Since $CA$ is tangent to $\omega$ and $AE \parallel XC$ so $\angle DAC= \angle AED= \angle DCX$. Therefore if $AD \cap XC=K$ then $\triangle KDC \sim \triangle KAC \; (\text{A.A})$. Hence, $KC^2=KD \cdot KA$. Similarly, we have $\angle KXD= \angle KAX$ so $KX^2=KD \cdot KA$. This follows $KC^2=KX^2=KD \cdot KA$ or $KD=KA$. Let $EH \cap \omega =H$. Since $EA \parallel XC$ so $$\angle EXC= \angle HEA= \angle HDA= 180^{\circ}- \angle HDK.$$ Therefore $H,K,D,X$ are concyclic. We obtain $\angle XHK= \angle XDK$. We also have $\angle XDK= \angle XHB (= \angle DBC+ \angle DCB)$ so $\angle XHB=\angle XHK$. From here we get $H,B,K$ are collinear. Note that $K$ is mispoint of $AD$ so $BK$ is the midline of triangle $XAC$ or $BK \parallel AC$. We get $\angle HKA \equiv BKA= \angle KAC= \angle ECX$. Since $HDKX$ is cyclic so $\angle HKA= \angle EXD$. Therefore, $\angle EXD= \angle ECX$. This follows that $\triangle EXD \sim \triangle ECX \; ( \text{A.A})$. Thus, $ED \cdot EC=EX^2$ or $EX$ is tangent to $\gamma$.

Here is my favorite solution out of the ones I got to grade. It was given by Viswanath and mathdebam. [asy][asy] size(8cm); defaultpen(fontsize(11pt)); pair A = dir(110); pair B = conj(A); pair C = 2/conj(A+B); pair X = 2*B-A; pair D = B*(2*B+3*A)/(2*A+3*B); pair E = B*(A+2*B)/(2*A+B); pair Y = A+X-C; draw(unitcircle, red+1); draw(circumcircle(X,B,C), dashed+blue); draw(A--C--B, red); draw(C--E, red); draw(X--A, heavymagenta); draw(E--X, blue); draw(A--D--B--E--cycle, orange+1); draw(circumcircle(B,X,E), dotted+heavygreen); draw(E--Y--X--C, grey+dashed); draw(B--Y, grey+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(210)); dot("$C$", C, dir(160)); dot("$X$", X, dir(-80)); dot("$D$", D, dir(220)); dot("$E$", E, dir(20)); dot("$Y$", Y, dir(Y-A)); [/asy][/asy] Using directed angles, $\angle BXC = \angle BDC = \angle BDE = \angle BAE$ so $\overline{AE} \parallel \overline{CX}$. Construct parallelogram $AYXC$. As $\angle BEY = \angle BEA = \angle BAC = \angle BXY$, quadrilateral $BEXY$ is cyclic. Thus $\angle XCB = \angle BYE = \angle BXE$ as desired. $\blacksquare$

Reflect $E$ about $B$ to get a parallelogram $EAE'X$. Then $CABE'$ is cyclic and the problem follows from angle chasing.

Nice problem, I have seen an extension Let $\omega$ be a circle and $C$ a point outside it; distinct points $A$ and $B$ are selected on $\omega$ so that $\overline{CA}$ and $\overline{CB}$ are tangent to $\omega$. The line passes through $C$ that intersects $(O)$ at $M,N$. Denote by $\gamma$ the circumcircle of triangle $CAN$. $AB$ cuts $\gamma$ again at $P$. $PM$ cuts $\gamma$ and $\omega$ again at $Q,R$, reps. Prove that $RA$ bisects $BQ$.

v_Enhance wrote: Let $\omega$ be a circle and $C$ a point outside it; distinct points $A$ and $B$ are selected on $\omega$ so that $\overline{CA}$ and $\overline{CB}$ are tangent to $\omega$. Let $X$ be the reflection of $A$ across the point $B$, and denote by $\gamma$ the circumcircle of triangle $BXC$. Suppose $\gamma$ and $\omega$ meet at $D \neq B$ and line $CD$ intersects $\omega$ at $E \neq D$. Prove that line $EX$ is tangent to the circle $\gamma$. Proposed by David Stoner $\angle DXC=\angle DBC=\angle BAD\ \ ... (1)$ let $\ell$ be the parallel to $AX$ from $D$, hence $(\ell, DB, DA, DX)=-1= (XD,XE, XC, XA)$ using $(1)$ we get $(\ell, DB, DA, DX) \sim (XD,XE, XC, XA)$ i.e. $\angle BDX=\angle EXA \Rightarrow EX$ is tangent to the circle $\gamma$.

buratinogigle wrote: Nice problem, I have seen an extension Let $\omega$ be a circle and $C$ a point outside it; distinct points $A$ and $B$ are selected on $\omega$ so that $\overline{CA}$ and $\overline{CB}$ are tangent to $\omega$. The line passes through $C$ that intersects $(O)$ at $M,N$. Denote by $\gamma$ the circumcircle of triangle $CAN$. $AB$ cuts $\gamma$ again at $P$. $PM$ cuts $\gamma$ and $\omega$ again at $Q,R$, reps. Prove that $RA$ bisects $BQ$. Since $\angle MBA= \angle MNA= 180^{\circ}- \angle ANC= \angle APC$ so $BM \parallel PC$. $BN \cap PC=K$. Since $CB$ is tangent to $\omega$ so $\angle CBK= \angle BAN= \angle NCP$. Hence, $\angle KC^2=KN \cdot KB$. Similarly, we have $\angle KPN= \angle KBP(= \angle NAC)$ so $KP^2=KN \cdot KB$. Therefore, $KP=KC$ or $K$ is the midpoint of $PC$. $BK \cap \gamma =L$. We have $\angle CLN= \angle NBP =( \angle NAC)$ so $CL \parallel BP$. Note that $K$ is the midpoint of $PC$, we obtain $K$ is the midpoint of $BL$. WLOG $M$ is between $R$ and $Q$. Since $BM \parallel PC$ so $\angle RPK= \angle RNB (= \angle RMB)$ so $RNKP$ is cyclic. We obtain $\angle PRK= \angle PNK$. On the other hand, $\angle PNK= \angle PRA(= \angle NCA+ \angle NAC)$. Thus, $\angle PRK= \angle PRA$. Hence, $R,K,A$ are collinear. Since $RNKP$ is cyclic so $\angle NKR= \angle NPR= \angle NLQ$. We get $KA \parallel LQ$. We also have that $K$ is the midpoint of $BL$ so $KA$ bisects $BQ$ or $RA$ bisects $BQ$.

Thank you very much dear shinichiman for your interest, here is my solution Let $BN$ intersect $\gamma$ again at $S$. We see $\angle PSN=\angle BAN=\angle NBS$ so $PS\parallel BC$. And $\angle CSN=\angle CAN=\angle ABN$ so $BP\parallel CS$. These imply $PBCS$ is paralleloram. This means the segments $PC$ and $SB$ intersect at midpoint $T$ of them. We have $\angle 180^\circ-\angle SQP=\angle PAS=\angle APC=\angle ANM=\angle QRA$. Hence, $SQ\parallel RA$. If $I$ is midpoint of $QB$ then $IT\parallel SQ$. Therefore, we will prove that $R,A,T$ are collinear. Indeed, apply Pascal theorem for six points $\left( \begin{array}{l} M\ B\ A \\ A\ R\ N \\ \end{array} \right)$ we deduce that $RA$ passes through $T$. We are done.

I solved this by first reflecting $E$ in the perpendicular bisector of $AX$ (which passes through $B$) to get $E'$, then proving $\angle BEX = \angle AE'B = \angle ABC$, and then finally angle chasing and power of point did it.

v_Enhance wrote: Let $\omega$ be a circle and $C$ a point outside it; distinct points $A$ and $B$ are selected on $\omega$ so that $\overline{CA}$ and $\overline{CB}$ are tangent to $\omega$. Let $X$ be the reflection of $A$ across the point $B$, and denote by $\gamma$ the circumcircle of triangle $BXC$. Suppose $\gamma$ and $\omega$ meet at $D \neq B$ and line $CD$ intersects $\omega$ at $E \neq D$. Prove that line $EX$ is tangent to the circle $\gamma$. Proposed by David Stoner Let $XD$ cut $(\gamma)$ at $F$; $F\neq D$. Since $CB$ is tangent to $\omega$ so $\angle CBD=\angle BFD$. And $C,D,B,X$ are concyclic so $\angle DBC=\angle DXC$. So $\angle DXC=\angle DFB$ so $CX\parallel BF$. Let $G$ be the intersection point of $BF$ and $CA$, $B$ is the midpont of $AX$ so $G$ is the midpoint of $AC$ (1). Since $\angle XCD=\angle ABD=\angle AEC$ so $AE \parallel CX$, so $AE\parallel BF$. In addition, $CA$ is tangent to $(\omega)$ so $\angle FAC=\angle FBA=\angle BAE$, $\angle AFG=\angle AEB$ so $\triangle AFG \sim \triangle AEB$ (2). From (1) and (2), we can prove that $\triangle CAF\sim \triangle XAE$ so $\angle CFA=\angle XEA$. Let $J$ be the intersection point of $CF,XE$ so $\angle JFA=\angle JEA$ so $J,A,E,F$ are concyclic. So $J\in (\omega)$. Apply Pascal theorem for six points $F,A,J,E,B,D$ so that $AF$ cuts $BE$ at a point on $CX$. Let $I$ be the intersection point of $AF,BE$ so $B$ is the midpoint of $EI$, $B$ is also the midpoint of $AX$ so $AIXE$ is a parrallelogram. So that $AF\parallel XE$ so $\angle EXA=\angle FAB=\angle XDB$ so $XE$ is tangent to $(\gamma)$

buratinogigle wrote: Nice problem, I have seen an extension Let $\omega$ be a circle and $C$ a point outside it; distinct points $A$ and $B$ are selected on $\omega$ so that $\overline{CA}$ and $\overline{CB}$ are tangent to $\omega$. The line passes through $C$ that intersects $(O)$ at $M,N$. Denote by $\gamma$ the circumcircle of triangle $CAN$. $AB$ cuts $\gamma$ again at $P$. $PM$ cuts $\gamma$ and $\omega$ again at $Q,R$, reps. Prove that $RA$ bisects $BQ$. $PC \cap QA \equiv U, QA \cap RB \equiv L, \angle APC=\angle ANM=\angle ABR \Rightarrow PC \parallel RB \Rightarrow \frac{PU}{UC}=\frac{BL}{LR}$, also $\angle PQA=\angle PCA \Rightarrow \triangle APC \cup U \sim \triangle ARQ \cup P \Rightarrow \frac{PU}{UC}=\frac{QP}{PR}=\frac{BL}{LR} \Rightarrow PL \parallel QB \Rightarrow QI=IB$

Beautiful but really easy for P3

A complex number solution can put this problem away in less than 1 hour if done correctly.

v_Enhance wrote: Let $\omega$ be a circle and $C$ a point outside it; distinct points $A$ and $B$ are selected on $\omega$ so that $\overline{CA}$ and $\overline{CB}$ are tangent to $\omega$. Let $X$ be the reflection of $A$ across the point $B$, and denote by $\gamma$ the circumcircle of triangle $BXC$. Suppose $\gamma$ and $\omega$ meet at $D \neq B$ and line $CD$ intersects $\omega$ at $E \neq D$. Prove that line $EX$ is tangent to the circle $\gamma$. Proposed by David Stoner darn , i should have tried this geo during ELMO, at least it would have fetch me a 7. solution = first angle chasing using cyc quad $ADBE,XBDC$ we have $\angle XCE=\angle ABD=\angle AED\Longrightarrow AE\parallel CX$ also $\angle XAC=\angle BEA$ and so $CXA\sim ABE$ thus $\frac{XC}{AB}=\frac{AC}{BE}=\frac{XA}{AE}$ using $XB=BA$ above is equivalent to $\frac{XA}{XC}=\frac{AE}{XB}$ and since $\angle AXC=\angle EAX$ we get $AXE\sim BXC$ and hence $\angle AXE=\angle BCX$ so that $EX$ is tangent to circle $\gamma$. so we are done

First note that $(AE) \parallel (CX)$ , indeed $XAE=BAE=BDE= \pi -BDC =BXC=AXC$ Call $Y$ the reflection of $E$ over $B$ ,we have $(XY) \parallel (AE) \parallel (CX) $ , implying $X,Y,C$ are collinear . Then $(YC) \parallel (AE) $ implies by Reim's Theorem that $(ABCY)$ is cyclic And thus $EXB=BAY=BCY=BCX$ , in other words $(EX)$ tangent to $ \gamma $

My solution is essentially similar to that of mathdebam. Let $T$ be the unique point such that $ACXT$ is a parallelogram. Note that $B$ is the midpoint of $AX$ and $CT$. Let $AT$ intersect the circumcircle of $ABE$ at $E'$. I shall show that $E'= E$. First, I shall show that $E'BXT$ is cyclic. To do this, note that $\angle AE'B = \angle CAX = \angle BXT$ (here is when I use the fact that CB and CA are tangents), which implies $\angle BXT + \angle BE'T = 180^{\circ}$, so $E'BXT$ is cyclic. Now, note that $\angle E'DB = \angle E'AB = \angle BXC = 180^{\circ} - \angle BDC$, so $E', D, C$ are collinear, implying $E'= E$. Now, note that $\angle EXB = \angle ETB = \angle BCX$, so $EX$ is tangent to the circumcircle of $BXC$, as desired.

and I use harmonic quadrilateral.

My solution: Let $O$ the midpoint of $CX$ and let $F\equiv OB\cap \omega$, $G\equiv OE\cap \omega$ $\Longrightarrow$ $\angle BXO=\angle BDE=\angle BGO$ hence $OBGX$ is cyclic. Let $\Phi$ be the inversion with center $O$ and radius $OC$. Since $\angle DAC=\angle DBA=\angle DCX$ $\Longrightarrow$ $OC^2=OD.OA=OB.OF=OG.OE$ $\Longrightarrow$ $\Phi(D)=A$, $\Phi(B)=F$ and $\Phi(G)=E$ $\Longrightarrow$ since $OBGX$ is cyclic we get $\Phi(OBGX)=\overline{XEF}$ hence we get $X$, $E$ and $F$ are collinear $\Longrightarrow$ since $CA\parallel OB$ we get $\angle CXB+\angle BCX=\angle CBA=\angle CAB=\angle BXO=\angle BFX+\angle BXF$ $\Longrightarrow$ $\angle BFX+\angle BXF=\angle CXB+\angle BCX$ but $\angle BXC=\angle BGO=\angle BXO$ hence $\angle BCX=\angle BXF$, hence we conclude that $EX$ is tangent to $\gamma$

No bary-bashes yet? We'll proceed from the viewpoint of a general triangle $EAB$, then find the conditions for $(BCDX)$ cyclic and $EX$ tangent to this circle; if they are the same we are done. Let $E=(1,0,0),A=(0,1,0),B=(0,0,1)$. Then $X=(0,-1,2)$. Now we'll compute $C$; it lies on the tangents at $A$ and $B$, hence $b^2x+e^2z=a^2x+e^2y=0$. This easily gives $C=(-e^2:a^2:b^2)$ which agrees with $C$ lying on the symmedian. Now let $D=(t:a^2:b^2)$ for some parameter $t$; plugging into the circumcircle equation immediately gives $t=\frac{-e^2}{2}$, so $D=(\frac{-e^2}{2}:a^2:b^2).$ Now let $(BCDX)$ have the standard form $e^2yz+a^2zx+b^2xy=(x+y+z)(ux+vy+wz)$. Plugging in $B$ gives $w=0$. Plugging in $X$ then gives $v=2e^2$, and now $D$ gives $u=4a^2$ (these are all nice linear equations in one variable, which allows us to win very quickly. So $(BXD)$ has equation \[e^2yz+a^2zx+b^2xy=(x+y+z)(4a^2x+2e^2y).\]Now plugging $C$ into this equation, we can quickly simplify to get \[2a^2+b^2=2e^2.\]So this is our condition on $\triangle EAB$. Now clearly the equation of $EX$ is $z=-2y$; we want to show that this intersects the circle exactly once. Plugging into the equation for $(BXD)$, we get \[e^2(-2y^2)-2a^2xy+b^2xy=(x-y)(4a^2x+2e^2y)\]\[e^2(-2y^2)-2a^2xy+b^2xy=4a^2x^2-2e^2y^2+(2e^2-4a^2)xy,\]which simplifies to \[x(4a^2x-y(2a^2+b^2-2e^2))=0.\]But since $2a^2+b^2-2e^2=0$ this has only one solution at $x=0$, so $EX$ is tangent to $(BCDX)$ at $X$.

Oh man, this is a really nice problem Note that $\angle AED = \angle ABD = \angle DCX$, so $AE\parallel CX$. Combining this with $AB = BX$ yields that $AYXC$ is a parallelogram, where $Y=AE\cap BC$. Hence $BC = BY = YA$, so $YX^2 = YB^2 = YE\cdot YA,$ meaning that $\triangle XYE\sim\triangle AYX$. But this in itself means that \begin{align*}\angle EXA &= \angle YXA - \angle YXE = \angle XAC - \angle YAX \\ &= \angle ABC - \angle AXC = \angle BCX,\end{align*}which establishes the desired tangency. $\blacksquare$

Inverse with $B$ then the result is trivial.

One of my favourite bary bash problems. Let's choose $ABE$ as referrence triangle since tangents are drawn to the circumcircle of $(ABE)$. $A=(1,0,0)$ $B=(0,1,0)$ and $E=(0,0,1)$. Since $B$ is midpoint of $AX$ we have $X=2B-A=(-1,2,0)$ It's obvious that $C=(a^2,b^2,-c^2)$. Intersecting $(AEB)$ and $CE$ we get $D=(2a^2,2b^2,-c^2)$.The circumcircle of $BCX$ has equation: $-a^2yz-b^2zx-c^2xy+(x+y+z)(2c^2x+4a^2z)=0$. If we put coordinates of $D$ in the equation of $(AEB)$ we will get $2a^2+b^2=2c^2$.Now let's assume that $EX\cap (AEB)=Y\ne X$.Here it's easy to get a contradiction. An easy extension: Prove that $EX=2EB$.

Solution: Let $AB\cap CE = N $ and $AD\cap CX = M $. We have $AE || CX $. Since, $P(C,N,D,E) $ forms a harmonic pencil, so, $M $ is the midpoint of $CX $. This gives $BM || AC $. So, $\angle BMD = \angle CAD = \angle ABD = \angle DCX $. Also, $\angle CXD = \angle CBD = \angle BAD $. So, $\Delta DXM$ ~ $\Delta XAM $. But, $\Delta XAM $ ~ $\Delta DEB $. This gives $\Delta DXM$ ~ $\Delta DEB $. Thus, $D $ is the center of the spiral similarity that takes $EB $ to $XM $, or, $BM $ to $EX $. So, $\angle EXD = \angle BMD = \angle DCX $ and hence the result follows.

The condition that $BC$ is tangent to $\omega$ is irrelevant, as the following solution shows. Let $Y$ be the reflection of $C$ in $B.$ Note that $ACXY$ is a parallelogram, because its diagonals bisect one another. By Reim's theorem for $\omega$ and $\gamma$ cut by $AX$ and $EC$, we infer that $AE \parallel XC.$ Therefore, $Y \in AE.$ Since $AC \parallel XY$, the converse of Reim's theorem for $\omega$ cut by $AX$ and $AY$ shows that $BXYE$ is cyclic. Therefore, $\measuredangle EXB = \measuredangle EYB = \measuredangle XCB$, so $EX$ is tangent to $\gamma.$

After inversion around $B$ with radius $BA=BX$ we get this problem. inverted wrote: Let $ABC$ be a isoceles triangle with $AB=AC$ and $M$ the midpoint of $BC$. A point $X$ is on $AC$ such that $AC=4AX$ and $MX \cap{AB} =Y$. Prove that $YX=YC$ - Which is trivial.

Let $E'$ be the reflection of $E$ over $B$, thus $E'XEA$ is a parallelogram. $\angle E'XB = \angle EAB = \angle EDB = \angle CXB$ so $\overline{X-E'-C}$ are collinear. $\angle BE'X = \angle BEA = \angle BAC$ so $BE'CA$ is cyclic. $\angle BXE = \angle BAE' = \angle BCE' = \angle BCX$. Thus $EX$ is tangent to $(BCX)$ at $X$.

Projective best geo ever Let $F$ be a point on $\gamma$ such that $DF \parallel AB$ and let $DF \cap AB=P_{\infty}$ and $AD \cap \gamma=G \ne D$ and $XE \cap AC=H$. By angle chase we have that: $$\angle GDF=\angle DAB=\angle DBC=\angle DGC \implies CG \parallel DF \implies CG \cap DF \cap AB=P_{\infty}$$By harmonic chase we have: $$-1=(A, X; B, P_{\infty}) \overset{D}{=} (G, X; B, F) \overset{P_{\infty}}{=} (D, X; B, C)$$Now by the conditions of the problem we have $ADBE$ harmonic, let $EX \cap \gamma=X'$. By harmonic chase: $$-1=(A, B; D, E) \overset{H}{=} (C, B; D, X')=(X', D; B, C)=(D, X'; B, C) \implies X=X'$$Meaning that $EX$ is tangent to $\gamma$ thus we are done.

Let $\omega$ and $(AXC)$ meet again at $F$, $H = AD \cap BF$, and $I = AB \cap DF$. By Radical Axes, we know $AF, BD, CX$ concur at some point $G$. Pascal's on $AADBBF$ yields $H \in CG$, so $G, C, H, X$ are all collinear. Now, Brocard's implies $GH$ is the polar of $I$ wrt $\omega$. Because $X \in GH$, we have $$-1 = (X, I; A, B) \overset{F}{=} (XF \cap \omega, D; A, B).$$But $ADBE$ is clearly harmonic, so $E = XF \cap \omega$, i.e. $X, E, F$ are collinear. Claim: $CAFX$ is harmonic. Proof. Let the $C$-symmedian of $ACX$ meet $(ACX)$ again at $F'$. Now, observe $$\angle CAB = CAX = \angle CF'X = \angle AF'B$$and $$\angle CBA = \angle CAB = \angle AF'B = \angle BF'A$$so $CA$ and $CB$ are tangent to $(ABF')$. But there exists exactly one circle tangent to $CA$ at $A$ and $CB$ at $B$, so $(ABF')$ and $\omega$ coincide, which gives $F' = F$, as desired. $\square$ Isogonality implies $\angle ACF = \angle BCX$, and we have $$\angle AFC = \angle AXC = \angle BXC$$so $CAF \sim CBX$. Hence, we know $$\angle EXC = \angle FXC = 180^{\circ} - \angle FAC = 180^{\circ} - \angle XBC$$which finishes. $\blacksquare$ Remarks: I guessed $CAFX$ would be harmonic because the setup of this question wrt $\triangle ACX$ is just like Swiss TST 2020/3. (Moreover, the second intersection between the $C$-symmedian of $ACX$ and $(ACX)$ must have some significance, and $F$ could provide said significance, as it also lies on $\omega$.) Also, $$\angle CFX = \angle CAX = \angle CAB = \angle AFB$$is an easier way to prove $CAFX$ is harmonic.

Same as #3; posting for storage. Since $$\measuredangle BAF=\measuredangle BDE=\measuredangle BDC=\measuredangle BXE$$so $\overline{AE}\parallel\overline{CX}$ and $AB=BX,$ there is a unique point $F$ such that $ACXF$ is a parallelogram. Also, $BEFX$ is cyclic as $$\measuredangle FXB=\measuredangle CAX=\measuredangle AEB=\measuredangle FEB.$$Since $\measuredangle EXB=\measuredangle EFB=\measuredangle XCF,$ we are done. $\square$

Similar to #5

Wictro wrote: After inversion around $B$ with radius $BA=BX$ we get this problem. inverted wrote: Let $ABC$ be a isoceles triangle with $AB=AC$ and $M$ the midpoint of $BC$. A point $X$ is on $AC$ such that $AC=4AX$ and $MX \cap{AB} =Y$. Prove that $YX=YC$ - Which is trivial. Or you could construct $C’$ Such that $ACXY$ Is a parallelogram and invert around $B$ With radius $BC$ which would also result in a trivial problem. I dont know how i missed the proof in #3 even though I was trying to prove $AE \parallel CX$ but after inverting everything became super easy. Way too easy for an ELMO #3

BVKRB- wrote: Way too easy for an ELMO #3 So easy that it was used as IMO p4 two years later

BVKRB- wrote: Way too easy for an ELMO #3 It's worth noting that in 2015 ELMO was a 5 problem test (instead of a 6 problem test split into two days). So this problem was a medium-difficulty question, and not a hard-difficulty problem like in most years. (Also, I believe that year the ELMO committee was explicitly trying to make the test more accessible to MOP rookies, but take that with a grain of salt as I could be misremembering.)

MatteD wrote: BVKRB- wrote: Way too easy for an ELMO #3 So easy that it was used as IMO p4 two years later I noticed the similarity, the same parallelogram picture and tangency, but was too lazy to see the P4 again. djmathman wrote: BVKRB- wrote: Way too easy for an ELMO #3 It's worth noting that in 2015 ELMO was a 5 problem test (instead of a 6 problem test split into two days). So this problem was a medium-difficulty question, and not a hard-difficulty problem like in most years. (Also, I believe that year the ELMO committee was explicitly trying to make the test more accessible to MOP rookies, but take that with a grain of salt as I could be misremembering.) Ohh! So that’s the reason! Thank you for informing So I never solved an actual P3

I'm gonna steal Evan's diagram: [asy][asy] size(8cm); defaultpen(fontsize(11pt)); pair A = dir(110); pair B = conj(A); pair C = 2/conj(A+B); pair X = 2*B-A; pair D = B*(2*B+3*A)/(2*A+3*B); pair E = B*(A+2*B)/(2*A+B); pair Y = A+X-C; draw(unitcircle, red+1); draw(circumcircle(X,B,C), dashed+blue); draw(A--C--B, red); draw(C--E, red); draw(X--A, heavymagenta); draw(E--X, blue); draw(A--D--B--E--cycle, orange+1); draw(circumcircle(B,X,E), dotted+heavygreen); draw(E--Y--X--C, grey+dashed); draw(B--Y, grey+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(210)); dot("$C$", C, dir(160)); dot("$X$", X, dir(-80)); dot("$D$", D, dir(220)); dot("$E$", E, dir(20)); dot("$Y$", Y, dir(Y-A)); [/asy][/asy] Define $Y=AE\cap CB$. Claim: Quadrilateral $BEYX$ is cyclic. Proof. First, note that $AY\parallel CX$ since $\measuredangle BAY=\measuredangle BAE=\measuredangle BDE=\measuredangle BDC=\measuredangle BXC$. Since $B$ is the midpoint $AX$, we discover that $ACXY$ is a parallelogram. Thus $\triangle BAC\cong \triangle BXY$ which implies that $YB=YX$ as $CA=CB$ by tangency. Now, observe $\measuredangle BEY=\measuredangle BEA=\measuredangle BAC=\measuredangle CBA=\measuredangle YBX=\measuredangle BXY$ implying the result. $\square$ To finish, just note $$\measuredangle XCB=-(\measuredangle CBX+\measuredangle BXC)=\measuredangle XBC+\measuredangle CXB=\measuredangle XBY+\measuredangle CDB=\measuredangle XEY+\measuredangle EDB=\measuredangle XEA+\measuredangle EAB=-(\measuredangle AEX+\measuredangle XAE)=\measuredangle AXE=\measuredangle BXE$$which finishes. $\blacksquare$

Complex numbers sketch: Define $(ABD)$ to be the unit circle. We have $c=2ab/(a+b)$ and $x=2b-a$. Find $d$ using $\angle DCB=\angle DXB$, i.e. $(c-b)/(x-b) \div (c-d)/(x-d)$ is real. This becomes $b(2b-a-d)/(2ab-ad-bd)$ is self-conjugate. Upon expansion we can solve for $d$ and we find $d=b(3a+2b)/(2a+3b)$. Now we find $E$ by using $|e|=1$ and $C\in \overline{ED}$, so $d+e-de\overline{c}=c$, i.e. $e=(c-d)/(1-d\overline{c})$. This is a pretty easy expansion and we can solve for $e$. Finally check $\angle EXB=\angle BDX$ which is a pretty short computation.