $$\angle ELD = \angle LDC + \angle LCD = \angle BDC + \angle ECD$$$$= \angle BEC + \angle ECD = \angle BCE + \angle ECD = \angle BCD$$as desired. $\square$ Now, it suffices to show $LK$ bisects $\angle ELD$. Claim: $K$ is the incenter of $ELD$. Proof. First, notice $$\angle DEK = \angle DER = \angle REC = \angle KEL.$$We're also given that $K$ is on the bisector of $\angle BDE$, which is equivalent to $\angle LDE$. The result follows easily. $\square$ The previous claim is obviously sufficient, so we're done. $\blacksquare$ Remark: This problem is just angle chase spam and recognizing incenters. In my opinion, the hardest part is probably noticing $\angle ELD = \angle BCD$. Note: $E$ lies on $(AHB)$ by orthocenter reflection relationships.

Note that $\measuredangle RBS=\measuredangle RCS=\measuredangle ABE,$ which means that $AB$ is the angle bisector of $\angle DBE$. As $BR=BE$, $AB$ is the perpendicular bisector of $RE$. Hence, $$\measuredangle LED=\measuredangle CBR= \measuredangle LSR=2\measuredangle LER,$$therefore $ER$ is the angle bisector of $\angle LED$. Hence, $K$ is the incenter of $\triangle LED$. Also note that $$\measuredangle BES=\measuredangle SRB=\measuredangle SCB\implies \measuredangle EDB=\measuredangle BDC$$and as $\measuredangle LED=\measuredangle CBD$, we get that $\triangle LED\sim\triangle CBD$. Hence, $\measuredangle DCB=\measuredangle DLE=2\measuredangle KLE$. We are done.

Difficult, but very nice, and of course uses the classic method of transferring angels between arcs of different circles. The problem can be split up into three main parts(note that angles are not directed in this sol): Claim: $\angle EBA=\angle ABD=\angle ECA=\angle ACD$. Proof: We have that $\angle EBA=\angle ECA=\angle ABD=\angle ACD$. Claim: $\triangle BRC$ and $\triangle BEC$ are isosceles. Proof: If we prove that $\triangle BRC$ is isosceles, in the process, we will establish that $\triangle BEC$ is isosceles. Notice that $$\angle BRC=\angle BSC=\angle BES+\angle BSE=\angle BRS+\angle ECA=\angle ECA+\angle BCE=\angle BCA=\angle BCR.$$ Of course, then because $\angle ECA=\angle EBA,$ we have that $\triangle BEC$ is isosceles. Claim: $SR||ED$(i.e. $K$ is the incenter of $\triangle ELD$) Proof: We have that $\angle LED=\angle DBC=\angle RSL.$ Now the result is easy to see using arcs and the fact that $\angle BEC=\angle BCE$.

Claim: $\overline{AB}$ bisects $\angle DBE.$ Proof. Notice $$\angle ABE=\angle ACE=\angle RBS.$$$\blacksquare$ Similarly, $\overline{AC}$ bisects $\angle DCE.$ Claim: $BE=BR=BC.$ Proof. Since $BE=BR,$ $\overline{AB}$ also bisects $\angle ESR.$ Hence, $$\angle RCB=\angle RSA=\angle ASE=\angle BRC.$$$\blacksquare$ Claim: $R$ is the incenter of $\triangle CDE.$ Proof. It suffices to prove $\overline{DB}$ bisects $\angle CDE.$ Indeed, $$\angle EDB=\angle ECB=\angle BEC=\angle BDC.$$$\blacksquare$ Claim: $K$ is the incenter of $\triangle DEL.$ Proof. We know $\overline{ER}$ bisects $\angle LED$ and $\overline{DK}$ bisects $\angle EDL.$ $\blacksquare$ Claim: $\angle ELD=\angle BCD.$ Proof. We see $$\angle ELD=\tfrac{1}{2}(\widehat{ED}+\widehat{BC})=\tfrac{1}{2}(\widehat{ED}+\widehat{EB})=\angle BCD.$$$\blacksquare$ Then, $2\angle ELK=\angle ELD=\angle BCD,$ as desired. $\square$

very nice problem! Claim 1:-$\overline{AB}$ bisects $\angle{DBE}$ and $\overline{AC}$ bisects $\angle{DCE}$ Proof. We notice that due to $\widehat{AE}$ we have $\angle{ACE}=\angle{ABE}$. Also $\angle{ACE}=\angle{ACS}=\angle{ABR}=\angle{ABD}$ hence we have $\overline{AB}$ bisects $\angle{DBE}$ in similar fashion we have $\overline{AC}$ bisects $\angle{DCE}$ $\square$ Claim 2:- $ED \parallel RS$ Proof. $\angle{LRS}=\angle{ECB}$ and since $BEDC$ is a cyclic quadrilateral we have $\angle{ECB}=\angle{EDB}$, which gives $ED \parallel RS$ $\square$ Claim 3:-$K$ is incenter of $\triangle{ELD}$ Proof. Since $\triangle{EBR}$ is isosceles and $\overline{AB}$ is angle bisector of $\angle{EBR}$ we have $S$ to be lying on $\overline{AB}$ which gives $\angle{SER}=\angle{SRE}=\angle{DER}$ , hence $KE$ is the angle bisector of $\angle{DEL}$ but also we have $KD$ bisecting $\angle{EDL}$ , we have that $K$ is the incenter of $\triangle{ELD}$ $\square$ Claim 4:- $R$ is the incenter of $\triangle{EDC}$ Proof.Since $ER$ bisects $\angle{CED}$ and $CR$ bisects $\angle{DCE}$ we have $R$ to be the incenter of $\triangle{EDC}$ Claim 5:- $\angle{ELD}=\angle{BCD}$ Proof. We denote $\angle{ECB}=2\beta$ and $\angle{DBC}=2\alpha$. Now Since $R$ is incenter of $\triangle{EDC}$ we have from Incenter-Excenter Lemma that $BE=BC$. Using this we have $\angle{BEC}=2\beta=\angle{EBD}=180^{\circ}-2(\alpha+\beta)$ and $\angle{EBD}=\angle{ECD}$ , so we have $\angle{BCD}=180^{\circ}-2(\alpha+\beta)$ also we have $\angle{SRB}=2\angle{\beta}$ and $\angle{CSR}=2\angle{\alpha}$ , so we have $\angle{SLR}=\angle{ELD}=180^{\circ}-2(\alpha+\beta)$ which gives $\angle{ELD}=\angle{BCD}$ $\square$ now since $KL$ is angle bisector of $\angle{ELD}$ we have $\angle{ELK}=\frac{1}{2}\cdot \angle{BCD}$ $\blacksquare$

We first make the following claims: Claim: $CA$ bisects $\angle DCE$ and $BA$ bisects $\angle EBR$. Proof: There is a spiral similarity at $C$ sending $SA \mapsto RD$. Hence $\angle ACS = \angle DCR$, and likewise the spiral similarity at $B$ sending $ES \mapsto AR$ gives us $\angle EBS = \angle ABR$. $\square$ Claim: $BE = BR = BC$. Proof: We have that \[\angle BCR = 180^{\circ} - \angle BEA = \angle ARB = \angle BRC, \]where $\angle BEA = 180^{\circ} - \angle ARB$ as $BA$ bisects $\angle EBR$. $\square$ We now observe that as \[\angle EDB = \angle ECB = \angle CEB = \angle CDB, \]that $R$ is the incenter of $\triangle CED$, so $DB$ bisects $\angle EDC$, as desired. $\blacksquare$

Claim 1. Point $K$ is the incenter of triangle $DEL$. Proof. In cyclic quadrilaterals $ACBE$ and $SRCB$ we have \[\angle ABD = \angle SBR = \angle SCR = \angle ECA = \angle EBA\]so $A$ is the midpoint of arc $DE$, that is $AE = AD$. Hence $\overline{CA}$ is an angle bisector of $\angle ECD$. In addition, $\overline{AB} \perp \overline{ER}$ so by congruent triangles if follows $AE = AR$. In conclusion, $AE=AR=AD$. By Incenter/Excenter Lemma $R$ must be the incenter of $\triangle CDE$ as $R$ lies on segment $CA$. As a result, $\overline{EK}$ is an angle bisector of $\angle DEL$. Furthermore, by assumption, $\overline{DK}$ is an angle bisector to $\angle LDE$, so $K$ is the incenter of $\triangle DEL$. Claim 2. Quadrilateral $ADLS$ is cyclic. Proof. As previously stated, $AD=AR$, so \[\angle ADL = \angle DRA = \angle BRC = \angle BSC = 180^{\circ} - \angle LSA\]proving our claim. Finish. In conclusion, due to the fact that $K$ is the incenter of $\triangle DEL$, it follows that \[ \angle ELK = \frac{1}{2} \angle SLD.\]But as $ADLS$ is cyclic, \[ \angle BCD = 180^{\circ} - \angle DAB = 180^{\circ} - \angle DAS = \angle SLD.\]Combining these two we see \[ \angle ELK = \frac{1}{2} \angle BCD\]and were done!

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