We use directed angles throughout the solution. Let $M$ be the midpoint of $\overline{BC}$. Now note that $\angle AFH = \angle AEH = \angle AXH = 90$. Thus $A,E,H,F,X$ are concyclic. Now $AX \parallel EF$ implies that $AXFE$ is an isosceles trapezoid. Claim: $MEC_1C$ is cyclic. Proof: Firstly, $\angle AEM = \angle CEM = \angle MCE = \angle BCA$. Also $\angle AEX = \angle AXE + \angle EAX = \angle AFE + \angle EFX = \angle BCA + \angle AEF = \angle BCA + \angle CBA$ $\angle XEM = \angle AEM + \angle XEA = \angle BCA + \angle ACB + \angle ABC =\angle ABC$ But $\angle C_1CM \equiv \angle C_1CB = \angle ABC = \angle XEM$. Thus $MEC_1C$ is cyclic as claimed. Similarly $MB_1BF$ is cyclic. Now $\angle B_1MB = \angle B_1FB = \angle XFA = \angle XEA = \angle C_1EC = \angle C_1MC$. Thus $B_1,M,C_1$ are collinear. Now $\angle B_1BC = \angle ACB \equiv \angle ECM = \angle MC_1C \equiv \angle B_1C_1C$. Here the second last equality holds as $EM = MC \implies$ they subtend equal angles. Thus $B,C,B_1,C_1$ are concyclic and we are done.

Let $M$ be the midpoint of $BC$, $H_A$ be the $A-Humpty$ point and $A' \equiv BB_1 \cap CC_1$ Note that $ACA'B$ is a parallelogram and that $A,H_A,M, A'$ are collinear. It is well known that $A,E,H_A,H,F$ are concyclic, and since $\angle HXA =\pi$, $X$ also lies on that circle. Moreover, it is well known by the properties of the $Humpty$ point that $FH_AMB$ and $CMH_AE$ are cyclic quadrilaterals. We will prove that $B_1$ lies on $(FH_AMB)$ and a similarly , a conclusion will follow that $C_1$ lies on $(CMH_AE)$ It is true that $\angle H_AFB_1=\angle XAH_A$. However, $\angle H_AFM=\angle H_ABM=\angle H_AAB$. Hence, $\angle MFB_1= \angle XAF= \angle AFE= \angle ACB= \angle CBA'$due to $XA || FE$ and $AC || BA'$ Thus , $FBB_1M$ is cyclic, or in other words $B_1$ lies on $(FH_AMB)$. Similar angle chase proves this for $C$. Therefore, by Power of Point, $A'B_1 \cdot A'B = A'M \cdot A'H_A = A'C \cdot A'C_1$ so $BB_1CC_1$ is cyclic, as desired

WLOG assume that $AB<AC$. The following Claim is the pith of the problem: Claim: $B_1C_1$ passes through the midpoint of $BC$. Proof: Firstly, note that $\angle AXH=\angle AFH=\angle AEH=90^\circ,$ and so $X \in (AFHE)$. Hence, $XAEF$ is a trapezoid. Let $M$ be the midpoint of $BC$. Note that $\angle FB_1B=180^\circ-\angle FBB_1-\angle BFB_1=\angle FAE-\angle XFA=$ $=\angle A-(\angle XFE-\angle XEF)=\angle A-\angle AEF+\angle AFE=$ $=\angle A-\angle B+\angle C=180^\circ-2\angle b=\angle FMB,$ and so $M \in (FBB_1)$. Similarly, $M \in (ECC_1)$. Hence, $\angle BMB_1=\angle BFB_1=\angle XFA=\angle XEA=\angle C_1EC=\angle C_1MC,$ which in turn implies that $M \in B_1C_1,$ as desired $\blacksquare$ To the problem, we may note that $\angle BB_1C_1=\angle BB_1M=180^\circ-\angle BFM=180^\circ-\angle B=\angle BCC_1,$ with the last equality being true as $AB$ is parallel to $CC_1$. Hence, $\angle BB_1C_1=\angle BCC_1,$ and so $B,C,B_1,C_1$ are concyclic, as desired.

This was my favorite in contest, here's what I submitted Edit : sniped

Define $A$, $B$ and $C$ as the respective angles of $\triangle{ABC}$ and $M$ be the midpoint of $BC$. Claim 1 - $CMEC_1$ is cyclic Proof - As $\measuredangle{AXH}=\measuredangle{AFH}=90^{\circ}$ and $AX||EF$, $X\in (AEHF)$ and $AXEF$ is an isosceles trapezoid. So, $\measuredangle{XEF} = \measuredangle{AFE} = \measuredangle{C}$. Also, $M$ is the center of $(BFEC)$, $\measuredangle{FMB} = 2\measuredangle{B}$ and $\measuredangle{EMC} = 2\measuredangle{C}$ so $\measuredangle{EMF} = 2\measuredangle{A}$ and $\measuredangle{FEM} = \measuredangle{A}$. Hence, $\measuredangle{C_1EM} = \measuredangle{B}$. Also, $\measuredangle{MCC_1}=\measuredangle{B}$ as $CC_1||AB$. Claim 2 - $EF || C_1M$ Proof - This follows as $\measuredangle{XEF} = \measuredangle{AFE} = \measuredangle{C}$ and $\measuredangle{MCE} = \measuredangle{MC_1E} = \measuredangle{C}$. Claim 3 - If $B_2=C_1M\cap (BFM)\neq M$, then $B_2\equiv B$ and so, $BFMB_1$ is cyclic. Proof - $$\measuredangle{BFB_2} = \measuredangle{BMB_2} = \measuredangle{C_1MC} = \measuredangle{C_1EC} = \measuredangle{AEX} = \measuredangle{AFX}$$So, $B_2 \in XF$. Also, $$\measuredangle{FBB_2} = \measuredangle{FMC_1} = \measuredangle{A}$$. So, $BB_2 ||AC$. Using Claim 1 and 3, we say that $$\measuredangle{BB_1C} = \measuredangle{BB_1M} = \measuredangle{B} = \measuredangle{BCC_1}$$Hence, $BB_1CC_1$ is cyclic

Claim (1): $AX$ tangent to $(ABC)$ and $X,A,E,H,F$ are cyclic. Proof: The first on is true because $EF$ is anti-paralle to $BC$ so $AO\perp EF$ and $EF//AX$ The second one we have $<AXH=<AEH=<AFH=90$ Let now $M$ be the midpoint of $ABC$ and $P=AM\cap (AEF)$ Claim (2): $ME,MF$ tangebt to $(AEF)$ Proof: $MB=MC=ME=MF$ so $<MFC=<MCF=<BAD=<BAH$. Claim (3): $MPFB$,$MPEC$ are cyclic Proof: Consider the invertion with center $A$ and power $AE*AC=AH*AD=AF*AB$ then $(AEF)\leftrightarrow BC\Rightarrow P\leftrightarrow M$ wich gives: $AE*AC=AH*AD=AF*AB=AM*AP$ wich gives the result Claim (4): $B_1\epsilon (BFPM),C_1\epsilon (ECMP)$ Proof: $<XC_1C=360-<C_1CF-<CFX-<FXC=270-<XFH-<A=270-<A-(180-<XAH)=90-<A+<XAH=90-<A+<C+90-<B=2<C=180-<CME$ Similar for $B_1$ Claim (5): $B_1,C_1,M$ are collinear Proof: $<B_1MP+<PMB_1=<PFX+<PEX=180$ Claim (6): $B_1,C_1,B,C$ are cycilc Proof: $<BCC_1=<MCC_1=<MEX=XFHP+<PEM=XFHPE=AEPF=<MFA=MB_1B=<BB_1C_1$ we use claim (2) and by $AX//EF$ and $(AXEF)$ cyclic we get $AF=XE$

Latexing my solution now: I am assuming orthic triangle config as well known (all the stuff given in EGMO). Also let angles of triangle $ABC$ be $A,B,C$ and the sides be $a,b,c$

Then clearly $A'CB \cong ABC$. It suffices to prove $A'B \cdot A"B_1 = A'C \cdot A'C_1$. Now look at the diagram attached below, and WLOG assume $\angle B > \angle C$ so that the diagram looks like this... Now because $\angle HXA=90^{\circ}=\angle HFA= \angle HEA, \ X \in (AEHF)$. Also $\angle XAF= \angle AFE = \angle ACB$ so $AX$ is just the tangent to $(ABC)$ So now at this point we see that if we can find $BB_1$ and similarly $CC_1$, we will be done because for instance $A'B_1=A'B-BB_1$ and then it will just be a matter of doing some trig manipulations... Now at this point we see that we know basically everything about orthic triangle, so we can probably just go ahead and find some angles, so let's do that. We want $BB_1$ and the only triangle that occurs is in in $\Delta BB_1F$ and as we already know $BF$ we just gotta find the angles of that triangle to solve this problem by law of sines. $\angle FBB_1 = \pi - A=B+C$ is clear due to parallel lines. All we gotta do now is find $\angle BFB_1$, but that is just $\angle XFA$, and $\angle XFA$ is easy to find because we know $\angle XAF=C$ (due to tangent) and $\angle AXF= \pi - \angle AEF = \pi - B$ (due to cyclic pentagon $AEHFX$) So we get that $\angle XFA = B-C$. So $\angle BFB_1 = B-C \implies \angle BB_1C = \pi - 2B$ So now all we gotta do is use sine rule and some trig, here it is: $\frac{BF}{\sin(\pi-2B)}=\frac{BB_1}{\sin(B-C)}$ And $BF=a \cdot \cos{B}$, So we get $BB_1 = \frac{a \cdot \cos{B} \cdot \sin(B-C)}{\sin(2B)}=\frac{a \cdot \sin(B-C)}{2 \sin{B}}$ Similarly, $CC_1=\frac{a \cdot \sin(B-C)}{2 \sin{C}}$ And we wanna prove $b(b-BB_1)=c(C+C_1)$ (This is important to note, it is $c$ + $CC_1$ not minus, i was tripping for a few minutes because of this... Now, replace all the side lengths by using $\frac{x}{\sin{X}}=2R$ So we wanna prove (after cancelling the $4R^2$ factor) $\sin^2{B} - \sin{A} \cdot \sin(B-C) = \sin^2{C} + \sin{A} \cdot \sin(B-C) \iff \sin^2{B} - \sin^2{C} = \sin(B+C) \cdot \sin(B-C)$ (bcoz $A = \pi - (B+C)$) but this is obvious, just expand the right thing and replace $\cos^2$ by $1 - \sin^2$ everywhere. Really fun, also the first problem I have ever solved on vacation : ) Trust the grind

Note that $AXFE$ is an isosceles trapezoid. Let $G= XF \cap AC$, now angle chase: $$\angle BB_1F= \angle FGA = \pi - 2\angle AEF = \pi-\angle 2B = \angle BMF \implies B_1 \in \odot(BMF)$$Similarly, $C_1\in \odot(CME)$. Let $A'= BB_1\cap CC_1$, then we just need to show that $A'$ lies on the radical axis of $\odot (BMF)$ and $\odot (CME)$(note that it lies on $AM$). This is easy, $AA'$ hits $(AEF)$ again at $R$, then we get $\angle FBM = \angle AEF = \angle ARF$. So $R\in \odot (BMF)$. Similarly it lies on $\odot (CME)$. So we're done. $\square$

How to introduce midpoint? You think like an illiterate person that since this ELMO, it must have something to do with midpoints. Here's how you solve geo! It's quite easy! Solution: We'll only add the midpoint of $\overline{BC}$ which we call $M$. Note that by the angle condition on $X$, we can easily say that $X \in \odot(AEFH)$ and $AXFE$ is an isosceles trapezoid. [asy][asy] import geometry; defaultpen(fontsize(13pt)); size(10cm); pair A = dir(120); pair B = dir(210); pair C = dir(330); pair H = orthocenter(A,B,C); pair E = foot(B,A,C); pair F = foot(C,A,B); pair M = (B+C)/2; pair X = 2*extension(bisectorpoint(E,F), (E+F)/2, A,foot(A,bisectorpoint(E,F), (E+F)/2)) - A; pair B1 = intersectionpoints(line(X,F), circumcircle(B,M,F))[1]; pair C1 = intersectionpoints(line(X,E), circumcircle(E,M,C))[0]; draw(A--B--C--A, purple); draw(circumcircle(M,E,C), orange); draw(circumcircle(M,F,B), orange); draw(C--F, purple); draw(B--E, purple); draw(C--C1, purple); draw(M--E, purple); draw(X--C1, purple + dashed); draw(X--B1, purple + dashed); draw(B1--C1, purple + dashed); draw(E--F, purple); draw(circumcircle(X,A,E), red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$H$", H, S); dot("$E$", E, 2*dir(75)); dot("$F$", F, dir(F)); dot("$M$", M, dir(M)); dot("$X$", X, dir(X)); dot("$B_1$", B1, dir(B1)); dot("$C_1$", C1, dir(C1)); [/asy][/asy] Claim: $C_1 \in \odot(CME)$ and $B_1 \in \odot(MFB)$. Proof: The proof will be by phantom points. We only show the first part since the other one follows by symmetry. Re-define $C_1$ to be a point such that $CC_1$ is parallel to $\overline{AB}$ and $C_1 \in \odot(CME)$. It now suffices to show that $X-E-C_1$ are collinear. This can be shown easily by an angle chase. \begin{align*} \measuredangle C_1EC & = \measuredangle C_1MC \\ & = \measuredangle EMC - \measuredangle EMC_1 \\ & = 2\measuredangle ACB - \measuredangle ECC_1 \\ & = 2\measuredangle ACB - \measuredangle CAB \\ & = 2\measuredangle ACB + \measuredangle BAC \\ & = \measuredangle ACB - \measuredangle ABC = \measuredangle AEX \end{align*}and thus the claim holds true. $\square$ We now show that $B_1-M-C_1$ are collinear. Note that \[\measuredangle EC_1M = \measuredangle ECM = \measuredangle EFA = \measuredangle XEF\]which shows that $EF \parallel MC_1$. By a similar argument, we have $EF \parallel MB_1$. This proves the collinearity. It's about time to finish it. We now have that \begin{align*} \measuredangle BB_1C_1 & = \measuredangle BB_1F + \measuredangle FB_1M \\ & = \measuredangle BMF + \measuredangle ABC \\ & = 2\measuredangle MCF + \measuredangle C_1CB \\ & = 2\measuredangle CBA + \measuredangle C_1CB \\ & = 2\measuredangle BCC_1 - \measuredangle BCC_1 = \measuredangle BCC_1 \end{align*}and the solution is complete. $\blacksquare$

One of my favorite angle-chasing problems. Let $Z = PX \cap AE,$ and let $M$ be the midpoint of $BC.$ Also denote by $\measuredangle$ a directed angle. We first note that $$\measuredangle HXA = \measuredangle HFA = \measuredangle HEA = 90^\circ,$$so $AXFHE$ is cyclic. Moreover, $AX \parallel EF,$ so $AXFE$ is an isosceles trapezoid. This means that $\triangle ZXA$ and $\triangle ZFE$ are isosceles. Now, we know that $$\measuredangle BB_1 F = \measuredangle BB_1 Z = \measuredangle AZB_1 = \measuredangle AXZ,$$since $AZ \parallel BB_1.$ We also know that by sum of angles $$\measuredangle AZX + \measuredangle ZXA + \measuredangle XAZ = 0.$$However, since $\measuredangle ZXA = \measuredangle XAZ,$ we get $$\measuredangle AZX + 2 \measuredangle XAZ = 0,$$so $$\measuredangle AZX = 2\measuredangle ZAX.$$On top of that, we know that $$\measuredangle XAX = \measuredangle AEF = \measuredangle CEF = \measuredangle CBF = \measuredangle CBA,$$since $AX \parallel FE$ and $BFEC$ is cyclic. Therefore, $$\measuredangle BB_1 F = 2\measuredangle CBA.$$However. we know that $M$ is the circumcenter of $BFEC,$ so $\triangle BMF$ is isosceles. Hence $$\measuredangle BMF + \measuredangle MFB + \measuredangle FBM = \measuredangle BMF + 2 \measuredangle ABC,$$so $\measuredangle BMF = 2 \measuredangle CBA = \measuredangle BB_1 F.$ This implies that $BB_1 MF$ is cyclic. Similarly, $CC_1 EM$ is also cyclic. Now, $$\measuredangle CMB_1 = \measuredangle BMB_1 = \measuredangle BFB_1 = \measuredangle AFX = \measuredangle AEX = \measuredangle CEC_1 = \measuredangle CMC_1,$$so $B_1, M, C_1$ are collinear. Finally, $$\measuredangle BB_1 C_1 = \measuredangle BB_1 M = \measuredangle BFM = \measuredangle MBF = \measuredangle CBA = \measuredangle BCC_1$$from our parallel lines, so $B,C,B_1,C_1$ are concyclic, as desired.

Let $A'=BB_1\cap CC_1$, $Y=AX\cap BB_1$, $Z=AX\cap CC_1$. Claim. $B_1$ and $C_1$ are the midpoints of $A'Y$ and $A'Z$, respectively. Proof. It suffices to show that $B_1$ is the midpoint of $A'Y$. Clearly, $AEFHX$ is cyclic. Let $K$ be the $A$-antipode in $(ABC)$. Then $AHA'K$ is a parallelogram, and thus $A'H\parallel AK\perp AX\perp XH$ so $X$, $H$, and $A'$ are colinear. Now let $F_1$ be a point on $(AEFHX)$ such that $XF_1\parallel AF$. Then Pascal on $AAEFF_1X$ gives $FF_1$ parallel to the tangent to $(AEFHX)$ at $A$ (since $AX\parallel EF$ and $AE\parallel XF_1$). Thus we have $$-1=(A,H;F,F_1)\overset{X}{=}(Y,A';B_1,\infty_{A'Y})$$which proves the claim. From the claim we get $B_1C_1\parallel YZ$ which is the tangent to $(ABC)$ at $A$ and is thus anti-parallel to $BC$ i.e. $B_1C_1$ and $BC$ are antiparallel so $BB_1CC_1$ is cyclic.

Let $Y := XF\cap AC$ and $Z := XE\cap AB$. We first observe that $AXFHE$ is a cyclic pentagon and $AXFE$ is an isosceles trapezoid (as $AX\parallel EF$), so that $FZ=ZE$ and $FY=YE$, that is, $YZ$ is the perpendicular bisector of $EF$. Since $AB\parallel CC_1$, we have $\angle BCC_1=180^\circ-\angle B$, so it suffices to show that $\angle BB_1C_1=180^\circ-\angle B$. We note that $\triangle BAC\sim \triangle FXE$ are similar since $\angle FXE=\angle A$ and $\angle XEF=\angle AFE=\angle C$. Thus, we have $\angle FB_1B=\angle FYA=\angle A-\angle AFY =\angle A+\angle ZFE-\angle XFE=\angle A+\angle C-\angle B$. As $\angle FYZ=90^\circ-\angle B=\angle FEH$ and $\angle YZA= 90^\circ-\angle C=\angle EFH$, by sine law we get \[\frac{YE}{ZF}=\frac{YF}{ZF}=\frac{\sin (180^\circ-\angle YZA)}{\sin \angle FYZ}=\frac{\sin \angle EFH}{\sin \angle FEH}=\frac{HE}{HF}=\frac{CE}{BF},\]where the last equality follows from $\triangle BHF\sim \triangle CHE$. By Thales' theorem, we have $\frac{AE}{EC}=\frac{ZE}{EC_1}\iff \frac{XF}{ZF}=\frac{EC}{EC_1}$ and $\frac{AF}{FB}=\frac{YF}{FB_1}\iff \frac{XE}{YE}=\frac{FB}{FB_1}$, so that \[\dfrac{XF}{FB_1}\cdot \dfrac{FB}{ZF}=\dfrac{XE}{EC_1}\cdot \dfrac{EC}{YE}\iff \dfrac{XF}{FB_1}=\dfrac{XE}{EC_1}.\]We then obtain that $EF\parallel B_1C_1$ and that $\angle FB_1C_1=\angle XFE=\angle B$. Therefore, we have $\angle BB_1C_1=\angle BB_1F+\angle FB_1C_1=\angle A+\angle C=180^\circ-\angle B$, as desired.

(Solution from actual ELMO submission) As stated earlier, $AEFX$ is a cyclic isosceles trapezoid. The key idea is to construct $M$ as the midpoint of $\overline{BC}$. Then the given angle conditions can be chased to show that $M\in(BFB_1),(CEC_1)$. Meanwhile, $AB\cdot AF=AC\cdot AE$. As a result, $\overline{AM}$ is the radical axis of those two circles. To finish, let $A'=2M-A=\overline{BB_1}\cap\overline{CC_1}$. This obviously lies on the radical axis mentioned earlier, so $A'B\cdot A'B_1=A'C\cdot A'C_1$, and we are done by power of a point converse. (comments on motivation for M to be written later)

Goofy aah solution: Define $T$ as the intersections of $BB_1$ and $CC_1$, and let $B_2$ and $C_2$ be $BB_1 \cap XA$ and $CC_1 \cap XA$, respecetively. Note that $A,X,F,H$ and $E$ are all cyclic on the circle with diameter $AH$. Also, since $\overline{AX} \parallel \overline{FE}$, it followss that $AXFE$ is a cyclic trapezoid. Some angle chasing gives us that $\triangle CBT \sim \triangle BCA$. Claim: $X$, $H$ and $T$ are collinear. Proof: we have that $$\overrightarrow {OT}=\overrightarrow{OC}+\overrightarrow{OB}-\overrightarrow{OA},$$we have that $$\overrightarrow{OH} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC},$$hence $$\overrightarrow{TH} = 2 \overrightarrow{OA}.$$Since $\overline{OA} \parallel \overline{HX}$, it follows that $\overline{HX} \parallel \overline{HT}$, hence $H$, $X$ and $T$ are collinear. More angle chasing now yields that $\triangle B_2 B_1 X$ and $\triangle C_2C_1 X$ are isosceles, hence $B_1$ and $C_1$ are the midpoints of $\overline{ B_2 T }$ and $\overline{ C_2 T} $, respectively. (Consider right triangles $\triangle TXB_2$ and $\triangle BXC_2$.) Even more angle chasing gives us that $B_2 BCC_2$ is cyclic, so $TB \cdot TB_2 = TC \cdot TC_2$. Dividing both sides by two gives us \[ TB \cdot TB_1 = TC \cdot TC_1,\]so $BB_1CC_1$ is cyclic.

Let $BB_1$ and $CC_1$ meet at $P$, which is seen to be the midpoint of $BC$. By a Spiral Similarity, we get $\triangle XB_1C_1 \sim \triangle ABC$, so $\angle ABC = \angle XB_1C_1$, which implies $BB_1PF$ is cyclic. Similarly, $CC_1EP$ is cyclic. However, it is well known that $(BFM)$ and $(CEM)$ meet at the $A$-humpty point $P_A$, from which we immediately get that $BB_1CC_1$ is cyclic via Radical Axis.

It's clear that $X$ lies on $\odot(AFHE)$, then $XAEF$ is isosceles trapezium with $XF = AE$. Let $L = B_{1}X \cap AC$ and as $XA \parallel FE$ $\Rightarrow$ $LF = LE$. By angle chasing \[\angle FLE = 2 \angle FXH\] Claim: $X$, $H$ and $BB_{1} \cap CC_{1} = P$ are collinear. Proof: $\angle PBC = \angle ACB$ and $\angle ABC = \angle BCP$ $\Rightarrow$ $\angle BPC = \angle BAC = 180^{\circ} - \angle BHC$ $\Rightarrow$ $BHCP$ is cyclic. Since $XEHF$ is cyclic too \[\angle BHP = \angle BCP = \angle ABC = \angle AEF \stackrel{XF = AE}{=} \angle XFE = \angle XHE \square \] Let $K = PX \cap AC$. Notice that \[ 2 \angle KXL = 2\angle FXH = \angle FLE = \angle XKL + \angle KXL \]then $LX = LK$ and \[\angle B_{1}PX = \angle XKL = \angle KXL = \angle FXH = \angle B_{1}XP \]so $B_{1}X = B_{1}P$. Analogously $C_{1}X = C_{1}P$ therefore $B_{1}XC_{1}P$ is a deltoid with $XP \perp B_{1}C_{1}$, this is, $B_{1}C_{1} \parallel FE \parallel XA$. Finally \[\angle B_{1}BC = \angle BCA = \angle AFE = \angle XEF = \angle XC_{1}B_{1} = \angle BC_{1}P = \angle B_{1}C_{1}C \]thereby $B$, $C$, $B_{1}$ and $C_{1}$ are concyclic $\blacksquare$ [asy][asy] import graph; size(10cm); pair A = (-2.72796,12.27769); pair B = (-6.14859,2.36593); pair C = (5.56774,3.20089); pair E = (0.65108,8.58048); pair F = (-4.64420,6.72512); pair H = (-2.27397,5.90713); pair X = (-4.31165,11.72279); pair B_1 = (-5.01667,1.12743); pair C_1 = (6.20965,5.06090); pair P = (2.14711,-6.71086); pair K = (-5.61156,15.43279); pair L = (-4.16976,13.85524); pen zzttqq = rgb(0.6,0.2,0); pen ccqqqq = rgb(0.8,0,0); pen fuqqzz = rgb(0.95686,0,0.6); pen qqwuqq = rgb(0,0.39215,0); draw((-4.12839,11.19976)--(-3.60536,11.38302)--(-3.78862,11.90605)--X--cycle, linewidth(2) + ccqqqq); draw((-0.55924,2.68922)--(-0.74250,3.21225)--(-1.26553,3.02899)--(-1.08227,2.50596)--cycle, linewidth(2) + ccqqqq); draw(arc(B,0.78376,-47.57435,4.07625)--B--cycle, linewidth(2) + qqwuqq); draw(arc(C_1,0.78376,-160.69061,-109.03999)--C_1--cycle, linewidth(2) + qqwuqq); draw(arc(C,0.78376,132.42564,184.07625)--C--cycle, linewidth(2) + qqwuqq); draw(arc(E,0.78376,147.65877,199.30938)--E--cycle, linewidth(2) + qqwuqq); draw(arc(C_1,0.78376,147.65877,199.30938)--C_1--cycle, linewidth(2) + qqwuqq); draw(arc(F,0.78376,19.30938,70.96000)--F--cycle, linewidth(2) + qqwuqq); draw(arc(K,0.78376,-70.69061,-47.57435)--K--cycle, linewidth(2) + qqwuqq); draw(arc(X,0.78376,86.19313,109.30938)--X--cycle, linewidth(2) + qqwuqq); draw(arc(L,0.78376,-93.80686,-70.69061)--L--cycle, linewidth(2) + qqwuqq); draw((-1.47352,7.83606)--(-1.65678,8.35909)--(-2.17981,8.17583)--(-1.99655,7.65280)--cycle, linewidth(2) + ccqqqq); draw(arc(P,0.78376,109.30938,132.42564)--P--cycle, linewidth(2) + qqwuqq); draw(arc(E,0.78376,-160.69061,-137.57435)--E--cycle, linewidth(2) + qqwuqq); draw(B--A, linewidth(2)); draw(B--C, linewidth(2)); draw(F--C, linewidth(2)); draw(B--E, linewidth(2)); draw(F--E, linewidth(2)); draw(P--X, linewidth(2) + linetype("4 4")); draw(P--C_1, linewidth(2) + zzttqq); draw(X--C_1, linewidth(2) + zzttqq); draw(X--A, linewidth(2)); draw(circle((-0.54467,6.35111), 6.87645), linewidth(2) + dotted); draw(circle((-2.50096,9.09241), 3.19335), linewidth(2) + ccqqqq); draw(X--K, linewidth(2) + linetype("4 4")); draw(K--A, linewidth(2)); draw(L--X, linewidth(2)); draw(E--A, linewidth(2)); draw(E--C, linewidth(2)); draw(B_1--F, linewidth(2) + fuqqzz); draw(F--X, linewidth(2) + fuqqzz); draw(B_1--C_1, linewidth(2)); draw(B--B_1, linewidth(2)); draw(B_1--P, linewidth(2) + fuqqzz); draw((-1.99655,7.65280)--L, linewidth(2) + linetype("4 4")); dot("$A$", A, dir((20.446, 72.656))); dot("$B$", B, dir((9.813, 26.646))); dot("$C$", C, dir((79.145, -38.562))); dot("$E$", E, dir((11.723, 21.755))); dot("$F$", F, dir((-195.489, -4.327))); dot("$H$", H, dir((11.623, 19.996))); dot("$X$", X, dir((-111.179, 41.931))); dot("$B_1$", B_1, dir((-137.342, -126.434))); dot("$C_1$", C_1, dir((88.105, 21.017))); dot("$P$", P, dir((-169.231, -84.572))); dot("$K$", K, dir((10.974, 21.014))); dot("$L$", L, dir((10.485, 22.015))); [/asy][/asy]

Let $A'$ be the point such that $ABA'C$ is a parallelogram, so $B-B_1-A'$ and $C-C_1-A'$. I claim that $A'$ is the reflection of $X$ over $\overline{B_1C_1}$. Clearly, $AEFHX$ is cyclic, so $AXFE$ is an isosceles trapezoid. Thus, $$\measuredangle FHX=\measuredangle FEX=\measuredangle AFE=\measuredangle BCA=\measuredangle CBA'=\measuredangle CHA'=\measuredangle FHA',$$since $BHCA'$ is cyclic ($\angle HBA=\angle HCA=90^\circ$), which implies that $X,H,A'$ are collinear. Further, this concyclicity also gives $$\measuredangle B_1A'H=\measuredangle BA'H=\measuredangle BCH=\measuredangle HAF=\measuredangle HXF=\measuredangle HXB_1.$$Similarly, we find that $\measuredangle C_1A'H=\measuredangle HXC_1$. Therefore, we have $B_1X=B_1A'$ and $C_1X=C_1A'$, hence $\triangle B_1C_1X \cong \triangle B_1C_1A'$, so $A'$ is indeed the reflection of $X$ over $\overline{B_1C_1}$: in particular, $\overline{XH} \perp \overline{B_1C_1}$. To finish, since we also have $\overline{XH} \perp \overline{AX}$ and $\overline{XH} \perp \overline{EF}$, we have $\overline{B_1C_1} \parallel \overline{EF}$. Finally, this implies that $$\measuredangle CC_1B_1=\measuredangle AFE=\measuredangle BCA=\measuredangle CBA'=\measuredangle CBB_1,$$hence $BCB_1C_1$ is cyclic as desired. $\blacksquare$ Remark: I received an unofficial 0.1 style score on this problem because the grader was apparently unwilling to read the bottom half of my page and notice that I had corrected my improper reasoning by, among other things, shuffling the order of my main arguments. I also forgot to state that $AEFHX$ was cyclic explicitly in my solution, but the grader apparently never noticed (just like me!)

We present a few solutions. In each, let \(A'=\overline{BB_1}\cap\overline{CC_1}\), so \(ABA'C\) is a parallelogram. First solution (author) Since \(\overline{A'H}\perp\overline{EF}\), we have \(X\), \(H\), \(A'\) collinear. But \[\measuredangle B_1XA'=\measuredangle FEH=\measuredangle FCB=\measuredangle XA'B_1,\]implying \(B_1X=B_1A'\). Similarly \(C_1X=C_1A'\), so \(\overline{B_1C_1}\perp\overline{XHA'}\). This means \(\overline{BC}\) and \(\overline{B_1C_1}\) are antiparallel in \(\angle A'\), so \(BB_1CC_1\) is indeed cyclic. [asy][asy] size(7cm); defaultpen(fontsize(10pt)); pen pri=darkgreen; pen pri2=heavygreen; pen sec=heavycyan; pen tri=lightblue; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair A,B,C,H,EE,F,X,B1,C1,Ap,M,P; A=dir(110); B=dir(215); C=dir(325); H=A+B+C; EE=foot(B,A,C); F=foot(C,A,B); X=foot(H,A,A+EE-F); B1=extension(X,F,B,B+C-A); C1=extension(X,EE,C,C+B-A); Ap=B+C-A; M=(B+C)/2; P=reflect(circumcenter(B,F,M),circumcenter(C,EE,M))*M; draw(A--Ap,tri+dashed); filldraw(circumcircle(B,F,M),tfil,tri); filldraw(circumcircle(C,EE,M),tfil,tri); draw(B1--C1,tri+dashed); draw(B1--X--C1,sec); filldraw(circumcircle(A,EE,F),sfil,sec+dashed); draw(X--Ap,pri2+dashed); draw(B--EE,pri2+linewidth(.4)); draw(C--F,pri2+linewidth(.4)); draw(B--Ap--C1,pri); fill(B--Ap--C--cycle,fil); filldraw(A--B--C--cycle,fil,pri); dot("\(A\)",A,N); dot("\(B\)",B,B); dot("\(C\)",C,C); dot("\(M\)",M,S); dot("\(E\)",EE,dir(60)); dot("\(F\)",F,dir(200)); dot(P); dot("\(A'\)",Ap,S); dot("\(H\)",H,S); dot("\(X\)",X,NW); dot("\(B_1\)",B1,SW); dot("\(C_1\)",C1,E); [/asy][/asy] Second solution (mine) Let \(M\) be the midpoint of \(\overline{BC}\). Since \(AEFX\) is an isosceles trapezoid and \(ME=MF\), \[\measuredangle B_1FM=\measuredangle XFM=\measuredangle MEA=\measuredangle ECM=\measuredangle B_1BM,\]so \(B_1\in(BMF)\). Similarly \(C_1\in(CME)\). But since \(AB\cdot AF=AC\cdot AE\), line \(AM\) is the radical axis of \((BMF)\) and \((CME)\). In particular, \(A'\) lies on this radical axis, so \(A'B\cdot A'B_1=A'C\cdot A'C_1\) as needed. Third solution (author) Let \(M\) be the midpoint of \(\overline{BC}\). Let \(\ell\) be the perpendicular bisector of \(\overline{EF}\) (so \(M\in\ell\)). Let \(B_2\) is the reflection of \(B_1\) in \(\ell\) and let \(M'\in\ell\) be the midpoint of \(\overline{B_1B_2}\). Since \(\overline{XF}\) and \(\overline{AE}\) are reflections in \(\ell\), we know \(B_2\) lies on \(\overline{AC}\). If \(M\ne M'\), this implies \(\ell=\overline{MM'}\parallel\overline{AC}\), which is absurd. Hence \(M\) is the midpoint of \(\overline{B_1B_2}\), i.e.\ \(\overline{B_1M}\perp\ell\). Similarly \(\overline{C_1M}\perp\ell\). Then \(\overline{B_1C_1}\parallel\overline{EF}\), implying \(\overline{BC}\) and \(\overline{B_1C_1}\) are antiparallel in \(\angle A'\), which gives the desired.

What math is this?

Similar to IMO Shortlist 2022 G2. Denote by K intersection of BB1 and CC1, it is enough to prove that KB1×KB=KC×KC1. By angle chasing we have FBB1M and EC1CM is cyclic (M is midpoint of BC). And from AF×AB=AE×AC we have AM is rad. axis of circles (FBB1M) and (EC1CM), and we know that A, M, K are collinear (because ABKC is parallelogram), then K lies on radical axis of (FBB1M) and (ECC1M), which gives us KB1×KB=KC1×KC, as desired.

Let $BB_1\cap CC_1=Y, YH\cap B_1C_1=K$ Note that $ABYC$ is a parallelogram $\Rightarrow \angle BHC=180^{\circ}-\angle BAC=180^{\circ} -\angle BYC\Rightarrow BHCY$ is cyclic We have $X\in (AH)$ and $AEFX$ is an isosceles trapezoid $\Rightarrow \angle XHF=\angle AFE=\angle ACB=\angle CBY=\angle CHY$ $\Rightarrow \overline{X,H,Y},$ also $\angle B_1YH=\angle BCH=\angle HEF=\angle B_1XH\Rightarrow B_1X=B_1Y,$ similarly $C_1X=C_1Y$ $\Rightarrow B_1C_1\perp XY\Rightarrow YB.YB_1=YK.YH=YC.YC_1\Rightarrow B,C,B_1,C_1$ are concyclic.

Let $M$ the midpoint of $BC$. $Z$ is the reflection of $A$ over $M$. $Y$ = $ AE \cap FX$. WLOG $AC \ge AB$ claim $1$: $AXFHE$ is cyclic and $AE = XF$. Proof: $\angle AXH = \angle AFH $ and $ AX \parallel EF$.$_\blacksquare$ claim $2$: $BB_1MF , CC_1EM$ are cyclic. Proof: $\angle FB_1B = B_1YC = 180 - 2\angle B = \angle FMB $ and similarly we get $CC_1EM$ cyclic.$_\blacksquare$ claim $3$: $AM$ is the radical axis of $(BB_1MF) , (CC_1EM)$ Proof: $AF.AB = AE.AC$ because $ EFBC$ is cyclic. $_\blacksquare$ Finish: $ Z \in AM \implies ZB_1.ZB=ZC_1.ZC$ and we are done.

Observe that $AXFE$ is an isosceles trapezoid as $H$ is the antipode of $A$ in $(AEF)$. A quick angle chase yields $\measuredangle BB_1F = -2\measuredangle ABC$ and $\measuredangle CC_1E = -2\measuredangle ACB$, so we have $$\frac{FB_1}{EC_1} = \frac{BF\cdot\frac{\sin\left(\angle A\right)}{\sin\left(2\angle B\right)}}{CE\cdot \frac{\sin\left(\angle A\right)}{\sin\left(2\angle C\right)}} = \frac{\frac{BC\cos\left(\angle B\right)}{2\cos\left(\angle B\right)\sin\left(\angle B\right)}}{\frac{BC\cos\left(\angle C\right)}{2\cos\left(\angle C\right)\sin\left(\angle C\right)}} = \frac{\sin\left(\angle C\right)}{\sin\left(\angle B\right)} = \frac{XF}{XE}$$ so $B_1C_1\parallel EF$. Hence, if $BB_1\cap CC_1 = A_1$ is the reflection of $A$ over the midpoint of $BC$, $BC$ and $B_1C_1$ are antiparallel in $\angle BA_1C$, as desired.

Might be a ridiculously long-lasting solution, but worth sharing (I hope). Let $T$ be the point such that $ABTC$ is a parallelogram, $M$ be the midpoint of the side $BC$, $O$ be the center of $(BAHC)$, and $K$ be the foot of the altitude from $H$ to $EF$. First, $X$ lies on the circle with diameter $AH$, which is $AFHE$. Now, take the inversion with center $H$ and radius $\sqrt{-HB.HC}$. Let $P'$ be the inverted version of $P$ for each point $P$ in the space. $90=\angle HKF= \angle K'F'H= \angle K'CH$ ; hence, $K' \in CC_1$. Similarly $K' \in BB_1$. Thus, $K'$=$T$, which means that $X,K,H,T$ are collinear. $\angle OBB_1=\angle OBT=\angle 90-TCB=\angle 90-ABC=\angle HAB=\angle HAF=\angle HXB_1=\angle OXB_1$. Thus, $B,X,O,B_1$ are concyclic. Similarly, $C,X,O,C_1$ are concyclic. $\implies TB_1.TB=TO.TX=TC_1,TC \implies B_1,B,C_1,C$ lie on a circle.

Solved with Kamatadu and distorteddragon1o4 It's easy to observe that $AXEF$ is an isosceles trapezoid.So from here we get that $\angle XFE=\angle AEF=\angle ABC$ and we also have $\angle FBB_1=\angle ABB_1=180-\angle BAC$. We have $$\angle EFB_1=180-\angle XFE=180-\angle B \implies \angle BFB_1=180-(180-\angle B+\angle C)=\angle B-\angle C $$This implies $\angle BB_1F=180-\angle BFB_1-\angle FBB_1=2\angle B=2\angle BMF$ where $M$ is midpoint of $BC$.Thus $BFMB_1$ is cyclic and similarly $CFMC_1$ is cyclic as well.Its well known that they meet at the Humpty point $H_A$.Now if $BB_1 \cap CC_1=A'$ it's well known that $A-H_A-M-A'$, finally by power of point we have, $$A'B.A'B_1=A'M.A'H_A=A'C.A'C_1 \implies BB_1CC_1\text{ is cyclic } \blacksquare$$

this is actually very nontrivial, very good problem! Let $M$ be the midpoint of $BC$. Through angle chasing, $B_1BMF$ and $C_1CME$ are cyclic. There is a point $K=(AEHF)\cap (BMF)\cap (CME)$. Verify that $K$ is the center of spiral similarity $XB_1\to AB$ and $XC_1\to AC$, hence $\triangle XB_1C_1\sim \triangle ABC$. Hence $B_1C_1\parallel EF$. Let $A'=BB_1\cap CC_1$ be the reflection of $A$ across $M$. Then $B_1C_1$ and $BC$ are antiparallels with respect to both $\angle BAC$ and $\angle BA'C$, so $B,C,B_1,C_1$ are concyclic. (motivation for M is the same as #17)

Let $M$ be the midpoint of $\overline{BC}$, and let $A' = \overline{BB_1} \cap \overline{CC_1}$ be the reflection of $A$ over $M$. Claim: $BB_1MF$ and $CC_1ME$ are cyclic Proof: Let $B_2 = \overline{XFB_1} \cap \overline{AE}$. Since $AXFE$ is an isosceles trapezoid as $X$ lies on $(AEHF)$, by symmetry we find that $AB_2X$ is isosceles with $AB_2 = XB_2$. Then, we get $$\measuredangle BB_1F = \measuredangle AB_2X = 2\measuredangle AXB_2 = 2\measuredangle AXF = 2\measuredangle AHF = 2\measuredangle MBF = \measuredangle BMF,$$so $BB_1MF$ is cyclic. Similarly, $CC_1ME$ is cyclic. Now, since $AB \cdot AF = AC \cdot AE$ from cyclic quadrilateral $BFEC$, we find that the radical axis of $(BB_1MF)$ and $(CC_1ME)$ is line $AM$. Therefore since $A'$ lies on this line, $A'B_1 \cdot A'B = A'C_1 \cdot A'C$, which finishes.

No midpoint solution just dropped. It is immediate the $X$ lies on the circle $(AH)$. We start off by proving the following key claim using trigonometry. Claim : Lines $\overline{EF}$ and $\overline{B_1C_1}$ are parallel. Proof : We start off with some angle chasing. Note that, \[\measuredangle CC_1E = \measuredangle CEC_1 + \measuredangle C_1CE = \measuredangle AEX + \measuredangle BAC = 2\measuredangle BCA\]A similar angle chase shows that $\measuredangle FBB_1 = 2\measuredangle CBA$. Now, we invoke our trigonometry. Applying the Sine Rule on $\triangle AEX$ and $\triangle EC_1C$, \[XE = \frac{AE \sin \angle XAE }{\sin \angle AXE}\]and \[EC_1 = \frac{EC \sin \angle C_1CE}{\sin \angle CC_1E}\]So, \[\frac{XE}{CC_1}= \frac{AE}{EC} \cdot \frac{\sin B \sin 2C}{\sin C \sin A}\]A similar calculation also yeilds, \[\frac{XF}{FB_1} = \frac{AF}{FB} \cdot \frac{\sin C \sin 2B}{\sin B \sin A}\]Thus, \begin{align*} \frac{XE}{EC_1} \div \frac{XF}{FB_1} &= \frac{AE \cdot FB}{AF \cdot EC} \cdot \frac{\sin^2B \sin 2C}{\sin ^2 C \sin 2B}\\ &= \left(\frac{AE}{\sin C} \cdot \frac{\sin B}{AF}\right) \cdot \frac{FB}{EC} \frac{\cos C}{\cos B}\\ &= \frac{FB}{EC} \cdot \frac{\cos C}{\cos B}\\ &=1 \end{align*}which indeed implies that $EF \parallel B_1C_1$ as claimed. Now, let $A' = \overline{BB_1} \cap \overline{CC_1}$. It is easy to see that $ABA'C$ is a parallelogram by definition. Now we also observe the following claim. Claim : Points $X$ , $H$ and $A'$ are collinear. Proof : Since $A'$ is the reflection of $A$ across the midpoint of $BC$, $A'$ lies on $(BHC)$. Thus, \[\measuredangle A'HB = \measuredangle A'CB = \measuredangle ABC = \measuredangle FEA = \measuredangle XAE = \measuredangle XHE\]which implies the claim. Now, let $T$ be the second intersection of circles $(BB_1H)$ and $(CC_1H)$. Note that, \[\measuredangle B_1TH + \measuredangle HTC_1 = \measuredangle HBB_1 + \measuredangle HCC_1 = \pi\]so points $B_1$ , $T$ and $C_1$ are collinear. Now, since $XH \perp AX \parallel EF \parallel B_1C_1$, $XH \perp B_1C_1$. But since $HT ]perp B_1C_1$ as well this implies that points $X$ , $H$ and $T$ must be collinear. This means $A'$ lies on the radical axis of circles $(BB_1H)$ and $(CC_1H)$, so \[AB_1\cdot AB = AT \cdot AH = AC_1 \cdot AC \]which implies that points $B$ , $C$ , $B_1$ and $C_1$ are concyclic, as desired.

Sketch: Let $\overline{BB_1}$ and $\overline{CC_1}$ intersect at $A'$. It suffices to show $\overline{B_1C_1} \parallel \overline{EF}$. This reduces to \[\frac{\operatorname{dist}(X,\overline{A'B})}{\operatorname{dist}(F,\overline{A'B})}=\frac{\operatorname{dist}(X,\overline{A'C})}{\operatorname{dist}(E,\overline{A'C})}.\]By angle chasing, we know $X$, $H$, and $A'$ are collinear. This gives \[\frac{\operatorname{dist}(X,\overline{A'B})}{\operatorname{dist}(H,\overline{A'B})}=\frac{\operatorname{dist}(X,\overline{A'C})}{\operatorname{dist}(H,\overline{A'C})}\]From $BHF \sim CHE$, $\overline{BH} \perp \overline{A'B}$, and $\overline{CH} \perp \overline{A'C}$, we have \[\frac{\operatorname{dist}(H,\overline{A'B})}{\operatorname{dist}(F,\overline{A'B})}=\frac{\operatorname{dist}(H,\overline{A'C})}{\operatorname{dist}(E,\overline{A'C})}.\]Multiplying these two equations gives us the desired result. $\blacksquare$

CyclicISLscelesTrapezoid wrote: Sketch: Let $\overline{BB_1}$ and $\overline{CC_1}$ intersect at $A'$. It suffices to show $\overline{B_1C_1} \parallel \overline{EF}$. This reduces to \[\frac{\operatorname{dist}(X,\overline{A'B})}{\operatorname{dist}(F,\overline{A'B})}=\frac{\operatorname{dist}(X,\overline{A'C})}{\operatorname{dist}(E,\overline{A'C})}.\]By angle chasing, we know $X$, $H$, and $A'$ are collinear. This gives \[\frac{\operatorname{dist}(X,\overline{A'B})}{\operatorname{dist}(H,\overline{A'B})}=\frac{\operatorname{dist}(X,\overline{A'C})}{\operatorname{dist}(H,\overline{A'C})}\]From $BHF \sim CHE$, $\overline{BH} \perp \overline{A'B}$, and $\overline{CH} \perp \overline{A'C}$, we have \[\frac{\operatorname{dist}(H,\overline{A'B})}{\operatorname{dist}(F,\overline{A'B})}=\frac{\operatorname{dist}(H,\overline{A'C})}{\operatorname{dist}(E,\overline{A'C})}.\]Multiplying these two equations gives us the desired result. $\blacksquare$ nice sketch!