Really hard problem to solve synthetically, or at least it took me two days. Congratulations to the few students who solved it on the exam. Let $X$, $Y$, $Z$ be midpoints of $EF$, $FD$, $DE$, and let $G$ be the Gregonne point. By radical axis on $(AEIF)$, $(DEF)$, $(AIC)$ we see that $B_1$, $X$, $B_2$ are collinear. Likewise, $B_1$, $Z$, $B_2$ are collinear, so lines $B_1B_2$ and $XZ$ coincide. Similarly, lines $C_1C_2$ and $XY$ coincide. In particular lines $B_1B_2$ and $C_1C_2$ meet at $X$. Note $G$ is the symmedian point of $DEF$, so it is well-known that $XG$ passes through the midpoint of $DK$. So we just have to prove $G$ lies on the radical axis. Construct parallelograms $GPFQ$, $GRDS$, $GTUE$ such that $P,R \in DF$, $S,T \in DE$, $Q,U \in EF$. As $FG$ bisects $PQ$ and is isogonal to $FZ$, we find $PQED$, hence $PQRU$, is cyclic. Repeating the same logic and noticing $PR$, $ST$, $QU$ not concurrent, all six points $PQRSTU$ are cyclic. Moreover, since $PQ$ bisects $GF$, we see that a dilation with factor $2$ at $G$ sends $PQ$ to $P', Q' \in AB$, say, with $F$ the midpoint of $P'Q'$. Define $R', S' \in BC$ similarly now and $T', U' \in CA$. Note that $EQPDS'$ is in cyclic too, as $\measuredangle DS'Q = \measuredangle DRS = \measuredangle DEF$. By homothety through $B$, points $B$, $P$, $X$ are collinear; assume they meet $(EQPDS')$ again at $V$. Thus $EVQPDS'$ is cyclic, and now \[ \measuredangle BVS' = \measuredangle PVS' = \measuredangle PQS = \measuredangle PTS = \measuredangle FED = \measuredangle XEZ = \measuredangle XVZ \]hence $V$ lies on $(BQ'S')$. Since $FB \parallel QP$, we get $EVFB$ is cyclic too, so $XV \cdot XB = XE \cdot XF$ now; thus $X$ lies on the radical axis of $(BS'Q')$ and $(DEF)$. By the same argument with $W \in BZ$, we get $Z$ lies on the radical axis too. Thus the radical axis of $(BS'Q')$ and $(DEF)$ must be line $XZ$, which coincides with $B_1B_2$; so $(BB_1B_2) = (BS'Q')$. Analogously, $(CC_1C_2) = (CR'U')$. Since $G = Q'S' \cap R'U'$, we need only prove that $Q'R'S'U'$ is cyclic. But $QRSU$ is cyclic, so we are done. Remark: The circle $(PQRSTU)$ is called the lemon circle of $ABC$. (It's actually Lemoine, but I like lemons better .)

Here's a computational way to finish with the same approach as v_Enhance. It suffices to show that the circumcircle of $BB_1B_2$ passes through $Q'$, and then the rest follows. This can be done with the coaxial lemma: Given two circles $\omega_1$ and $\omega_2$ and a constant $c$, the locus of points $P$ such that $\text{p}(P,\omega_1)=c\text{p}(P,\omega_2)$ is a circle $\omega$ coaxial to $\omega_1$ and $\omega_2$. We apply this to the circumcircle of $AIC$, call it $\omega_1$, and the incircle, call it $\omega_2$. If we show that $\frac{\text{p}(C,\omega_1)}{\text{p}(C,\omega_2)}=\frac{\text{p}(Q',\omega_1)}{\text{p}(Q',\omega_2)}$, then that means that $C$ and $Q'$ both lie on a circle coaxial to $\omega_1$ and $\omega_2$, so $CQ'C_1C_2$ is cyclic. The rest follows from length chasing. Let $C'$ be the second intersection of the circumcircle of $AIC$ with $AB$. Note that $BC=BC'$. Note that $\text{p}(B,\omega_1)=BA\cdot BC'=ac$ and $\text{p}(B,\omega_2)=BF^2=(s-b)^2$. Suppose that $BQ'=t$. We want to show that $\frac{\text{p}(Q',\omega_1)}{\text{p}(Q',\omega_2)}=\frac{Q'A\cdot Q'C'}{Q'F^2}=\frac{(a-t)(c-t)}{(s-b-t)^2}=\frac{ac}{(s-b)^2}$. Expanding, we get $(s-b)^2t^2-(s-b)^2(a+c)t+(s-b)^2ac=act^2-2ac(s-b)t+ac(s-b)^2$, or $y^2t-y^2(x+2y+z)=(x+y)(y+z)t-2y(x+y)(y+z)$ after the Ravi Substitution and some cancellations. This in turn becomes $(xy+yz+xz)t=y(xy+yz+2xz)$, so $t=\frac{y(xy+yz+2xz)}{xy+yz+xz}=y+\frac{2xyz}{xy+yz+xz}$. To finish, we can easily verify that $BQ'=y+\frac{2xyz}{xy+yz+zx}$ with barycentric coordinates.

Here's a hybrid solution which doesn't require any significant synthetic observation and from start to finish takes less than half an hour: Let $UVW$ be the medial triangle of $\triangle DEF$; then $UV\equiv C_1C_2, UW\equiv B_1B_2$. Then since $B_1U\cdot UB_2=C_2U\cdot UC_1$, it follows that $U$ lies on the radical axis of both circles. By a well-known lemma, it suffices to show that $L$, the Gergonne point of $\triangle ABC$, lies on the radical axis of both circles. Let the line through $L$ parallel to $DE$ intersect $AC$ and $B$ at $L_1$ and $L_2$ respectively; we claim that pentagon $CL_1C_1C_2L_2$ is cyclic. To do this, we use barycentric coordinates with respect to $\triangle DEF$; we want to show that the radical axis of $\odot(CL_1L_2)$ and $\odot(DEF)$ is of the form $k(x+y-z)=0$, where $k$ is some constant. This would imply that $$\odot(CL_1L_2)\equiv a^2yz+b^2zx+c^2xy=k(x+y-z)(x+y+z)$$which is clear by subtracting the equations of both circles. Since $C=(-a^2:b^2:c^2)$, we easily get that $k=\frac{-(abc)^2}{(a^2+b^2+c^2)(a^2+b^2-c^2)}$. Compute $L_1\equiv (a^2+2b^2:-b^2:c^2), L_2\equiv (-a^2:2a^2+b^2:c^2)$; plugging in $L_1$ to our circle equation gives $$a^2(-b^2)(c^2)+b^2(c^2)(a^2+2b^2)+c^2(a^2+2b^2)(-b^2)=\frac{-(abc)^2}{(a^2+b^2+c^2)(a^2+b^2-c^2)}(a^2+2b^2-b^2+c^2)(a^2+2b^2-b^2-c^2)$$and it is easy to verify that both sides are equal to $-(abc)^2$, so $L_1\in \odot(CC_1C_2)$, and by symmetry, $L_2\in\odot(CC_1C_2)$, as desired. Define $L_3$ and $L_4$ similarly with respect to $\odot(BB_1B_2)$. To finish, the midpoint of $\overline{LL_1} \in AC$, as it has barycentric coordinates $(a^2+b^2:0:c^2)$, and so each of the midpoints of $\overline{LL_1}, \overline{LL_2}, \overline{LL_3}, \overline{LL_4}$ are concyclic on the Lemoine circle of $\triangle DEF$. Hence quadrilateral $L_1L_2L_3L_4$ is concyclic by a homothety at $L$ with ratio $2$, so $L$ lies on the radical axis of $\odot(BB_1B_2)$ and $\odot(CC_1C_2)$, as desired.

I messed around in Geogebra and found the following generalization: Let $ABC$ be a triangle, and let $P$ be a point with pedal triangle $A_0B_0C_0$. Suppose that $AA_0$, $BB_0$, and $CC_0$ concur at $Q$. The circumcircle of $A_0B_0C_0$ intersects the circumcircle of $PBC$ at $A_1$ and $A_2$, the circumcircle of $PCA$ at $B_1$ and $B_2$, and the circumcircle of $PAB$ at $C_1$ and $C_2$. Then $Q$ is the radical center of the circumcircles of $AA_1A_2$, $BB_1B_2$, and $CC_1C_2$. Anyone want to try?

ABCDE wrote: I messed around in Geogebra and found the following generalization: Let $ABC$ be a triangle, and let $P$ be a point with pedal triangle $A_0B_0C_0$. Suppose that $AA_0$, $BB_0$, and $CC_0$ concur at $Q$. The circumcircle of $A_0B_0C_0$ intersects the circumcircle of $PBC$ at $A_1$ and $A_2$, the circumcircle of $PCA$ at $B_1$ and $B_2$, and the circumcircle of $PAB$ at $C_1$ and $C_2$. Then $Q$ is the radical center of the circumcircles of $AA_1A_2$, $BB_1B_2$, and $CC_1C_2$. Lemma (well-known) : Given a $ \triangle ABC $ and two points $ P, $ $ Q. $ Let $ \triangle DEF $ be the cevian triangle of $ P $ WRT $ \triangle ABC $ and $ AQ, $ $ BQ, $ $ CQ $ cuts $ EF, $ $ FD, $ $ DE $ at $ X, $ $ Y, $ $ Z, $ respectively. Then $ PX, $ $ BZ, $ $ CY $ are concurrent. Back to the main problem : Let $ AP, $ $ BP, $ $ CP $ cuts $ B_0C_0, $ $ C_0A_0, $ $ A_0B_0 $ at $ P_a, $ $ P_b, $ $ P_c, $ respectively. Clearly, $ P_b $ is the radical center of $ \odot (BP), $ $ \odot (APB) $ and the pedal circle $ \Omega_P $ of $ P $ WRT $ \triangle ABC $ $ \Longrightarrow $ $ P_b $ lies on the radical axis $ C_1C_2 $ of $ \Omega_P, $ $ \odot (APB), $ so if $ CP_b $ cuts $ \odot (CC_1C_2) $ again at $ Z, $ then we get $ CP_b $ $ \cdot $ $ ZP_b $ $ = $ $ BP_b $ $ \cdot $ $ PP_b $ $ \Longrightarrow $ $ Z $ $ \in $ $ \odot (BPC). $ Analogously, we can prove $ P_a $ lies on $ B_1B_2, $ $ C_1C_2, $ and $ Y $ $ \equiv $ $ BP_c $ $ \cap $ $ \odot (BB_1B_2) $ lies on $ \odot (BPC), $ so the radical axis $ \mathcal{R}_a $ of $ \odot (BB_1B_2), $ $ \odot (CC_1C_2) $ passes through $ P_a $ and the intersection of $ BY, $ $ CZ, $ hence from Lemma $ \Longrightarrow $ $ Q $ $ \in $ $ \mathcal{R}_a. $

v_Enhance wrote: Let $ABC$ be a triangle with incenter $I$, and whose incircle is tangent to $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ at $D$, $E$, $F$, respectively. Let $K$ be the foot of the altitude from $D$ to $\overline{EF}$. Suppose that the circumcircle of $\triangle AIB$ meets the incircle at two distinct points $C_1$ and $C_2$, while the circumcircle of $\triangle AIC$ meets the incircle at two distinct points $B_1$ and $B_2$. Prove that the radical axis of the circumcircles of $\triangle BB_1B_2$ and $\triangle CC_1C_2$ passes through the midpoint $M$ of $\overline{DK}$. Proposed by Danielle Wang My solution : Lemma : (well-known) Let triangle $ABC$. The tangent line at $A$ of $(ABC)$ intersects the line passes through $B$ and perpendicular to $AB$ at $P$, the line passes through $C$ perpendicular to $AC$ at $Q.M\equiv PC\cap QB$. Draw $AH\perp BC.A'$ is the symmectric of $A$ through the midpoint of $BC$. Then $M,H,A'$ are collinear. ______________________________________ Back to the main problem : Let $M,N$ be the midpoints of $DE,DF.\left \{ A_1;A_2 \right \}\equiv MN\cap (I)$. From $NA_1.NA_2=NF.ND=ND^2=NB.NI\implies B,A_1,I,A_2$ are concyclic. Similarly, $C,A_2,I,B$ are concyclic implies $\left \{ A_1;A_2 \right \}\equiv (BIC) \cap (I)$. Similarly $B_1B_2,C_1C_2$ be the averages of $\triangle DEF$. Let $P$ be the midpoint of $EF$. Then $P\equiv B_1B_2\cap C_1C_2$. From $PB_1.PB_2=PC_1.PC_2$ so $P$ lies on the radial axis of $(BB_1B_2)$ and $(CC_1C_2)$. From $MB_1.MB_2=MA_1.MA_2$ so $M$ lies on the radial axis of $(BIC)$ and $(BB_1B_2)$ i.e $BM$ is the radial axis of $(BB_1B_2)$ and $(BIC)$. Similarly $CN$ is the radial axis of $(CC_1C_2)$ and $(BIC)$. Then $BM,CN$ and the radial axis of $(BB_1B_2)$ and $(CC_1C_2)$ are concurents. Let $X\equiv BM\cap CN$ then $PX\equiv $ the radial axis of $(BB_1B_2)$ and $(CC_1C_2)$. According to the lemma $PX$ passes through the projection of $D$ on $MN$ i.e $PX$ passes through the midpoints of $DK.\blacksquare$

baopbc wrote: My proof. Lemma (well-known) Let triangle $ABC$. The tangent line at $A$ of $(ABC)$ intersects the line passes through $B$ and perpendicular to $AB$ at $P$, the line passes through $C$ perpendicular to $AC$ at $Q.M\equiv PC\cap QB$. Draw $AH\perp BC.A'$ is the symmectric of $A$ through the midpoint of $BC$. Then $M,H,A'$ are collinear. Dear mathlinkers, I have the same idea with you. But this Lemma isn't easy with me. I have proved it by using Steiner line for cyclic quadirlateral $PQCB$ Can you give me your solution for this Lemma?

olympia1234 wrote: Dear mathlinkers, I have the same idea with you. But this Lemma isn't easy with me. I have proved it by using Steiner line for cyclic quadirlateral $PQCB$ Can you give me your solution for this Lemma? Proof : Easy to see that $A'$ be the orthocenter of $\triangle BNC$ implies $A'$ lies on the radial axis of $(BQ)$ and $(PC)$. From $P,B,C,Q$ are concyclic so $MP.MC=MQ.MB$ implies $M$ lies on the radial axis of $(BQ)$ and $(PC)$. Let $PX,QY\perp BC$. From $\frac{HB}{HC}=\frac{AP}{AQ}=\frac{HX}{HY}\implies HB.HY=HC.HX$ implies $H$ lies on the radial axis of $(BQ)$ and $(PC)$ which implies $M,H,A'$ are collinear.

Here is another solution based on Allen Liu's paper (edited now that I know how to prove the lemma). First, we prove the lemma as suggested by TelvCohl. Note that $\triangle DEF$ is the cevian triangle of the Gregonne point $G$ and $\triangle XYZ$ is another cevian triangle. Set $V = \overline{XY} \cap \overline{AB}$, $W = \overline{XZ} \cap \overline{AC}$, and $T = \overline{BW} \cap \overline{CV}$. Then one can show points $G$, $X$, $T$ are collinear (see here for a proof of this completely projective fact). Now, point $V$ is the radical center $(CC_1C_2)$, $(ABC)$ and $(DEF)$. To see this, let $V' = \overline{ED} \cap \overline{AB}$; then $(FV';AB)$ is harmonic, and $V$ is the midpoint $FV'$, and thus $VA \cdot VB = VF^2 = VC_1 \cdot VC_2$. So in fact $\overline{CV}$ is the radical axis of $(ABC)$ and $(CC_1C_2)$. Similarly, $\overline{BW}$ is the radical axis of $(ABC)$ and $(BB_1B_2)$; thus $T = \overline{BW} \cap \overline{CV}$ is the radical center of $(ABC)$, $(BB_1B_2)$, $(CC_1C_2)$. Since by the lemma $X$, $G$, $T$ are collinear. But as in the first solution we know $X$, $G$, $M$ are collinear, as needed.

TelvCohl wrote: Lemma (well-known) : Given a $ \triangle ABC $ and two points $ P, $ $ Q. $ Let $ \triangle DEF $ be the cevian triangle of $ P $ WRT $ \triangle ABC $ and $ AQ, $ $ BQ, $ $ CQ $ cuts $ EF, $ $ FD, $ $ DE $ at $ X, $ $ Y, $ $ Z, $ respectively. Then $ PX, $ $ BZ, $ $ CY $ are concurrent. Now I see the lemma I mentioned in my post is exactly the one here (up to an application of Cevian nest). Telv, do you happen to know the proof?

v_Enhance wrote: TelvCohl wrote: Lemma (well-known) : Given a $ \triangle ABC $ and two points $ P, $ $ Q. $ Let $ \triangle DEF $ be the cevian triangle of $ P $ WRT $ \triangle ABC $ and $ AQ, $ $ BQ, $ $ CQ $ cuts $ EF, $ $ FD, $ $ DE $ at $ X, $ $ Y, $ $ Z, $ respectively. Then $ PX, $ $ BZ, $ $ CY $ are concurrent. Now I see the lemma I mentioned in my post is exactly the one here (up to an application of Cevian nest). Telv, do you happen to know the proof? From Cevian nest theorem we get $ \triangle XYZ $ is the cevian triangle of $ \triangle DEF, $ so notice $ \triangle DEF $ is the cevian triangle of $ A $ WRT $ \triangle PCB $ we conclude that (by Cevian nest theorem) $ PX, $ $ BZ, $ $ CY $ are concurrent.

This can also be done with pure complex bash. It appears worse than it actually is. Take $D=1, E=e, F=f$, with $DEF$ the unit circle. Then $A=\frac{2ef}{e+f}, B=\frac{2f}{f+1}, C=\frac{2e}{e+1}$. Then its easy to see that circle $(AIB)$ has equation $$z\overline{z} - \frac{2ef}{(e+f)(f+1)}\overline{z}-\frac{2f}{(e+f)(f+1)}z=0$$On the other hand, the incircle has equation $z\overline{z}=1$. Thus the radical axis of $(CC_1C_2)$ and $(DEF)$ has equation $$1=\frac{2ef}{(e+f)(f+1)}\overline{z}+\frac{2f}{(e+f)(f+1)}z$$For some constant $j$, it follows that $(CC_1C_2)-j(DEF) = \frac{2ef}{(e+f)(f+1)}\overline{z}+\frac{2f}{(e+f)(f+1)}z-1$. Indeed, the equation is true for $C_1, C_2$; thus we just need to plug in $z=C = \frac{2e}{e+1}$ to find $j$. Thus the equation of $(CC_1C_2)$ is $$0 = jz\overline{z} + \frac{2ef}{(e+f)(f+1)}\overline{z}+\frac{2f}{(e+f)(f+1)}z - j - 1 \quad (1)$$and after plugging in $z =C$, we get that $j = \frac{(e+1)^2}{(e-1)^2}\left[\frac{8ef}{(e+f)(e+1)(f+1)}-1\right]$. Similarly, the equation of $(BB_1B_2)$ can be given by $$0=lz\overline{z}+\frac{2ef}{(e+f)(e+1)}\overline{z}+\frac{2e}{(e+f)(e+1)}z - l - 1\quad (2)$$where $l = \frac{(f+1)^2}{(f-1)^2}\left[\frac{8ef}{(e+f)(e+1)(f+1)}-1\right]$. Now the Radical Axis can be given by $(1)\cdot \frac{(f+1)^2}{(f-1)^2} - (2)\cdot \frac{(e+1)^2}{(e-1)^2}$, which reduces to $$\frac{2ef\overline{z}}{e+f}\left[\frac{(f+1)}{(f-1)^2} - \frac{(e+1)}{(e-1)^2}\right] + \frac{2z}{e+f}\left[\frac{f(f+1)}{(f-1)^2} - \frac{e(e+1)}{(e-1)^2}\right] =\frac{(f+1)^2}{(f-1)^2}-\frac{(e+1)^2}{(e-1)^2} $$Taking $2z = 2m = 1+k$, we have $$\frac{ef(1+\overline{k})}{e+f}\left[\frac{(f+1)}{(f-1)^2} - \frac{(e+1)}{(e-1)^2}\right] + \frac{1+k}{e+f}\left[\frac{f(f+1)}{(f-1)^2} - \frac{e(e+1)}{(e-1)^2}\right] = \frac{(f+1)^2}{(f-1)^2}-\frac{(e+1)^2}{(e-1)^2} $$Recalling that $ef\overline{k} + k = e+f$ since $K\in EF$, we have $$\frac{ef}{e+f}\left[\frac{(f+1)}{(f-1)^2} - \frac{(e+1)}{(e-1)^2}\right] + \frac{1}{e+f}\left[\frac{f(f+1)}{(f-1)^2} - \frac{e(e+1)}{(e-1)^2}\right]$$$$ + \frac{e+f-k}{e+f}\left[\frac{(f+1)}{(f-1)^2} - \frac{(e+1)}{(e-1)^2}\right] + \frac{k}{e+f}\left[\frac{f(f+1)}{(f-1)^2} - \frac{e(e+1)}{(e-1)^2}\right] = \frac{(f+1)^2}{(f-1)^2}-\frac{(e+1)^2}{(e-1)^2} $$$$\Longrightarrow$$$$\frac{(e+1)(f+1)}{e+f}\left[\frac{f}{(f-1)^2}-\frac{e}{(e-1)^2}\right] + \frac{f+1}{(f-1)^2} -\frac{e+1}{(e-1)^2}$$$$+ \frac{1/2(1+e+f-ef)}{e+f}\left[\frac{f+1}{f-1} - \frac{e+1}{e-1}\right] = \frac{(f+1)^2}{(f-1)^2}-\frac{(e+1)^2}{(e-1)^2}$$$$\Longrightarrow$$$$(e+1)(f+1)[f(e-1)^2-e(f-1)^2] + 1/2(1+e+f-ef)(e-1)(f-1)[(f+1)(e-1)-(f-1)(e+1)] $$$$= (e+f)[f(f+1)(e-1)^2-e(e+1)(f-1)^2]$$which is true after a bit of expanding! Thus $M$ lies on the radical axis, as desired. $\square$

hi i solved it too

It's a difficult, but very nice problem. Define $A_1,A_2$ in a similar fashion. Let $X,Y,Z$ be the midpoints of $\triangle EDF$, $w$ the incircle $(DEF)$ and $\Omega=(BIC)$. An inversion with respect to $w$ shows that $X,Z,B_1,B_2$ are collinear, etc. So we have that $Z$ is on the radical axis of $( BB_1B_2),(CC_1C_2)$. Also $XA_1*XA_2=XB_1*XB_2$, so $BX$ is the radical axis of $(BB_1B_2),\Omega$. If $W = BX \cap CY$, then $W$ is then the radical center of $\Omega,( BB_1B_2),(CC_1C_2)$. It remains to prove that $W,M,Z$ are collinear. Let $I_a$ be the A-excenter, and $R,S,T$ be the intersections of $EF$ and $CI$, $EF$ and the external bisectors at $B,C$ respectively. We easily see that $\angle IRT =\angle FBI$, so $R$ lies on the circle $(IFBD)$. Applying Pascal's theorem to $BBIRFD$ yields that $S,Y,C$ are collinear. Similarly $B,X,T$ are collinear. Since the sides of $\triangle DYX, \triangle I_aS,T$ are parallel, $W$ is the center of similarity of those two triangles. Since $M,Z$ are the corresponding feet of altitudes in the two triangles, we are done.

Long but really straight-forward!

So, after 364 days, I (along with EulerMac), found an interesting solution, remote of any computation. It's surprising that no solution I glanced at so far, had this idea or a variant. Presenting in all it's glory: USA TSTST 2016/6 wrote: Let $ABC$ be a triangle with incenter $I$, and whose incircle is tangent to $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ at $D$, $E$, $F$, respectively. Let $K$ be the foot of the altitude from $D$ to $\overline{EF}$. Suppose that the circumcircle of $\triangle AIB$ meets the incircle at two distinct points $C_1$ and $C_2$, while the circumcircle of $\triangle AIC$ meets the incircle at two distinct points $B_1$ and $B_2$. Prove that the radical axis of the circumcircles of $\triangle BB_1B_2$ and $\triangle CC_1C_2$ passes through the midpoint $M$ of $\overline{DK}$. Proposed by Danielle Wang [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(16.208cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 15.96, ymin = -5.24, ymax = 6.3; /* image dimensions */ /* draw figures */ draw(circle((4.377845724572135,1.5267857385789958), 1.672360600101097), linewidth(1.2)); draw((3.5,4.58)--(2.16,-0.18), linewidth(1.2)); draw((2.16,-0.18)--(8.64,-0.08), linewidth(1.2)); draw((8.64,-0.08)--(3.5,4.58), linewidth(1.2)); draw((2.7680564481537795,1.97996171135223)--(5.501116483990229,2.7657582071217), linewidth(1.2)); draw((5.501116483990229,2.7657582071217)--(4.403650685985681,-0.1453757610187395), linewidth(1.2)); draw((4.403650685985681,-0.1453757610187395)--(2.7680564481537795,1.97996171135223), linewidth(1.2)); draw(circle((3.738733530951814,1.0349446851921345), 1.992106962446594), linewidth(1.2)); draw(circle((6.030750812906075,0.9036585236700474), 2.7885059461186454), linewidth(1.2)); draw((4.377845724572136,1.5267857385789965)--(8.64,-0.08), linewidth(1.2) + dotted); draw((4.377845724572136,1.5267857385789965)--(2.16,-0.18), linewidth(1.2) + dotted); draw((4.377845724572136,1.5267857385789965)--(3.5,4.58), linewidth(1.2) + dotted); draw(circle((5.391638619285755,0.4118174702831861), 3.2853821823026204), linewidth(1.2) + linetype("2 2")); draw((4.4455702544254985,3.19777447726835)--(3.3256018156999305,0.22695008242401804), linewidth(1.2)); draw((3.633289771710591,3.02425908052166)--(5.634704131641809,0.42356458153532317), linewidth(1.2)); draw((4.315931664173603,5.538816852403914)--(2.16,-0.18), linewidth(1.2) + dotted); draw((4.315931664173603,5.538816852403914)--(8.64,-0.08), linewidth(1.2) + dotted); draw((3.0256680604752417,2.895059677509067)--(5.748110286939079,3.6778033909435877), linewidth(1.2)); draw((3.714357686463116,2.2520377679538224)--(4.403650685985681,-0.1453757610187395), linewidth(1.2)); draw((4.315931664173603,5.538816852403914)--(3.94445371025816,-0.9465116876948286), linewidth(1.2) + linetype("4 4")); draw((3.0256680604752417,2.895059677509067)--(5.354204655864398,-0.13070671827369756), linewidth(1.2) + dotted); draw((4.796812384878383,3.4042907172114276)--(3.4530967161069643,-0.1600448037637814), linewidth(1.2) + dotted); /* dots and labels */ dot((3.5,4.58),dotstyle); label("$A$", (3.58,4.78), NE * labelscalefactor); dot((2.16,-0.18),dotstyle); label("$B$", (1.92,-0.64), NE * labelscalefactor); dot((8.64,-0.08),dotstyle); label("$C$", (8.74,-0.34), NE * labelscalefactor); dot((4.403650685985681,-0.1453757610187395),linewidth(3.pt) + dotstyle); label("$D$", (4.34,-0.42), NE * labelscalefactor); dot((4.403650685985681,-0.1453757610187395),linewidth(3.pt) + dotstyle); dot((5.501116483990229,2.7657582071217),linewidth(3.pt) + dotstyle); label("$E$", (5.58,2.88), NE * labelscalefactor); dot((2.7680564481537795,1.97996171135223),linewidth(3.pt) + dotstyle); label("$F$", (2.52,1.76), NE * labelscalefactor); dot((4.134586466072005,2.372859959236965),linewidth(3.pt) + dotstyle); label("$N$", (4.22,2.36), NE * labelscalefactor); dot((4.952383584987955,1.3101912230514803),linewidth(3.pt) + dotstyle); dot((3.5858535670697305,0.9172929751667454),linewidth(3.pt) + dotstyle); dot((4.4455702544254985,3.19777447726835),linewidth(3.pt) + dotstyle); label("$C_2$", (4.48,2.84), NE * labelscalefactor); dot((3.3256018156999305,0.22695008242401804),linewidth(3.pt) + dotstyle); label("$C_1$", (3.,0.04), NE * labelscalefactor); dot((3.633289771710591,3.02425908052166),linewidth(3.pt) + dotstyle); label("$B_2$", (3.46,2.68), NE * labelscalefactor); dot((5.634704131641809,0.42356458153532317),linewidth(3.pt) + dotstyle); label("$B_1$", (5.66,0.1), NE * labelscalefactor); dot((4.377845724572136,1.5267857385789965),linewidth(3.pt) + dotstyle); label("$I$", (4.4,1.18), NE * labelscalefactor); dot((3.3549555132659936,2.989734848603373),linewidth(3.pt) + dotstyle); label("$B'$", (3.2,3.2), NE * labelscalefactor); dot((5.748110286939079,3.6778033909435877),linewidth(3.pt) + dotstyle); label("$C'$", (5.82,3.8), NE * labelscalefactor); dot((3.0256680604752417,2.895059677509067),linewidth(3.pt) + dotstyle); label("$P$", (2.7,3.02), NE * labelscalefactor); dot((4.796812384878383,3.4042907172114276),linewidth(3.pt) + dotstyle); label("$Q$", (5.,3.2), NE * labelscalefactor); dot((5.354204655864398,-0.13070671827369756),linewidth(3.pt) + dotstyle); label("$R$", (5.28,-0.46), NE * labelscalefactor); dot((3.4530967161069643,-0.1600448037637814),linewidth(3.pt) + dotstyle); label("$S$", (3.28,-0.52), NE * labelscalefactor); dot((4.315931664173603,5.538816852403914),linewidth(3.pt) + dotstyle); label("$H$", (4.4,5.66), NE * labelscalefactor); dot((3.714357686463116,2.2520377679538224),linewidth(3.pt) + dotstyle); label("$K$", (3.54,1.86), NE * labelscalefactor); dot((4.059004186224398,1.0533310034675414),linewidth(3.pt) + dotstyle); label("$M$", (4.16,0.86), NE * labelscalefactor); dot((4.169360499748456,2.979951256793683),linewidth(3.pt) + dotstyle); dot((3.94445371025816,-0.9465116876948286),linewidth(3.pt) + dotstyle); dot((4.085605790850003,1.5177465734158),linewidth(3.pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] (Solution with EulerMac) Note that midpoint $N$ of $\overline{EF}$ is radical center of $\odot(AI), \odot(AIC), \odot(I)$; by similar reasoning, we conclude $\overline{B_1B_2}$ is the $E$-midline and $\overline{C_1C_2}$ is the $F$-midline in $\triangle DEF$. As an immediate corollary, $N$ lies on the radical axis of $\odot(BB_1B_2)$ and $\odot(CC_1C_2)$ (henceforth $\omega_B, \omega_C$ respectively). Let $\overline{BA}, \overline{BC}$ meet $\omega_B$ at $P, R$ respectively; define $Q, S$ similarly. Let $H$ be the orthocenter of $\triangle BIC$; $B'=\overline{HB} \cap \omega_B, C'=\overline{HC} \cap \omega_C$. Claim: $P, Q, B', C'$ are collinear. (Proof) Note that $\measuredangle RPB'=\measuredangle RBB'=\measuredangle DFE$ and $\measuredangle SQC'=\measuredangle DEF$. Also, $\measuredangle (PR, QS)=\measuredangle EDF$ since $\overline{PR} \parallel \overline{DF}$ and $\overline{QS} \parallel \overline{DE}$. To conclude, observe that $$\measuredangle RPB'+\measuredangle SQC'+\measuredangle (PR, QS)=180^{\circ},$$hence the claim is true. $\blacksquare$ Consequently, $B, C, B', C'$ are concyclic (say, on $\gamma$); indeed, $\measuredangle CC'Q=\measuredangle CSQ=\measuredangle CBB'$ as $\overline{SQ} \parallel \overline{BB'}$. Apply radical axis theorem to $\omega_B, \omega_C$ and $\gamma$, to get that $H$ lies on the radical axis of $\omega_B, \omega_C$. However, it's no secret that $H, M, N$ are collinear, so we are done! $\blacksquare$ Remark It seems this proof does rely on the Gergonne proof implicitly.

Clean complex after showing that $B_1B_2$ and $C_1C_2$ are midlines of $DEF$ and that the midpoint of $EF$ lies on the radical axis: Set $DEF$ on the unit circle in the complex plane. As $B_1B_2\parallel DF$, $b_1b_2=df$. As $\frac{d+e}2$ lies on $B_1B_2$, we have that $b_1+b_2=\frac{d+e}2+\overline{\frac{d+e}2}df=\frac{(d+e)(e+f)}{2e}$. Note that the midpoint of the $D$-altitude is given by $\frac{d+\frac{d+e+f-\frac{ef}d}2}2=\frac{3d^2+de+df-ef}{4d}$. We have that $B=\frac{2df}{d+f}$ and that the circumcenter of $BB_1B_2$ is given by \[\frac{\begin{vmatrix}b&b\overline{b}&1\\b_1&b_1\overline{b_1}&1\\b_2&b_2\overline{b_2}&1\end{vmatrix}}{\begin{vmatrix}b&\overline{b}&1\\b_1&\overline{b_1}&1\\b_2&\overline{b_2}&1\end{vmatrix}}=\frac{\begin{vmatrix}\frac{2df}{d+f}&\frac{4df}{(d+f)^2}&1\\b_1&1&1\\b_2&1&1\end{vmatrix}}{\begin{vmatrix}\frac{2df}{d+f}&\frac2{d+f}&1\\b_1&\frac1{b_1}&1\\b_2&\frac1{b_2}&1\end{vmatrix}}\]as $B_1,B_2$ lie on the unit circle. Expansion by minors gives the numerator as $\frac{4df}{(d+f)^2}(b_2-b_1)+(b_1-b_2)=(b_1-b_2)\cdot\frac{(d-f)^2}{(d+f)^2}$ and the denominator as $\frac{2df}{d+f}\cdot\frac{b_2-b_1}{b_1b_2}+\frac2{d+f}(b_2-b_1)+\frac{(b_1-b_2)(b_1+b_2)}{b_1b_2}$ $=(b_1-b_2)\cdot\left(\frac{b_1+b_2}{b_1b_2}-\frac2{d+f}-\frac{2df}{b_1b_2(d+f)}\right)$ $=(b_1-b_2)\cdot\left(\frac{(d+e)(e+f)}{2def}-\frac4{d+f}\right)=(b_1-b_2)\cdot\frac{(d+e)(e+f)(d+f)-8def}{2def(d+f)}$. Hence, the circumcenter of $BB_1B_2$ is $\frac{2def}{(d+e)(e+f)(f+d)-8def}\cdot\frac{(d-f)^2}{d+f}$. Similarly, the circumcenter of $CC_1C_2$ is $\frac{2def}{(d+e)(e+f)(f+d)-8def}\cdot\frac{(d-e)^2}{d+e}$. It suffices to show that the vector between the circumcenters, $\frac{2def}{(d+e)(e+f)(f+d)-8def}\left(\frac{(d-e)^2}{d+e}-\frac{(d-f)^2}{d+f}\right)$ is perpendicular to the vector between the midpoint of $EF$ and the midpoint of the $D$-altitude, $\frac{3d^2+de+df-ef}{4d}-\frac{e+f}2=\frac{3d^2-de-df-ef}{4d}$. Throwing away real factors such as 4 and $\frac{2def}{(d+e)(e+f)(f+d)-8def}$, it suffices to show that $\frac{\frac{(d-e)^2}{d+e}-\frac{(d-f)^2}{d+f}}{\frac{3d^2-de-df-ef}{d}}$ is the negative of its conjugate. We have that the expression is $\frac{\frac{(f-e)(3d^2-de-df-ef)}{(d+e)(d+f)}}{\frac{3d^2-de-df-ef}{d}}=\frac{d(f-e)}{(d+e)(d+f)}$, and the result is now obvious. Note: We use here the incredibly clean identity that if $x,y,z$ are points with $y,z$ on the unit circle, then the circumcenter of $XYZ$ is $\frac{yz(1-x\overline{x})}{y+z-x-\overline{x}yz}$.

v_Enhance wrote: Note $G$ is the symmedian point of $DEF$, so it is well-known that $XG$ passes through the midpoint of $DK$. Do you have a proof of this lemma without Barycentric Coordinate?

MarkBcc168 wrote: v_Enhance wrote: Note $G$ is the symmedian point of $DEF$, so it is well-known that $XG$ passes through the midpoint of $DK$. Do you have a proof of this lemma without Barycentric Coordinate? https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2013-04/lets_talk_about_symmedians.pdf You can find the proof here. There is 2 proofs but my proof was using menelaus 2 times.

Let $\omega$ be the incircle, $\omega_b = (BB_1B_2)$ and $\omega_c = (CC_1C_2)$. Let $N_a, N_b, N_c, R_b, R_c$ denote the midpoints of $\overline{EF}, \overline{FD}, \overline{DE}, \overline{B_1B_2}, \overline{C_1C_2}$ respectively. Let $T_b = BI \cap \omega_b \neq B$, and $T_c = CI \cap \omega_c \neq C$. Claim 1: $BI$ is the perpendicular bisector of $\overline{B_1B_2}$. $CI$ is the perpendicular bisector of $\overline{C_1C_2}$. Proof: By incentre-excentre, $M_b$ is the centre of $(AIB)$, so $M_b$ lies on the perpendicular bisector of $\overline{B_1B_2}$. Similarly, $\overline{B_1B_2}$ is a chord in $\omega$, so $I$ does too. Since $M_b - I - B$, $BI$ is the perpendicular bisector of $\overline{B_1B_2}$. An analogous proof suffices for $CI$. $\square$ Claim 2: $B_2 - N_c - N_a - B_1$, $C_2 - N_b - N_a - C_1$. Proof: By radical centre on $\omega, (AIC), (AFIE)$, we find that $B_1B_2, AI, EF$ concur. Thus, $N_a \in B_1B_2$. We also have $B_1B_2 \perp BI$ by Claim 1, yet $BI \perp DF \parallel N_aN_b$ since $BI$ is the perpendicular bisector of $\overline{DF}$ and $N_aN_b$ is the $E$-midline in $\triangle DEF$. Thus, $B_1B_2 \parallel N_aN_b$ and $B_2 - N_c - N_a - B_1$ holds. Analogously, $C_2 - N_b - N_a - C_1$. $\square$ Consider $P(X) = Pow_{\omega_b}(X) - Pow_{\omega_c}(X)$. We show $P(M) = 0$ via linearity. Note that by Claim 2, $N_b$ is the radical centre of $(BDIF), \omega, (AIB)$. Thus, \begin{align*} Pow_{\omega_b}(N_b) &= N_bB \cdot N_bT_b \\ Pow_{\omega_C}(N_b) &= N_bC_1 \cdot N_bC_2 = Pow_{(AIB)}(N_b) \\ &= Pow_{(BDIF)}(N_b) = N_bI \cdot N_bB \\ \Rightarrow P(N_b) &= N_bB \cdot (N_bT_b - N_bI) \\ &= N_bB \cdot IT_b \end{align*}and analogously, \[P(N_c) = - N_cC \cdot IT_c.\]$M$ lies on $\overline{N_bN_c}$. Hence, by linearity it suffices to show: \begin{align*} MN_c \cdot P(N_b) + MN_b \cdot P(N_c) &= 0 \\ \iff MN_c \cdot N_bB \cdot IT_b &= MN_b \cdot N_cC \cdot IT_c. \end{align*}Note that $\angle DN_cM = \angle DEK = 90 - \hat{B}/2 = \angle BDN_b$, and $\angle N_cMD = 90 = \angle DN_bB$. Thus, $\triangle DN_cM \sim \triangle BDN_b$ and $\frac{MN_c}{MD} = \frac{N_bD}{N_bB} \Rightarrow MN_c \cdot N_bB = MD \cdot N_bD = MD \cdot DF / 2$. Hence, by symmetry it suffices to show: \[DF \cdot IT_b = DE \cdot IT_c.\] Now a few computations: \begin{align*} R_bT_b \cdot R_bB &= Pow_{\omega_b}(R_b) \\ &= R_bB_1 \cdot R_bB_2 = R_bB_1^2 \\ &= IB_1^2 - IR_b^2 \\ \therefore R_bT_b &= \frac{r^2 - IR_b^2}{R_bB} \\ \Rightarrow IR_b + R_bT_b &= IR_b \cdot (1 - \frac{IR_b}{R_bB}) + \frac{r^2}{R_bB} \\ \Rightarrow IT_b &= IR_b \cdot \frac{IB}{R_bB} + \frac{r^2}{R_bB} \\ IB \cdot IN_b &= ID^2 \\ \therefore IT_b &= IR_b \cdot \frac{r^2}{IN_b \cdot R_bB} + \frac{r^2}{R_bB} \end{align*}By symmetry, it now suffices to show: \begin{align*} DF \cdot (IR_b \cdot \frac{r^2}{IN_b \cdot R_bB} + \frac{r^2}{R_bB}) &= DE \cdot (IR_c \cdot \frac{r^2}{IN_c \cdot R_cC} + \frac{r^2}{R_cC}) \\ DF \cdot R_cC \cdot IN_c \cdot (IR_b + IN_b) &= DE \cdot R_bB \cdot IN_b \cdot (IR_c + IN_c) \\ \iff \frac{DE}{DF} \cdot \frac{R_bB}{R_cC} \cdot \frac{IN_b}{IN_c} &= \frac{R_bN_b}{R_cN_c}. \end{align*}However, note that $\angle N_cR_cN_b = 90 = \angle N_cR_bN_b$, so $R_bR_cN_bN_c$ cyclic and $R_bR_c$ anti-parallel to $N_bN_c$. Furthermore, $IN_b \cdot IB = ID^2 = IN_c \cdot IC$, so $N_bBCN_c$ cyclic and $BC$ anti-parallel to $N_bN_c$. Thus, $R_bR_c \parallel BC$, and $\frac{R_bB}{R_cC} \cdot \frac{IN_b}{IN_c}= \frac{IB}{IC} \cdot \frac{IN_b}{IN_c} = 1$. Furthermore, $\triangle DEF$ shares base $\overline{DE}$ with $\triangle DR_cE$, and has double the height since $R_c$ lies on the $F$-midline. Thus, $DE \cdot R_cN_c = 2 \cdot [DER_c] = [DEF] = 2 \cdot [DFR_b] = DF \cdot R_bN_b$. $\blacksquare$

Let $\ell$ be the radical axis of $(BB_1B_2)$ and $(CC_1C_2)$. Claim: The midpoint $M_{EF}$ of $EF$ lies on $\ell$. Proof. Note that $AFIE$ is cyclic, and thus we consider the radical center of the three circles $(AFIE)$, $(AIB)$, and the incircle of $\triangle ABC$. Clearly these are coaxial with common radical axis of $\overline{AI}$. Since $M_{EF}$ lies on $AI$, it must be the radical center of the circles, and so it also lies on the radical axis of $(AIB)$ and the incircle, i.e.\ $\overline{C_1C_2}$. Analogously, $M_{EF} \in \overline{B_1B_2}$, so $M_{EF} = \overline{B_1B_2} \cap \overline{C_1C_2}$. Thus, considering the radical center of $(AIB)$, $(AIC)$, and the incircle, $M_{EF}$ lies on $\ell$, and we conclude. Now consider the Gergonne point $G$ of $\triangle ABC$. Claim: $G$ lies on $\ell$. Proof. By the prior claim, the midpoint $M_{DE}$ of $DE$ is precisely the intersection $\overline{B_1B_2} \cap \overline{IC}$, and similarly $M_{DF} = \overline{C_1C_2} \cap \overline{IB}$. Thus, considering the radical center of $(BIC)$, $(BB_1B_2)$, $(CC_1C_2)$, $\overline{BZ}$, $\overline{CY}$, and $\ell$ concur. To that end, use cevian nest on $\triangle CGB$, $\triangle DEF$, $\triangle M_{EF}M_{DF}M_{DE}$ to realize that $\overline{GX}$, $\overline{BZ}$, $\overline{CY}$, and $\ell$ concur, which implies that $G$ lies on $\ell$, as desired. Note that $G$ is also the Lemoine point of $\triangle DEF$, so we are done by Iran lemma.

Nice! Here's a sketch for a clean bary solution. So by radax and a couple of times we get that the midpoint $M$ of $EF$ gives that $M$ is on $B_1B_2$ and $C_1C_2$. Then by bary we get that the midpoint of the altitude is collinear with $M$ and the symmedian point of $DEF$ which I label $K$. Now let the coefficients $u$, $v$, and $w$ in the barycentric circle formula for $\omega_B$ be $b_1$, $b_2$, $b_3$ (wrt the intouch triangle). Define similarly for $A$ and $C$. Then since $M$ is on the radax of $\omega_B$ and the incircle, we get $b_2+b_3=0$. Similarly $b_1+b_2=0$ by repeating the argument symmetrically for $A$. Now the circle is of the form $a^2yz+b^2xz+c^2xy=k(x+y+z)(x-y+z)$ for some constant $k$. Now we can just compute $k$ because of the fact that $B=(a^2:-b^2:c^2)$ is on the circle. From here we can directly compute the power of $K$ as $\frac{-4a^2b^2c^2}{(a^2+b^2+c^2)^2}$ and take the W by symmetry.

A certain individual has claimed that this problem follows by Cayley-Bacharach. I have not verified this, but apparently one of the cubics is the Thomson cubic of $DEF$.

Holy hell. Solution path is basically almost the same as Anant Mudgal's but I lost my train of thought and ended up proving something else which also finished the problem when instead I could've just gotten a cyclic quadrilateral. Oops :sob: Firstly from inversion, we obtain that the $B_1B_2$ and $C_1C_2$ are the $E$ and $F$ midlines of $\triangle DEF$. Let $(BB_1B_2)$ intersect $AB$ and $AC$ again at points $X_1$ and $X_2$ respectively. Define $Y_1$ and $Y_2$ similarly but for $(CC_1C_2)$. We let $P$ and $Q$ lie on $(BB_1B_2)$ and $(CC_1C_2)$ such that $BQ \parallel DE$ and $CQ \parallel DF$. Remark then that $BP \cap CQ$ is the orthocenter of $BIC$, which is very reminiscent of a certain Iran problem... (here's where i started diverging) In any case, observe that \[ \providecommand{\dang}{\measuredangle} \dang PX_1X_2 = \dang PBX_2 \dang EDC \qquad \dang QY_1Y_1 = \dang QCY_2 = \dang FDB, \]and moreover $X_1X_2 \parallel DF$ and $Y_1Y_2 \parallel DE$ by symmetry. From this we observe that the quadrilateral formed by the lines $PX_1$, $QY_1$, $X_1X_2$, and $Y_1Y_2$ must be a triangle and hence $PQX_1Y_1$ is a line. It's also not hard to angle chase that \[ \providecommand{\dang}{\measuredangle} \dang Y_1X_1X_2 = \dang PX_1X_2 = \dang PBX_2 = \dang Y_1Y_2X_2 \]which shows that $X_1X_2 \cap Y_1Y_2$ lies on the radical axis. The fun part now: $\textbf{Claim.}$ We have that $BE \cap X_1X_2$ is the Gergonne point of $\triangle ABC$. $\textit{Proof.}$ Let $B_1B_2$ intersect $AB$ again at a point $T$. Observe that $T$ is the radical center of the three circles $(BB_1B_2)$, $(DEF)$, and $(F)$ where $F$ is a point circle, and so $TA^2 = TX_2 \cdot TB$. Let $F'$ be the reflection of $F$ over $T$, whereby we have by orthogonal circles \[ -1 = (B, X_1; F, F') \overset{DF \cap X_1X_2}{=} (B, X_1X_2 \cap BE, DF \cap BE, E) \overset{D}{=} (D, ?, F, E), \]where the $?$ is the harmonic conjugate of $D$ with respect to $\triangle DEF$ on the circumcircle. It follows then that $BE \cap X_1X_2$ is the symmedian point of $\triangle DEF$, or equivalently the Gergonne point of $\triangle ABC$. We are done by the $D$-Schwatt line in $\triangle DEF$.

PoP saves the day Let $\omega = (DEF), \omega_B = (AIC)$, $\omega_C = (AIB)$, $t_B = (BB_1B_2)$ and $t_C = (CC_1C_2)$, $AE=AF=x$, $BF=BD=y$, $CD=CE=z$. The idea is to avoid having to deal with $B_1B_2$ by appealing to the fact that $\omega, \omega_B$ and $t_B$ are coaxial. Observe that the reflections of $A$ and $C$ across $BI$ also lie on $\omega_B$ and so on, which lets us compute all the required powers easily. Since $\text{Pow}(B,\omega) = y^2$ and $\text{Pow}(B, \omega_B) = (y+x)(y+z)$, we have \[ \text{Pow}(X,t_B) = \frac{ (y+x)(y+z) \text{Pow}(X,\omega) - y^2 \text{Pow}(X, \omega_B) }{ (y+x)(y+z) - y^2 } = \frac{ (y+x)(y+z) \text{Pow}(X,\omega) - y^2 \text{Pow}(X, \omega_B) }{ \sum xy } \]since it has leading coefficient $1$ when viewed as a quadratic function, and also vanishes when $X = B, B_1, B_2$. Hence define the linear function \[ f(X) = \text{Pow}(X,t_B) - \text{Pow}(X,t_C) = \frac{ (y^2 - z^2) \text{Pow}(X,\omega) - y^2 \text{Pow}(X, \omega_B) + z^2 \text{Pow}(X,\omega_C) }{ \sum xy } \]Now we just need to compute $f(M)$. It is easy to show that $FK : KE = y : z$. Now \[ f(E) = \frac{ 0 - y^2 (-xz) + z^2 (xy) }{ \sum xy } = \frac{ xyz (y+z) }{ \sum xy }, f(F) = \frac{ xyz (-y-z) }{ \sum xy }\]so \[ f(K) = \frac{xyz}{\sum xy} \left( \frac{y(y+z)+z(-y-z)}{y+z} \right) = \frac{xyz}{\sum xy} (y - z) \]and on the other hand, \[ f(D) = \frac{ 0 - y^2 (xz) + z^2 (xy) }{\sum xy} = \frac{xyz}{\sum xy} (-y + z) \]so $f(M) = (f(K) + f(D))/2 = 0$ and we are done.

Let the medial triangle of contact triangle $\triangle DEF$ be $\triangle M_1M_2M_3$. We will show that the radical axis of $(BB_1B_2)$ and $(CC_1C_2)$ contains the Gergonne point $G$ of $\triangle ABC$ and the midpoint $M_1$ of $EF$. By Schwatt line, $GM_1$ also contains the midpoint of $DK$. Note that $AD$, $BE$ and $CF$ concur at G, and that $M_1 = AI \cap EF$. By Rad Axis on $(AFIE)$, $(ACI)$ and $(DEF)$, we get that $AI$, $C_1C_2$, and $EF$ concur at $M_1$ and similarly $B_1B_2$ also passes through $M_1$. Then $\text{Pow}_{(DEF)}(M_1) = M_1B_1 \cdot M_1B_2 = M_1C_1 \cdot M_1C_2$ so $M_1$ lies on the radical axis of $(BB_1B_2)$ and $(CC_1C_2)$. Notice that $\text{Pow}_{(AIC)}(M_3) = M_3B_2 \cdot M_3B_1 = M_3I \cdot M_3C$ so $BM_3$ is the radical axis of $(BIC)$ and $(BB_1B_2)$. Similarly, $CM_2$ is the radical axis of $(BIC)$ and $(CC_1C_2)$, so $CM_2$, $BM_3$ and the radical axis in question concur. Notice that $BE$, $CF$ and $DA$ concur at $G$, $DM_1$, $EM_2$, and $FM_2$ concur so by Cevian Nest $GM_1$, $BM_2$ and $CM_3$ concur on the desired radical axis. Thus, both $M_1$ and $G$ lie on the radical axis, so we are done.

First we claim that $\overline{C_1C_2}$ passes through the midpoint of $\overline{EF}$, say $N$. Note that by radax on $(AEF)$, $(ABC_1C_2)$ and $(DEF)$ we find that $\overline{C_1C_2}$, $\overline{AI}$ and $\overline{EF}$ concur, so $C_1C_2$ passes through $N$. Similarly so does $\overline{B_1B_2}$. Now by power of a point we find that $$B_1N \cdot B_2N = AN \cdot IN = C_1N \cdot C_2N$$and hence $N$ lies on the radax of $(BB_1B_2)$ and $(CC_1C_2)$. Now define $G = \overline{BE} \cap \overline{CF}$ as the Gergonne point of $\triangle AB C$. We claim that $G$ also lies on the radax of $(BB_1B_2)$ and $(CC_1C_2)$. To see this we will use barycentric coordinates. Set $\triangle DEF$ as reference and note that then $G$ is the symmedian of $(DEF)$. Then we have, \begin{align*} G &= (a^2 : b^2 : c^2)\\ B &= (a^2 : -b^2 : c^2)\\ \end{align*}Lets say now that $(BB_1B_2)$ has equation, \begin{align*} (BB_1B_2)& \colon -a^2yz - b^2xz - c^2xy + (u_1x + v_1y + w_1z)(x+y+z) = 0\\ (DEF)& \colon -a^2yz - b^2xz - c^2xy = 0 \end{align*}Then their pairwise radical axes has equation, \begin{align*} B_1B_2 &\colon u_1 x + v_1y + w_1z = 0\\ \end{align*}But also $B_1B_2$ is just the $E$ midline so we have, \begin{align*} u_1x + v_1y + w_1z = k(x - y + z) \end{align*}for some constant $k$. Thus we can parametrize, \begin{align*} (BB_1B_2) \colon -a^2yz - b^2xz - c^2xy + k_1(x - y + z)(x + y + z) &= 0\\ \end{align*}Then we can actually compute this constant! Plugging in $B$ into the first equation we have, \begin{align*} k_1 &= -\frac{a^2b^2c^2}{(a^2 + b^2 + c^2)(a^2 - b^2 + c^2)} \end{align*}Thus we have the equation of $(BB_1B_2)$ is given by, \begin{align*} -a^2yz - b^2xz - c^2xy - \frac{a^2b^2c^2}{(a^2 + b^2 + c^2)(a^2 - b^2 + c^2)}(x + y + z)(x - y + z) &= 0 \end{align*}Now to compute $\text{Pow}_{(BB_1B_2)}(G)$ we will plug in $G = \left(\frac{a^2}{s}, \frac{b^2}{s}, \frac{c^2}{s} \right)$ where $s = a^2 + b^2 + c^2$. We find, \begin{align*} \text{Pow}_{(BB_1B_2)}(G) &= -3a^2b^2c^2/s^2 - \frac{a^2b^2c^2}{(a^2 + b^2 + c^2)(a^2 - b^2 + c^2)}(a^2 + b^2 + c^2)(a^2 - b^2 + c^2)/s^2\\ &= \frac{-4a^2b^2c^2}{s^2} \end{align*}which is symmetric in $B$ and $C$ so we are done. Thus $\overline{GN}$ is the radical axis of $(BB_1B_2)$ and $(CC_1C_2)$. Now we check that $N = (0: 1 : 1)$, $G = (a^2 : b^2 : c^2)$ and, $K = (a^2: S_C :S_B)$ the midpoint of the $D$ altitude are collinear. Indeed note that $\overline{NG}$ has equation, \begin{align*} (c^2 - b^2)x + a^2(y - z) = 0\\ \end{align*}Then plugging in $K$ we have, \begin{align*} a^2(c^2 - b^2) + a^2(S_C - S_B) &= 0\\ \iff S_C - S_b &= b^2 - c^2\\ \iff 0 &= 0 \end{align*}and thus we are done. $\square$ Remark: The really neat trick in this solution was figuring out a way to get the circles equation by noting that we had the equation of the radax of our desired circle and a known circle, as well as a point on the desired circle. Then we can actually compute really nicely the equation of the circle using radax up to a constant, and then finish on that. Main takeaway is we can determine the ratio of constants for some circle, given another circle and the radax equation.

Let $G$ be the Gergonne point of $\triangle ABC,$ which by definition is the Symmedian point of $\triangle DEF.$ Let $X,Y,Z$ be the midpoints of sides $\overline{EF},\overline{FD},\overline{DE},$ respectively. Thus, we need only show that $\overline{XG}$ is on the radical axis as $MXG$ is the $D$-Schwatt line of $\triangle DEF.$ Let $\overline{XZ}\cap \overline{AC}=W,$ and $\overline{XY} \cap \overline{AB}=V.$ Then let $\overline{BW} \cap \overline{CV}=T.$ Claim: $\overline{GXT}$ is collinear. Proof: First, by Cevian Nest on triangles $\triangle GBC, \triangle DEF, \triangle XYZ,$ we get that $\overline{BZ}\cap \overline{CY}$ is on $\overline{XG}.$ By Cevian nest again, $\overline{AX}, \overline{BY}, \overline{CZ}$ are concurrent, so $\triangle BYV$ and $\triangle CZW$ are perspective by Descargues. Finally, $\triangle CYV$ and $\triangle BZW$ are perspective from the same point that $\triangle BYV$ and $\triangle CZW$ is, so by Descargues $T,X,\overline{BZ} \cap \overline{CY}$ is collinear, so $\overline{XGT}$ is collinear. \end{proof} Note that $\overline{AI}\cap \overline{EF}=X.$ As $(AEIF)$ is concyclic, note that $X$ must be the radical center of $(AEIF),(DEF),(AIC),(AIB).$ Note that by Incenter Excenter lemma, $\overline{BI}$ must have the center of $(AIC)$ on it. Thus, as $I$ is the center of $(DEF),$ $\overline{BI}\perp \overline{B_1B_2},$ and as $\overline{XY}\| \overline{DF},$ we have that $Z \in \overline{B_1B_2},$ and similarly for $Y.$ Therefore as $W=\overline{B_1B_2}\cap \overline{AC},$ then $W$ must be the radical center of $(BB_1B_2),(ABC),(AIC),$ so $\overline{BW}$ goes through both of $(ABC)\cap (BB_1B_2).$ Same thing for $V$ gets that $\overline{CV}$ goes through both of $(ABC) \cap (CC_1C_2).$ Thus, as $T$ lies on the radical axis of $(ABC), (BB_1B_2)$ and $(ABC),(CC_1C_2),$ it must be the radical center of $(ABC),(BB_1B_2),(CC_1C_2),$ so it must also be on the radical axis of $(BB_1B_2), (CC_1C_2),$ which finishes.$\blacksquare$

Let $M,N,P$ be the midpoints of $EF,FD,DE$ and let $X,Y$ be the midpoints of $B_1B_2$ and $C_1C_2$. Note that $B_1,P,M,B_2$ and $C_1,N,M,C_2$ are collinear by inversion about the incircle, so $M$ has equal power WRT $(BB_1B_2)$ and $(CC_1C_2)$. Now, by linearity of power of a point we have $$\text{pow}_{(BB_1B_2)}(N) - \text{pow}_{(DEF)}(N) = \frac{XN}{XB}\cdot (\text{pow}_{(BB_1B_2)}(B) - \text{pow}_{(DEF)}(B)) = \frac{XN}{XB}(-BD^2).$$ Hence $$\text{pow}_{(BB_1B_2)}(N) - \text{pow}_{(CC_1C_2)}(N) = \frac{XN}{XB}(-BD^2) - ND\cdot NF + NC_1\cdot NC_2 = \frac{XN}{XB}(-BD^2).$$ Then $$\frac{\text{pow}_{(BB_1B_2)}(N) - \text{pow}_{(CC_1C_2)}(N)}{\text{pow}_{(BB_1B_2)}(P) - \text{pow}_{(CC_1C_2)}(P)} = \frac{\frac{XN}{XB}(-BD^2)}{\frac{YP}{YC}(-CD^2)} = \frac{\tan{\angle DEF}}{\tan{\angle DFE}}$$ since $XY\parallel BC$. Hence the midpoint of $DK$ also lies on the radical axis by linearity of power of a point, as desired.

I would like to introduce this paper written by my student and myself on this interesting and challenging problem. I have always regarded this as one of the model problems when discussing power, radical axis and radical center. Thank you very much for your attention.

Let $M,N,P$ be midpoints of $EF,FD,DE$. $PD\cdot PE=PI\cdot PC$ so $P$ lies on radical axis of $(I)$ and $(AIC)$, so we get that $B_2,P,B_1$ collinear. We get the same thing for other midpoints. $MC_1\cdot MC_2=MB_1\cdot MB_2$ so $M$ lies on radical axis of $(BB_1B_2),(CC_1C_2)$. Now let $N_1=CN\cap (CC_1C_2),P_1=BP\cap (BB_1B_2),Q=CN\cap BP$. $NN_1\cdot NC=NC_1\cdot NC_2=ND\cdot NF=NB\cdot NI$ so $BN_1IC$ cyclic. Analogously $BIP_1C$ cyclic, hence $BN_1IP_1C$ cyclic. $QN_1\cdot QC=QP_1\cdot QB$ so $Q$ also lies on radical axis of $(BB_1B_2),(CC_1C_2)$. It remains to show that $MQ$ bisects $DK$. Let $R$ be midpoint of $DK$, $U=MB\cap NP,V=MC\cap NP,R'=MQ\cap NP,J=MI\cap NP$. By DDIT on $M$ and $INQP$ we get that there is an involution which swaps $(MB,MC),(MN,MP),(MI,MQ)$. Projecting it through $M$ onto $NP$ we get that $(U,V),(N,P),(J,R')$ are involutive pairs which is equivalent to $(MUV),(MNP),(MJR')$ are coaxial. $NPMK$ is cyclic because it is $\triangle DEF$ nine point circle. Now let $G=EF\cap BC$, since $(B,C;G,D)=-1$ and $\angle GKD=90$ we get that $\angle FKB=\angle CKE$. $UV\parallel EF\implies MU=MV$ so $KMVU$ cyclic. Hence $KMJR'$ is cyclic because of coaxial circles. This means $\angle R'KM=\angle MJR=90$ so $R'\in DK$. As $NP$ is a midline $R'K=R'D$ so $R'=R$ as desired.

Let the medial triangle of $\triangle DEF$ be $\triangle M_aM_bM_c$. Define $A_1A_2$ similarly because I like to define the shape from $A$. Claim: $A_1A_2$ is $D-$midline in $\triangle DEF$ Proof Define $M_bM_c \cap BC=Y_a$. From $Y_aB.Y_aC=Y_aD^2=Y_aM_b.Y_aM_c$, $Y_a$ lies on the radical axis of $(DEF),(ABC)$ Radical axis theorem on $(DEF),(ABC),(BIC)$ gives $Y_a-A_1-A_2$. Claim:$M_a$ passes through radical axis $(BB_1B_2),(CC_1C_2)$ Proof obvious. Let $CM_b\cap BM_c=Q$ Claim: $Q$ passes through radical axis $(BB_1B_2),(CC_1C_2)$ Proof Define $CM_b\cap (CC_1C_2)=B'$ and $BM_c \cap (BB_1B_2)=C'$ $CM_b.M_bB'=M_bC_1.M_bC_2=M_bF.M_bD=M_bB.M_bI$ so $B,I,C,B'$ cyclic. Similarly $C'$ lies on this circle too. Define midpoint of $DK$ is $Y$ Final Claim:$Y-Q-M_a$ DDIT on $M_a$ to $BM_bCM_c$ and project it to $M_bM_c$ Let $M_aB\cap M_bM_c=X_b$ and $M_aC \cap M_bM_c=X_c$ From $-1=(M_aF,M_aD;M_aB,M_aC)$ we get $M_aD$ passes through midpoint of $B'C'$. So the involution on line is reflection to midpoint of $M_bM_c$. One of the pairs of involution is $(M_aI),(M_aQ)$ so $Y-Q-M_a$.

If $N$ is the midpoint of $EF$ then $NE\cdot NF=NA\cdot NI$ so $N$ is the radical center of the incircle, $(AIB)$ and $(AIC)$ so $N=B_1B_2\cap C_1C_2$ and since $B_1B_2C_1C_2$ cyclic $N$ is on the radical axis of $(BB_1B_2)$ and $(CC_1C_2)$. Let $G$ be the gergonne point and choose $R,S\in BC,P,T\in AB,Q,U\in AC$ with $PR\cap QS\cap TU=G$ and $PR\parallel DF,QS\parallel DE,TU\parallel EF$. Now $PG\parallel FD,GQ\parallel DE,PF\cap GD\cap QE=A$ so $\triangle PGQ,\triangle FDE$ are homothetic and $PQ\parallel EF$. Similarly $ST\parallel DF,RU\parallel DE$ so $FP=DR=EU=FT=DS=EQ$. In particular the perpendicular bisectors of $PQ,QU,UR,RS,ST,TP$ all pass through $I$, so $(PQRSTU)$ is cyclic. Now notice $(AIC)$ passes through the reflections $A',C'$ of $A,C$ over $BI$. Let $L=BE\cap DF$ and $X=AD\cap(DEF)$ and we have \[\frac{\frac{PF^2}{PA\cdot PC'}}{\frac{BF^2}{BA\cdot BC'}}=\frac{\frac{RD^2}{RA'\cdot RC}}{\frac{BD^2}{BA'\cdot BC}}=\frac{\frac{QE\cdot UE}{AQ\cdot CU}}{\frac{BD\cdot BF}{BA\cdot BC}}=\frac{\frac{GD}{AG}\cdot\frac{GF}{CG}}{\frac{[FBD]}{[ABC]}}=\frac{\frac{[DGF]}{[AGC]}}{\frac{[FBD]}{[ABC]}}=\frac{[DGF]}{[FBD]}\cdot\frac{[ABC]}{[AGC]}=\frac{LG}{BL}\cdot\frac{BE}{GE}=-(G,B;L,E)\stackrel D=-(X,D;F,E)=1,\]so forgotten coaxiality lemma gives $(BPR)$ is coaxial to the incircle and $(AIC)$ and thus is the same as $(BB_1B_2)$. Similarly, $QS$ is the radical axis of $(CC_1C_2)$ and $(PQRS)$, so the radical center $G$ of $(BB_1B_2),(CC_1C_2),(PQRS)$ lies on the radical axis of $(BB_1B_2),(CC_1C_2)$ as well. By Schwatt line of the intouch triangle, $N,G,M$ are collinear, so $M$ lies on the radical axis as desired.

Let $\omega$ be the incircle. Let $\omega_B = (BB_1B_2)$ and $\omega_C = (CC_1C_2)$. Let $A', B', C'$ be the midpoints of $EF$, $FD$, and $DE$, respectively. By inversion about $I$, $B_1$ and $B_2$ lie on $A'C'$, while $C_1$ and $C_2$ lie on $A'B'$. Let $T$ be the projection from $B$ onto $A'C'$. Then, $T_B$ is on the radical axis of $\omega_B$ and $\omega$. Thus, \[\text{Pow}_\omega(T_B) - \text{Pow}_{\omega_B}(T_B) = 0.\] Let $A'$, $B'$, and $C'$ denote angles $\angle C'A'B'$, $\angle A'B'C'$, and $\angle B'C'A'$. Let $r$ be the radius of $\omega$. By linpop, because $B_1B_2$ is the radical axis of $\omega$ and $\omega_B$, we have \begin{align*} \frac{\text{Pow}_\omega(M) - \text{Pow}_{\omega_B}(M)}{\text{Pow}_\omega(B) - \text{Pow}_{\omega_B}(B)} &= \frac{[MB_1B_2]}{[BB_1B_2]}\\ &= \frac{MC'\sin C'}{BT}\\ &= \frac{r\sin^2 C'\cos B'}{r\frac{\sin^2 B'}{\cos B'} + r\sin A'\sin C'}\\ &= \frac{\sin^2 C'\cos^2 B'}{\sin^2 B' + \sin A'\sin C'\cos B'}\\ &= \frac{\sin^2 C'\cos^2 B'}{\sin^2 B' + (\sin A'\sin C'-\cos A'\cos C')\cos B'+\cos A'\cos B'\cos C'}\\ &= \frac{\sin^2 C'\cos^2 B'}{\sin^2 B' - \cos(A'+C')\cos B'+\cos A'\cos B'\cos C'}\\ &= \frac{\sin^2 C'\cos^2 B'}{1+\cos A'\cos B'\cos C'}\\ \frac{\text{Pow}_\omega(M) - \text{Pow}_{\omega_B}(M)}{\text{Pow}_\omega(B) - \text{Pow}_{\omega_B}(B)} &= \frac{\sin^2 C'\cos^2 B'}{1+\cos A'\cos B'\cos C'}\\ \text{Pow}_\omega(M) - \text{Pow}_{\omega_B}(M) &= r^2 \frac{\sin^2 B'}{\cos^2 B'}\frac{\sin^2 C'\cos^2 B'}{1+\cos A'\cos B'\cos C'}\\ \text{Pow}_\omega(M) - \text{Pow}_{\omega_B}(M) &= r^2 \frac{\sin^2 C'\sin^2 B'}{1+\cos A'\cos B'\cos C'}. \end{align*}So, \[\text{Pow}_{\omega_B}(M) = \text{Pow}_\omega(M) - r^2 \frac{\sin^2 C'\sin^2 B'}{1+\cos A'\cos B'\cos C'} = \text{Pow}_{\omega_C}(M).\]