Nice problem (although easy for a USA TST P6): [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -14.75, xmax = 11.45, ymin = -3.79, ymax = 7.45; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen wwzzff = rgb(0.4,0.6,1); pen ffxfqq = rgb(1,0.4980392156862745,0); pen qqccqq = rgb(0,0.8,0); /* draw figures */ draw((-4.781523029254206,2.683377195388904)--(6.871149068786946,5.218374928896586), linewidth(0.4) + wrwrwr); draw(circle((0.7015597483382036,1.4355753715441475), 2.546602274766395), linewidth(0.4) + blue); draw((-4.781523029254206,2.683377195388904)--(2.0614153440904808,-0.7175562789934126), linewidth(0.4) + wrwrwr); draw((2.0614153440904808,-0.7175562789934126)--(6.871149068786946,5.218374928896586), linewidth(0.4) + wrwrwr); draw((-0.7268440288168109,-0.6727049217884695)--(-0.66,3.58), linewidth(0.4) + wrwrwr); draw((-7.217533164119876,-0.5682970536602701)--(-2.76,-0.64), linewidth(0.4) + wrwrwr); draw((-7.217533164119876,-0.5682970536602701)--(-0.09465334849420755,0.9812603385617286), linewidth(0.4) + wrwrwr); draw(circle((-0.9294856026840693,-10.516443266922067), 11.768821318012856), linewidth(0.4) + linetype("2 2") + wwzzff); draw((-7.217533164119876,-0.5682970536602701)--(-0.66,3.58), linewidth(0.4) + wrwrwr); draw(circle((-3.943883513499975,1.5047383027747945), 3.8846624103269236), linewidth(0.4) + ffxfqq); draw((-0.66,3.58)--(-2.76,-0.64), linewidth(0.4) + qqccqq); draw((-2.76,-0.64)--(4.7,-0.76), linewidth(0.4) + qqccqq); draw((4.7,-0.76)--(-0.66,3.58), linewidth(0.4) + qqccqq); draw((-3.9387665820599382,1.505851473169865)--(3.0875794770264293,0.545579304049496), linewidth(0.4) + wrwrwr); /* dots and labels */ dot((-0.66,3.58),dotstyle); label("$A$", (-0.69,3.87), NE * labelscalefactor); dot((-2.76,-0.64),dotstyle); label("$B$", (-3.15,-1.15), NE * labelscalefactor); dot((4.7,-0.76),dotstyle); label("$C$", (4.93,-1.19), NE * labelscalefactor); dot((-0.09465334849420755,0.9812603385617286),linewidth(4pt) + dotstyle); label("$I$", (0.15,0.57), NE * labelscalefactor); dot((6.871149068786946,5.218374928896586),linewidth(4pt) + dotstyle); label("$I_B$", (7.13,5.43), NE * labelscalefactor); dot((-4.781523029254206,2.683377195388904),linewidth(4pt) + dotstyle); label("$I_C$", (-5.33,2.89), NE * labelscalefactor); dot((-1.8356826708257379,1.2174376805311355),linewidth(4pt) + dotstyle); label("$C_1$", (-2.35,1.71), NE * labelscalefactor); dot((3.0875794770264293,0.545579304049496),linewidth(4pt) + dotstyle); label("$B_1$", (3.09,-0.03), NE * labelscalefactor); dot((2.0614153440904808,-0.7175562789934126),linewidth(4pt) + dotstyle); label("$A_1$", (2.21,-1.05), NE * labelscalefactor); dot((-0.7268440288168109,-0.6727049217884695),linewidth(4pt) + dotstyle); label("$X$", (-0.71,-1.25), NE * labelscalefactor); dot((-7.217533164119876,-0.5682970536602701),linewidth(4pt) + dotstyle); label("$D$", (-7.99,-0.45), NE * labelscalefactor); dot((-3.9387665820599382,1.505851473169865),linewidth(4pt) + dotstyle); label("$M$", (-4.53,1.51), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] From USA TST for EGMO 2019, Problem 5, we get that $I$ lies on $B_1C_1$, and that $\odot (AB_1A_1C_1) \cap BC=X$ is the foot of the $A$-altitude. Let $I_BI_C \cap BC=T$, where $I_B$ and $I_C$ are the $B$ and $C$-excenters. Then, from that same solution, we have that $I_B,B_1,A_1$ and $I_C,C_1,A_1$ are collinear. Thus, $$-1=(T,A;I_B,I_C) \overset{A_1}{=} (X,A;B_1,C_1) \Rightarrow AB_1XC_1 \text{ is harmonic.}$$Suppose the tangents to $\odot (AB_1A_1C_1)$ at $A$ and $X$ meet on $B_1C_1$ at a point $M$. Then, as $I$ is the foot of the $A$-internal angle bisector of $\angle B_1AC_1$, so we get that $\odot (AIX)$ is the $A$-Apollonius circle of $\triangle AB_1C_1$. Also, as $\angle DIA=\angle DXA=90^{\circ}$, so we have that $D$ also lies on this circle centered at $M$, i.e. $MD^2=MA^2=MB_1 \cdot MC_1$. Hence, done. $\blacksquare$

math_pi_rate wrote: easy for a USA TST P6 I agree with you. What did you all think of the problem? When I originally came up with it I thought it was genius, but now I think it's really ugly and not any fun to try and solve. Many others mentioned they thought the problem was nice, so perhaps it's just my aversion to configuration-based geometry at work here...

@above I believe that the configuration is really nice and rich, and problem is also fine. However, considering at TST level, the problem could have been more difficult. Cause after finding that $A_1$ is the Bevan point of $\triangle ABC$ (again not that difficult), the above solution is quite easy to find (Probably saying this because I personally found P1 harder than this ).

$\qquad \qquad$

Here is a sketch of a relatively straightforward solution which doesn't require much thinking after guessing that $A_1$ should be the Bevan point of $\triangle ABC$.

Btw here's another nice property of the given configuration: Let $\omega$ be the circumcircle of $\triangle AC_1D$. Suppose the tangents to $\omega$ at $A$ and $D$ meet at $K$, and let $AB_1 \cap \omega=X, DB_1 \cap \omega=Y,B_1K \cap AD=Z$. Then $MXYZ$ is cyclic, where $M$ is the midpoint of $AD$.

Well, my first drawing confused me, so I will consider $D$ to be be a point on the tangent at $A$ wrt $(AC_1B_1)$ satisfying $\angle AID=90$. Good drawing really helps you see the main steps of the problem( I did 3 sketches with some obtuse triangles to see that $AA_1$ was a diameter, or that $HI$ was the angle bisector of $C_1HB_1$ which was implying the harmonic quad). Let $A_2,B_2,C_2$ be the excenters, let $H$ be the foot of the altitude from $A$. Since $ABC$ is the orthic triangle of $A_2B_2C_2$, by Cevian Nests we have $A_2A_1,B_2B_1,C_2C_1$ concurrent in $X$. As both $X$ and $A_1$ lie on $A_2A_1$ and $AB_1XC_1A_1$ is cyclic with diameter $AX$, if $X,A$ differ, then $AA_1\perp AX$ which implies $AA_1\parallel BC$, which is impossible. So $X=A_1$, so $AC_1HA_1B_1$ is cyclic.By Pappus, $I$ is on $C_1B_1$. Note that since $ABC$ is the orthic triangle, it follows that $A_1(C_1,B_1,H,A)=-1$, so $AB_1HC_1$ is harmonic. It follows that $\angle AHI=\angle BHI -\angle BHA=90-\frac{A}{2}-\angle AB_1C_1=90-\angle C_1AI-\angle DAC_1=\angle ADI$, so $DHIA$ is cyclic, implying that $\angle DHA=90$ and $D$ being on $BC$. Let $M$ be $AD\cap C_1B_1$. By the properties of harmonic quadrilaterals, it follows that $MH$ will be also tangent, so $MH=MA=MD$ since $\angle DHA=90$. We can easily finish by power of point from here. Another fun observation is that $B_2C_2$ meets $XI$ on the circle.

Anyone bashed this successfully without guessing any geometrical claims? I got a degree 7 equation to verify subject to a deg 3 one and was too tired to verify it...

Solution avoids using \(A_1\) to be the bevan (because I couldn't find that). \(\overline{B_1,I,C_1-}\) Proof- Let \(M_B, M_C\) denote the midpoints of small \(\overarc{AC}, \overarc{AB}\) resp. Let \(M_BB_1 \cap \odot ABC = T \Rightarrow \Delta ATC\sim \Delta A_1BC_1 \Rightarrow \angle CAT = \angle BA_1C_1\) Similarly, let \(M_CC_1 \cap \odot ABC = T' \Rightarrow \angle BAT' = \angle CA_1B_1 \therefore \angle CAT + \angle BAT' = \angle A \Rightarrow T= T'.\) Now applying Pascal's on cyclic hexagon \((ABM_BTM_C)\) we get \(\overline{B_1,I,C_1}\) \(\quad \square\) \(AD\) is tangent to \(\odot AC_1B_1-\) Proo - In any \(\triangle\) let \(E,F\) be the incircle touch points on \(AC, AB\) resp. then \(AD, \odot AEF, \odot ABC\) are concurrent. Thus the result is obvious for any arbitrary \(\Delta \quad\square\) Let \(B_1C_1 \cap AD= M\). Since \(AI\) is the angle bisector and \(AM\) the tangent we have \(MA= MI = MD\) but \(\angle MAC_1 = \angle MB_1A \Rightarrow C_1\) is the Humpty point of \(\Delta AB_1D \Rightarrow AD\) is tangent to \(\odot DB_1C_1. \quad\blacksquare\)

Official solutions post (also at http://web.evanchen.cc/problems.html): We present two solutions. First solution using spiral similarity (Ankan Bhattacharya) First, we prove the part of the problem which does not depend on the condition $A B_1 A_1 C_1$ is cyclic. Lemma: Let $ABC$ be a triangle and define $I$, $D$, $B_1$, $C_1$ as in the problem. Moreover, let $M$ denote the midpoint of $\overline{AD}$. Then $\overline{AD}$ is tangent to $(AB_1C_1)$, and moreover $\overline{B_1 C_1} \parallel \overline{IM}$. Proof. Let $E$ and $F$ be the tangency points of the incircle. Denote by $Z$ the Miquel point of $BFEC$, i.e.\ the second intersection of the circle with diameter $\overline{AI}$ and the circumcircle. Note that $A$, $Z$, $D$ are collinear, by radical axis on $(ABC)$, $(AFIE)$, $(BIC)$. [asy][asy] size(11cm); pair A = dir(130); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair E = foot(I, C, A); pair B_1 = A+C-E; pair F = foot(I, A, B); pair C_1 = A+B-F; pair T = foot(I, B, C); pair Z = foot(A, I, foot(T, E, F)); filldraw(A--B--C--cycle, invisible, blue); draw(circumcircle(A, E, F), orange); pair D = extension(A, Z, B, C); draw(arc(dir(270), C, B), orange); draw(D--B, blue); filldraw(A--B_1--C_1--cycle, invisible, blue); pair M = midpoint(A--D); filldraw(A--D--I--cycle, invisible, red); draw(I--M, red); filldraw(Z--E--F--cycle, invisible, deepgreen); filldraw(unitcircle, invisible, blue); draw(incircle(A, B, C), dotted+blue); draw(circumcircle(A, B_1, C_1), dashed+red); dot("$A$", A, dir(A)); dot("$B$", B, dir(225)); dot("$C$", C, dir(315)); dot("$I$", I, dir(270)); dot("$E$", E, dir(30)); dot("$B_1$", B_1, dir(B_1)); dot("$F$", F, dir(F)); dot("$C_1$", C_1, dir(C_1)); dot("$Z$", Z, dir(Z)); dot("$D$", D, dir(D)); dot("$M$", M, dir(M)); /* TSQ Source: !size(11cm); A = dir 130 B = dir 210 R225 C = dir 330 R315 I = incenter A B C R270 E = foot I C A R30 B_1 = A+C-E F = foot I A B C_1 = A+B-F T := foot I B C Z = foot A I foot T E F A--B--C--cycle 0.1 lightcyan / blue circumcircle A E F orange D = extension A Z B C !draw(arc(dir(270), C, B), orange); D--B blue A--B_1--C_1--cycle 0.1 deepgreen / blue M = midpoint A--D A--D--I--cycle 0.1 lightred / red I--M red Z--E--F--cycle 0.1 deepgreen / deepgreen unitcircle 0.1 lightcyan / blue incircle A B C dotted blue circumcircle A B_1 C_1 dashed red */ [/asy][/asy] Then the spiral similarity gives us \[ \frac{ZF}{ZE} = \frac{BF}{CE} = \frac{AC_1}{AB_1} \]which together with $\measuredangle FZE = \measuredangle FAE = \measuredangle BAC$ implies that $\triangle ZFE$ and $\triangle AC_1B_1$ are (directly) similar. (See IMO Shortlist 2006 G9 for a similar application of spiral similarity.) Now the remainder of the proof is just angle chasing. First, since \[ \measuredangle DAC_1 = \measuredangle ZAF = \measuredangle ZEF = \measuredangle AB_1C_1 \]we have $\overline{AD}$ is tangent to $(AB_1C_1)$. Moreover, to see that $\overline{IM} \parallel \overline{B_1C_1}$, write \begin{align*} \measuredangle (\overline{AI}, \overline{B_1C_1}) &= \measuredangle IAC + \measuredangle AB_1C_1 = \measuredangle BAI + \measuredangle ZEF = \measuredangle FAI + \measuredangle ZAF \\ &= \measuredangle ZAI = \measuredangle MAI = \measuredangle AIM \end{align*}the last step since $\triangle AID$ is right with hypotenuse $\overline{AD}$, and median $\overline{IM}$. $\blacksquare$ Now we return to the present problem with the additional condition. [asy][asy] unitsize(100); pair A, B, C, I, A1, B1, C1, D, E, F, J, K, Z, M; B = dir(193); C = reflect((0, 1), (0, -1)) * B; I = intersectionpoints(circle(dir(270), abs(dir(270)-B)), -B--(-C))[1]; A = 2 * foot(origin, I, dir(270)) - dir(270); D = extension(B, C, I, rotate(90, I) * A); E = foot(I, C, A); F = foot(I, A, B); A1 = B + C - foot(I, B, C); B1 = C + A - E; C1 = A + B - F; M = (A + D)/2; Z = foot(I, A, D); K = 2 * foot(circumcenter(A, B1, C1), A, I) - A; draw(unitcircle); filldraw(circumcircle(A, B1, C1), invisible, red); filldraw(circumcircle(A, E, F), invisible, heavycyan+dashed); draw(incircle(A, B, C), heavycyan+dotted); filldraw(circumcircle(A, B, C), invisible, cyan); filldraw(circumcircle(B, I, C), invisible, heavycyan+dashed); draw(arc(circumcenter(D, B1, C1), B1, D), heavyred); draw(A--D, heavyred); draw(A--B--C--cycle, heavymagenta); draw(D--B, lightmagenta); draw(A--I--D, heavycyan); draw(I--K, heavyred); draw(M--B1, dashed+heavyred); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(10)); dot("$I$", I, dir(-I)); dot("$A_1 = V$", A1, dir(A1)); dot("$B_1$", B1, dir(10)); dot("$C_1$", C1, dir(130)); dot("$D$", D, dir(D)); dot("$E$", E, dir(40)); dot("$F$", F, dir(170)); dot("$M$", M, dir(130)); dot("$Z$", Z, dir(Z)); // dot("$K$", K, dir(K-circumcenter(A, B1, C1))); clip(box((-3, -0.4), (1.2, 1.25))); [/asy][/asy] Claim: Given the condition, we actually have $\angle AB_1A_1 = \angle AC_1A_1 = 90^{\circ}$. Proof. Let $I_A$, $I_B$ and $I_C$ be the excenters of $\triangle ABC$. Then the perpendiculars to $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ from $A_1$, $B_1$, $C_1$ respectively meet at the so-called Bevan point $V$ (which is the circumcenter of $\triangle I_A I_B I_C$). Now $\triangle AB_1C_1$ has circumdiameter $\overline{AV}$. We are given $A_1$ lies on this circle, so if $V \ne A_1$ then $\overline{AA_1} \perp \overline{A_1V}$. But $\overline{A_1V} \perp \overline{BC}$ by definition, which would imply $\overline{AA_1} \parallel \overline{BC}$, which is absurd. $\blacksquare$ Claim: Given the condition the points $B_1$, $I$, $C_1$ are collinear (hence with $M$). Proof. By Pappus theorem on $\overline{I_B A I_C}$ and $\overline{BA_1C}$ after the previous claim. $\blacksquare$ To finish, since $\overline{DMA}$ was tangent to the circumcircle of $\triangle AB_1C_1$, we have $MD^2 = MA^2 = MC_1 \cdot MB_1$, implying the required tangency. Remark: The triangles satisfying the problem hypothesis are exactly the ones satisfying $r_A = 2R$, where $R$ and $r_A$ denote the circumradius and $A$-exradius. Remark: If $P$ is the foot of the $A$-altitude then this should also imply $AB_1PC_1$ is harmonic. Second solution by inversion and mixtilinears (Anant Mudgal) As in the end of the preceding solution, we have $\angle AB_1A_1=\angle AC_1A_1=90^{\circ} \quad\text{and}\quad I \in \overline{B_1 C_1}$. Let $M$ be the midpoint of minor arc $BC$ and $N$ be the midpoint of arc $\widehat{BAC}$. Let $L$ be the intouch point on $\overline{BC}$. Let $O$ be the circumcenter of $\triangle ABC$. Let $K=\overline{AI} \cap \overline{BC}$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(12cm); 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., xmax = 6., ymin = -3., ymax = 4.; /* image dimensions */ /* draw figures */ draw(circle((2.7,0.6), 2.7658633371878665), linewidth(0.4)); draw((1.892579421398536,1.2)--(3.5074205786014643,0.), linewidth(0.4) + dotted); draw((1.445226775603362,3.064861893765504)--(-3.109814760356735,0.), linewidth(0.4)); draw((-3.109814760356735,0.)--(2.7,-2.1658633371878664), linewidth(0.4)); draw((0.8894305730917524,2.690894150918487)--(3.9547732243966385,-1.864861893765504), linewidth(0.4) + linetype("2 2")); draw((1.445226775603362,3.064861893765504)--(0.,0.), linewidth(0.4)); draw((1.445226775603362,3.064861893765504)--(5.4,0.), linewidth(0.4)); draw((5.4,0.)--(-3.109814760356735,0.), linewidth(0.4)); draw((-3.5571674061519136,1.864861893765503)--(2.7,3.3658633371878666), linewidth(0.4)); draw((1.445226775603362,3.064861893765504)--(2.7,-2.1658633371878664), linewidth(0.4)); draw((-3.5571674061519136,1.864861893765503)--(-3.109814760356735,0.), linewidth(0.4)); draw((-3.109814760356735,0.)--(1.892579421398536,1.2), linewidth(0.4)); draw((1.892579421398536,1.2)--(1.892579421398536,0.), linewidth(0.4) + dotted); draw((2.7,3.3658633371878666)--(2.7,-2.1658633371878664), linewidth(0.4)); draw((0.8894305730917524,2.690894150918487)--(0.8894305730917507,-1.4908941509184854), linewidth(0.4) + linetype("2 2")); draw((2.7,3.3658633371878666)--(0.8894305730917507,-1.4908941509184854), linewidth(0.4) + linetype("2 2")); draw((-3.5571674061519136,1.864861893765503)--(4.217574445111032,0.9163536869921585), linewidth(0.4)); draw((1.445226775603362,3.064861893765504)--(0.8894305730917507,-1.4908941509184854), linewidth(0.4) + linetype("2 2")); draw((3.5074205786014643,0.)--(4.217574445111032,0.9163536869921585), linewidth(0.4) + dotted); draw((0.6380284948350587,1.353053551156462)--(3.5074205786014643,0.), linewidth(0.4) + dotted); draw((0.8894305730917524,2.690894150918487)--(2.7,-2.1658633371878664), linewidth(0.4) + linetype("2 2")); /* dots and labels */ dot((0.,0.),linewidth(4.pt) + dotstyle); label("$B$", (-0.32,-0.42), NE * labelscalefactor); dot((2.7,0.6),linewidth(4.pt) + dotstyle); label("$O$", (2.78,0.76), NE * labelscalefactor); dot((2.7,-2.1658633371878664),linewidth(4.pt) + dotstyle); label("$M$", (2.78,-2.), NE * labelscalefactor); dot((1.892579421398536,1.2),linewidth(4.pt) + dotstyle); label("$I$", (1.5,1.32), NE * labelscalefactor); dot((1.445226775603362,3.064861893765504),linewidth(4.pt) + dotstyle); label("$A$", (1.52,3.22), NE * labelscalefactor); dot((-3.109814760356735,0.),linewidth(4.pt) + dotstyle); label("$D$", (-3.22,-0.46), NE * labelscalefactor); dot((1.892579421398536,0.),linewidth(4.pt) + dotstyle); label("$L$", (1.72,-0.42), NE * labelscalefactor); dot((4.217574445111032,0.9163536869921585),linewidth(4.pt) + dotstyle); label("$B_1$", (4.3,1.08), NE * labelscalefactor); dot((0.6380284948350587,1.353053551156462),linewidth(4.pt) + dotstyle); label("$C_1$", (0.2,1.02), NE * labelscalefactor); dot((-3.5571674061519136,1.864861893765503),linewidth(4.pt) + dotstyle); label("$G$", (-3.48,2.02), NE * labelscalefactor); dot((3.5074205786014643,0.),linewidth(4.pt) + dotstyle); label("$V$", (3.38,-0.4), NE * labelscalefactor); dot((2.7,3.3658633371878666),linewidth(4.pt) + dotstyle); label("$N$", (2.78,3.52), NE * labelscalefactor); dot((0.8894305730917524,2.690894150918487),linewidth(4.pt) + dotstyle); label("$Z$", (0.36,2.5), NE * labelscalefactor); dot((0.8894305730917507,-1.4908941509184854),linewidth(4.pt) + dotstyle); label("$T$", (0.46,-1.78), NE * labelscalefactor); dot((3.9547732243966385,-1.864861893765504),linewidth(4.pt) + dotstyle); label("$A'$", (4.04,-1.7), NE * labelscalefactor); dot((-0.8322939923766871,1.5324309468827513),linewidth(4.pt) + dotstyle); label("$X$", (-0.98,1.74), NE * labelscalefactor); dot((5.4,0.),linewidth(4.pt) + dotstyle); label("$C$", (5.38,-0.36), NE * labelscalefactor); dot((2.1804415825316923,0.),linewidth(4.pt) + dotstyle); label("$K$", (2.2,-0.4), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim: We have $\angle (\overline{AI}, \overline{B_1C_1})=\angle IAD$. Proof. Let $Z$ lie on $(ABC)$ with $\angle AZI=90^{\circ}$. By radical axis theorem on $(AIZ), (BIC),$ and $(ABC)$, we conclude that $D$ lies on $\overline{AZ}$. Let $\overline{NI}$ meet $(ABC)$ again at $T \ne N$. Inversion in $(BIC)$ maps $\overline{AI}$ to $\overline{KI}$ and $(ABC)$ to $\overline{BC}$. Thus, $Z$ maps to $L$, so $Z, L, M$ are collinear. Since $BL=CV$ and $OI=OV$, we see that $MLIN$ is a trapezoid with $\overline{IL} \parallel \overline{MN}$. Thus, $\overline{ZT} \parallel \overline{MN}$. It is known that $\overline{AT}$ and $\overline{AA_1}$ are isogonal in angle $BAC$. Since $\overline{AV}$ is a circumdiameter in $(AB_1C_1)$, so $\overline{AT} \perp \overline{B_1C_1}$. So $\measuredangle ZAI=\measuredangle NMT=90^{\circ}-\measuredangle TAI=\measuredangle (\overline{AI}, \overline{B_1C_1})$. $\blacksquare$ Let $X$ be the midpoint of $\overline{AD}$ and $G$ be the reflection of $I$ in $X$. Since $AIDG$ is a rectangle, we have $\measuredangle AIG=\measuredangle ZAI=\measuredangle (\overline{AI}, \overline{B_1C_1})$, by the previous claim. So $\overline{IG}$ coincides with $\overline{B_1C_1}$. Now $\overline{AI}$ bisects $\angle B_1AC_1$ and $\angle IAG=90^{\circ}$, so $(\overline{IG}; \overline{B_1C_1})=-1$. Since $\angle IDG=90^{\circ}$, we see that $\overline{DI}$ and $\overline{DG}$ are bisectors of angle $B_1DC_1$. Now $\angle XDI=\angle XID \implies \angle XDC_1=\angle XID-\angle IDB_1=\angle DB_1C_1$, so $\overline{XD}$ is tangent to $(DB_1C_1)$.

Solution. As usual, let $I_X$ be the $X$-excenter of $\bigtriangleup ABC$. It's known that $I_AA_1,I_BB_1$ and $I_CC_1$ concur at the circumcenter $K$ of $\bigtriangleup I_AI_BI_C$. Since $\angle AC_1I_C=90^\circ=\angle I_BB_1A$, $K$ lies on $(AB_1A_1C_1)$. Moreover, $A_1$ and $K$ lie on the same side with respect to $B_1C_1$, therefore that $K=A_1$. Let's start with the following result, which doesn't depend of the cyclicity of $AC_1A_1B_1$. Lemma 1. Let $ABC$ a non-isosceles triangle and $A_1$ as defined in the original problem. Let $M$ be the midpoint of $\widehat{BAC}$ and $H=\overline{MI}\cap \overline {BC}$. Then $AMA_1H$ is cyclic. Proof. $T=\overline{MI}\cap (ABC)$ is the $A$-mixtilinear intouch point. Lines $AT$ and $AA_1$ are known to be the reflections of each other across the internal bisector of $\angle BAC$. Hence, $$\angle AMH=\angle AMT=\angle ACB+\angle BAT=\angle ACA_1+\angle A_1AC=\angle AA_1H$$as required. $\blacksquare$ Now, let's prove the following. Claim 1. $B_1, I $ and $C_1$ lie on a same line. Proof. Observe that $M$ is the second point of intersection of $(ABC)$ and $(AB_1A_1C_1)$, so it carries $\overline{BC}$ to $\overline{C_1B_1}$ and thus $MB_1=MC_1$. Define $H$ to be the $A$-foot of altitude on $BC$, which clearly lies on $(AB_1A_1C_1)$. By lemma 1, $H,\ I$ and $M$ are collinear. Now, $-1=(B,A;I_C,I_B)\overset{A}{=}(H,A;C_1,B_1)$ implying that $AC_1HB_1$ is a harmonic quadrilateral, so the internal bisectors of $\angle C_1HB_1$ and $\angle C_1AB_1$ meet at a point $I'$ on $C_1B_1$. Because $M$ is the midpoint of $\widehat{C_1AB_1}$ we conclude that $I'$ is the intersection point of $AI$ and $MH$, which forces $I=I'$ to happen. Hereby, $C_1,\ I$ and $B_1$ are collinear. $\blacksquare$ Claim 2. $DI$ bisects $\angle C_1DB_1$. Proof. Let $F=\overline{I_CI_B}\cap\overline{C_1B_1}$. We have $(F,I;C_1,B_1)=-1$ so $\angle FHI=90^\circ$ and $AIHF$ is cyclic. Further, $\angle AID=90^\circ=\angle AHD$, so $AIHD$ is inscribed. We infer that $AIHDF$ is a cyclic pentagon, implying that $D$ lies on the $A$- Apollonius circle of $\bigtriangleup AB_1C_1$, from which the result follows. $\blacksquare$ Finally, note that $\angle IDF=\angle IHF=90^\circ$, then, $$\angle ADC_1=90^\circ-\angle ADF-\angle C_1DI=90^\circ-\angle AIF-\angle B_1DI=\angle C_1ID-\angle B_1DI=\angle C_1B_1D$$thus, $DA$ is tangent to $(B_1DC_1)$. $\blacksquare$

Here s another solution. It seems a little longer, and not so straight-forward maybe, but I think it reveals more interesting properties of the configuration. Firslty, take $Q$ the second intersection of $(AB_1C_1)$ and $BC$. Notice that the perpendiculars raised from $A_1,B_1,C_1$ onto $BC,CA,AB$ meet at some point. Recall that if $P$ and $P'$ are isogonal conjugates WRT $ABC$ the projections of $P,P'$ to the sides of the triangle are concyclic. Conversely, taking the limit case, The perpendiculars raised from $A$ onto $AB$, $A$ onto $AC$ and $Q$ onto $BC$ meet at some point, that must be $A$! Hence, $Q$ is the foot of the altitude from $A$ to $BC$, and $(AB_1C_1)$ is the circle of diameter $AA_1$. Also note that the midpoin $M$ of $\overarc{BAC}$ in $(ABC)$ generally lays on circle $(AB_1C_1)$. Denote now $I_a,I_b,I_c$ the three center of the excircles, noticing that $A_1, I_a, B_1$, respectively $A_1, C_1, I_c$ are collinear. Now, from quadrilateral $B_1A_1C_1M$, as $MB_1=MC_1$ it follows that $(A_1M$ bisect $\angle{B_1A_1C_1}$ but $A_1M\perp{I_bI_c}$, so $A_iI_bI_c$ is isosceles, and also $m(\angle{I_bA_1I_c})=2m(\angle{I_bI_aI_c})$ so $A1$ is the Bevan point of $ABC$. It follows that $O$ is the midpoint of $A_1I$. Taking $M'$ the midpoint of $\overarc{BC}$ in circle $(ABC)$ it follows that $MA_1M'I$ is a parallelogram. To prove $M,I,Q$ collinear, it suffices to prove $\frac{AQ}{AI}=\frac{MM'}{MI}$, which reduces to $2\sin{\frac{A}{2}}\cos{\frac{B}{2}}\cos{\frac{C}{2}}=1$ (which i think is a nice way to characterise such triangles). To prove this, using Pappus for $B,A_1,C$ and $I_c, A,I_b$ it follows that $B_1,I,C_1$ are collinear. Using now a theorem(whose name i do not know in English) it follows that $BA_2\cdot{\frac{CB_1}{AB_1}}+CA_2\cdot{\frac{BC_1}{C_1A}}=BC\cdot{\frac{A_2I}{IA}}$, where $A_2\in{ BC\cap{AI}}$, which,after some computations, finally yields the desired $2\sin{\frac{A}{2}}\cos{\frac{B}{2}}\cos{\frac{C}{2}}=1$, and $M,I,Q$ collinear. As $M$ is the midpoint of $I_bI_c$ and $ID||I_bI_c$ it follows that $(II_b,II_c;IM,ID)=-1 \Rightarrow (II_b\cap{BC},II_c\cap{BC};IM\cap{BC},ID\cap{BC})=-1 \Rightarrow (B,C;Q,D)=-1$ Now, take $B_2\in{DC_1\cap{AC}}, C_2\in{DB_1\cap{AB}}, R\in{C_2B_2\cap{C_1B_1}}, S\in{AB\cap{DR}}, T\in{AC\cap{DR}}, U\in{AQ\cap{DB_1}}, V\in{AQ\cap{DC_1}}$. $|angle{AQD}$ is right and $(D,U;C_2,B_1)+(D,V;C_1,B_2)=-1$ so $(QA$ bisects both $\angle{C_2QB_1}$ and $C_1QB_2$, which implies that $AB_2QC_2$ is cyclic. This means $Q$ is the Miguel point of $C_2C_1B_1B_2$, but them $QRC_1C_2, QRB_2B_2$ are cyclic as well. We have that $(D,R;S,T)=-1$ so $D,AR\cap{BC};B,C)=-1$ hence $R\in{AQ}$. Using now the power of $A$ WRT $CRC_1, QRB_1$ it follows that $C_1B_2B_1C_2$ is cyclic. Doing some angle chasing, the desired tangency is equivalent to $B_2C_2||AD$, whyich however i was not able to prove staight-forwar (, so, even more $B_2C_2||AD$ is equivalent to $AD$ tangent to $(AC_1B_1)$. let $N\in{B_1C_1\cap{AD}}$, then $(N,R;C_1,B_1)=-1$ and it is enough to prove that $AR$ is symmedian in $AC_1B_1$, which is equivalent to $(\frac{AC_1}{AB_1})^2=\frac{C_1R}{B_1R}$. But $\frac{C_1R}{RB_1}=\frac{AC_1\sin{QAC_1}}{AB_1\sin{QAB_1}}$, so tha tangency actually resumes to $\frac{\cot{\frac{B}{2}}}{\cot{\frac{C}{2}}}=\frac{\cos{B}}{\cos{C}}$, which comes down, once again, to $2\sin{\frac{A}{2}}\cos{\frac{B}{2}}\cos{\frac{C}{2}}=1$, which, after laborious work, ends the proof.

Following are equivalent: 1. $r_{a}=2R$; 2. $\frac{a}{p-a}=\frac{b}{p-b}+\frac{c}{p-c}$; 3. $\frac{1}{p}+\frac{1}{p-a}=\frac{1}{p-b}+\frac{1}{p-c}$; 4. $O$ is midpoint of $IA_{1}$; and other facts could be driven from figure of problem which had been noticed by other matlinkers!

Let $X$ be the antipode of $A$ on $(AB_1C_1)$. Let the incircle of $\triangle ABC$ touch $\overline{CA}$ and $\overline{AB}$ at $E$ and $F$, respectively, and let $J$ be the Miquel point of $BCEF$. Furthermore, let $M$ be the midpoint of $\overline{AD}$. [asy][asy] size(10cm); defaultpen(fontsize(10pt)); pen pri=royalblue+linewidth(0.5); pen sec=rgb(41, 207, 255)+linewidth(0.5); pen tri=springgreen+linewidth(0.5); pen qua=chartreuse+linewidth(0.5); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pen qfil=invisible; pair B, C, L, I, A, A1, B1, C1, EE, F, J, D, M, K, IB, IC; B=dir(190); C=dir(350); L=dir(270); I=intersectionpoint(arc(L, length(B-L), 90, 180, CCW), (-B) -- (-C)); A=intersectionpoint(I -- (I+100*(I-L)), circle((0, 0), 1)); A1=B+C-foot(I, B, C); B1=foot(A1, A, C); C1=foot(A1, A, B); EE=foot(I, A, C); F=foot(I, A, B); J=intersectionpoints(circumcircle(A, B, C), circumcircle(A, EE, F))[1]; D=extension(A, J, B, C); M=(A+D)/2; K=dir(90); IB=extension(B, I, A, K); IC=extension(C, I, A, K); draw(A -- B -- C -- A, pri); filldraw(circumcircle(A, B, C), fil, pri); draw(IB -- A1 -- IC -- IB, sec); filldraw(circumcircle(A, B1, C1), tfil, tri); draw(B -- D -- A, pri); filldraw(circumcircle(A, EE, F), sfil, sec); filldraw(circumcircle(B, I, C), qfil, qua); draw(B -- IB, tri+dashed); draw(C -- IC, tri+dashed); draw(M -- B1, pri); draw(D -- I -- J, sec); draw(A -- L, pri); clip((L + (100, -1/8)) -- (L+(-100, -1/8)) -- (-100, 100) -- (100, 100) -- cycle); dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$L$", L, S); dot("$I$", I, NE); dot("$A_1$", A1, SE); dot("$B_1$", B1, NE); dot("$C_1$", C1, SW); dot("$E$", EE, NE); dot("$F$", F, NW); dot("$J$", J, dir(150)); dot("$D$", D, SW); dot("$M$", M, dir(195)); dot("$K$", K, N); dot("$I_B$", IB, NE); dot("$I_C$", IC, W); [/asy][/asy] We claim that $\overline{AA_1}$ is a diamter of $(AB_1A_1C_1)$. Note that $X$ is the Bevan point of $\triangle ABC$, so $\overline{XA_1}\perp\overline{BC}$. Furthermore, if $X\ne A_1$, then $\overline{XA_1}\perp\overline{AA_1}$, which would require that $A\in\overline{BC}$, which is absurd. By Pappus' Theorem on $\overline{BA_1C}$ and $\overline{I_CAI_B}$, $I$ lies on $\overline{B_1C_1}$, and by the Radical Axis Theorem on $(AI)$, $(ABC)$, and $(BIC)$, $J$ lies on $\overline{AD}$. Since $\triangle JBF\sim\triangle JCE$, $$\frac{AC_1}{AB_1}=\frac{JF}{JE}=\frac{JB}{JC},$$and also $\measuredangle C_1AB_1=\measuredangle BAC=\measuredangle BJC$, so $\triangle AC_1B_1\sim\triangle JBC$. This implies that $$\measuredangle DAB=\measuredangle JAF=\measuredangle JEF=\measuredangle JCB=\measuredangle AB_1C_1,$$so $\overline{AD}$ is tangent to $(AB_1C_1)$. Moreover, $$\measuredangle MIA=\measuredangle IAM=\measuredangle IAC_1+\measuredangle C_1AM=\measuredangle B_1AI+\measuredangle C_1B_1A=\measuredangle B_1IA,$$whence $M$ lies on $\overline{B_1IC_1}$. Hence, $MD^2=MA^2=MB_1\cdot MC_1$, and the desired result follows. $\square$

I think I'm getting worse at synthetic geometry as I wasn't able to solve this one. However, I can still bash and that worked out It turns out that this problem can be bashed mostly with barycentric coordinates, with only some basic synthetic insights. It's easy to see that $(AB_1C_1)=(AV)$ where $V$ is the Bevan point, so either $V=A_1$, or $\angle AA_1V=90^\circ$. But $I_A,A_1,V$ collinear, so we have $\angle AA_1I_A=90^\circ$. But $I_AA_1\perp BC$, so we have $BC\parallel AA_1$, which is nonsense. Thus, $V=A_1$, so the Bevan point is on $BC$. Thus, we have $I_BB_1$, $I_CC_1$, and $BC$ concur. In barycentric coordinates, let $V=(0:y:z)=I_CC_1\cap BC$. We see then that \[\begin{vmatrix}0 & y & z \\ a & b & -c \\ s-a & s-b & 0\end{vmatrix}=0,\]which gives $V=(0:s(a-b):c(s-a))$. Similarly, we get $V=(0:b(s-a):s(a-c))$, so for these to be equal, we must have \[s^2(a-b)(a-c)=bc(s-a)^2.\]We see that \begin{align*} s^2(a-b)(a-c)=bc(s-a)^2 &\iff (a+b+c)^2(a-b)(a-c) = bc(s-a)^2 \\ &\iff bc\left[(a+b+c)^2-(b+c-a)^2\right]+a^2(a+b+c)^2 = a(b+c)(a+b+c)^2 \\ &\iff bc[2(b+c)][2a]=a(a+b+c)^2(b+c-a) \\ &\iff 4bc(b+c)=(a+b+c)^2(b+c-a) \\ &\iff \boxed{bc(b+c)=2s^2(s-a)}. \end{align*}We now compute $D=(0,y,z)$. We can use the perpendicularity criterion for $AI\perp ID$, but that's ugly. A better way is to define $K\in BC$ such that $AK\parallel ID$. We see that $AK$ is the external angle bisector which is parameterized by $(t:b:-c)$ (an easy way to see this is that $I_C=(a:b:-c)$ is on it), so $K=(0:b:-c)=\left(0,\frac{b}{b-c},\frac{c}{b-c}\right)$. We have $AK\parallel ID$, so \[[AKI]=[AKD]\implies \frac{1}{(b-c)(a+b+c)}\begin{vmatrix}1 & 0 & 0\\ 0 & b & -c \\ a & b & c\end{vmatrix}=\frac{1}{b-c}\begin{vmatrix}1 & 0 & 0\\ 0 & b & -c \\ 0 & y & z\end{vmatrix}.\]This implies $bz+cy=\frac{2bc}{a+b+c}$, which combined with $y+z=1$ can be solved to give \[\boxed{D=\left(0,\frac{b(s-c)}{b-c},-\frac{c(s-b)}{b-c}\right)=(0:b(s-c):-c(s-b))}.\]Let $E$ be the midpoint of $AD$, which by the above is given by \[E=(s(b-c):b(s-c):-c(s-b)).\]The first big claim is that $E,B_1,C_1$ are collinear. This is equivalent to \[\begin{vmatrix}s(b-c) & b(s-c) & -c(s-b) \\ s-a & s-b & 0 \\ s-a & 0 & s-c\end{vmatrix}=0,\]which is equivalent to \[\Gamma:=s(b-c)(s-b)(s-c)-(s-a)b(s-c)^2+(s-a)c(s-b)^2=0.\]We have \begin{align*} \Gamma=0 &\iff s(b-c)(s-b)(s-c)=(s-a)[b(s-c)^2-c(s-b)^2] \\ &\iff s(b-c)(s-b)(s-c)=(s-a)(b-c)(s^2-bc) \\ &\iff s(s-b)(s-c)=(s-a)(s^2-bc) \\ &\iff s^3-s^2(b+c)+sbc=s^3-sbc-as^2+abc \\ &\iff bc(2s-a)=s^2(b+c-a) \\ &\iff bc(b+c)=2s^2(s-a), \end{align*}which as we saw before, is directly equivalent to $(AB_1A_1C_1)$ cyclic. Thus, $E,C_1,B_1$ are collinear. The next big claim is that $AD$ is tangent to $(AB_1C_1)$. Doing this directly is difficult, but after performing a $\sqrt{bc}$ inversion, it is quite susceptible to barycentric coordinates. Let $D'$ be on $BC$ such that $AD$ and $AD'$ are isogonal. Given that $D=(0:b(s-c):-c(s-b))$, we have that \[D'=(0:b(s-c):-c(s-c)),\]recalling that $(0:y:z)\mapsto(0:b^2/y:c^2/z)$. Let $B_1'\in AB$ be the $\sqrt{bc}$ inverse of $B_1$, and let $C_1'\in AC$ be the $\sqrt{bc}$ inverse of $C_1$. We have that \[AB_1\cdot AB_1'=bc,\]so $AB_1'=\frac{bc}{s-c}$. Thus, \[BB_1'=c-\frac{bc}{s-c}=c\frac{s-b-c}{s-c}=-\frac{c(s-a)}{s-c},\]so \[B_1'=(BB_1':AA_1':0)=(a-s:b:0),\]and similarly $C_1'=(a-s:0:c)$. Note that $(a-s)+b=c-s$ and $(a-s)+c=b-s$. The fact that $AD$ is tangent to $(AB_1C_1)$ is equivalent to $AD'\parallel B_1'C_1'$ under the $\sqrt{bc}$ inversion, which is equivalent to \[[AD'B_1']=[AD'C_1'].\]This is equivalent to \[\frac{1}{c-s}\begin{vmatrix}1 & 0 & 0\\ 0 & b(s-b) & -c(s-c) \\ a-s & b & 0\end{vmatrix}=\frac{1}{b-s}\begin{vmatrix}1 & 0 & 0\\ 0 & b(s-b) & -c(s-c) \\ a-s & 0 & c\end{vmatrix},\]which is immediate upon expansion as they are both equal to $-bc$. Note that we proved $DA$ tangent to $(AB_1C_1)$ independent of the problem condition, so it's true for any $\triangle ABC$. Thus, we have $EB_1C_1$ collinear where $E$ is the midpoint of $AD$, and we have $DA$ tangent to $(AB_1C_1)$. Thus, \[EA^2=EB_1\cdot EC_1,\]so $ED^2=EB_1\cdot EC_1$ as $EA=ED$. This directly implies $ED$ tangent to $(DB_1C_1)$, or $AD$ tangent to $(DB_1C_1)$, as desired.

yayups wrote: I think I'm getting worse at synthetic geometry as I wasn't able to solve this one. It's TST6, don't be too hard on yourself you'll get some practice in June anyways.

This problem generalizes a little Quote: Let $ABC$ be a triangle and let $A_1$ be a point on $\overline{BC}$. Let $B_1$ and $C_1$ be the projections of $A_1$ onto $\overline{AC}$ and $\overline{AB}$, respectively. Let $B_2$ be the reflection of $B_1$ over the perpendicular bisector of $\overline{AC}$ and define $C_2$ similarly. Let the radical axis of $\odot(ABC)$ and $\odot(A_2B_2C_2)$ meet $\overline{BC}$ at $D$. Given that $\overline{AA_1}, \overline{BB_1}, \overline{CC_1}$ are concurrent, prove that the circumcircle of $\triangle B_1C_1D$ is tangent to $\overline{AD}$.

Nice problem with lots of good properties. We list some of them and prove them: Claim 1. Let $\omega = (AA_1B_1C_1)$. We have that $AA_1$ is a diameter of $\omega$. (Hence if $AH\perp BC$ at $H$, then $H$ also lies on $\omega$.) Claim 2. Let $K$ be the tangency point of the $A$-mixtilinear circle, then $KH$ bisects $\angle BKC$. This immediately implies that $K,H,I$ are collinear. Claim 3. Let $\gamma=(AID)$. Then $\gamma$ and $\omega$ are orthogonal. Claim 4. Let $M$ be the midpoint of $AD$, then $I,M,B_1,C_1$ are collinear. This implies the desired conclusion. Proof of Claim 1. Let $I'$ be the reflection of $I$ over the circumcenter of $ABC$ (i.e. the Bevan point). Then $I'C_1\perp AB$, $I'B_1\perp AC$, $I'A_1\perp BC$. Thus $I'$ lies on $\omega$, forcing $I'=A$. Proof of Claim 2. Observe $\angle KBC=\angle KAC=\angle BAA_1=\angle C_1B_1A_1$, so $\triangle A_1B_1C_1\sim\triangle KBC$. Then $$\frac{BH}{HC} = \frac{BA}{AC}\cdot\frac{\cos B}{\cos C} = \frac{\sin C}{\sin B} \cdot \frac{CA_1}{BA_1}=\frac{B_1A_1}{A_1C_1}=\frac{BK}{KC}.$$Proof of Claim 3. Let $J$ be the midpoint of arc $BAC$. We know that $K,H,I,J$ are collinear, so $\angle AHI=\angle AHJ$. Thus $\angle A_1AI=\frac12 \angle A -\angle A_1AC= \angle AC_1J=\angle AHI$, meaning that $AA_1$ is tangent to $\gamma$. Proof of Claim 4. By angle chasing, $\angle MIA = \frac{\pi}{2} - \angle IAA_1=\angle AIC_1 = \pi - \angle AIB_1$.

Label all of the following points: The excenters $I_B$ and $I_C$. The midpoint $O$ of $\overline{AD}$ (so the center of $(AID)$). The foot $E$ of $A$ onto $\overline{BC}$. The point $X = AM_A \cap B_1C_1$. The $A$-mixtillinear touchpoint $T$. In typical ``American geometry" fashion, we now prove many lemmas. Claim 1. Quadrilateral $AM_AB_1C_1$ is cyclic. Proof. Take the spiral similarity at $M_A$ sending $BC_1 \to CB_1$, which is actually a rotation as $BC_1 = CB_1$. So $BM_A = CM_A$ and the result follows. $\square$ Claim 2 (EGMO TST 2019/5). Circle $(AB_1A_1C_1)$ has diameter. In addition, $B_1, I, C_1$ are collinear. Proof. Lines $I_BB_1$ and $I_CC_1$ meet at the Bevan point $V$ with $VA_1 \perp BC$. However, this implies $V=A_1$, as desired. The second part follows from Pappus on $\overline{I_CAI_B}$ and $\overline{BA_1C}$. $\square$ Claim 3. Point $E$ lies on $\overline{TIM_A}$. Proof. Note \[\measuredangle AM_AT = \measuredangle TAI + \measuredangle IAC + \measuredangle CBA = \measuredangle IAA_1 + 90^\circ + \measuredangle EAI = \measuredangle AM_AE,\]as desired. $\square$ Claim 4. Line $AA_1$ is tangent to $(AID)$. Proof. Note $(AB_1A_1C_1)$ and $(AID)$ both pass through $E$. By the first concylicitiy, it suffices to show $\measuredangle AEI = \measuredangle A_1AI \measuredangle IAT$, or $\measuredangle IAE = \measuredangle ATI = \measuredangle ATM_A$, which follows from simple angle chasing (since both of these are known angles in terms of the angles of the triangle). $\square$ Claim 5. Point $O$ lies on $\overline{B_1IC_1}$. Proof. It suffices to show lies $AM_A$ and $B_1C_1$ meet on $(AID)$, or $\angle AXI = \measuredangle AEI = \measuredangle IAT$, or $AT \perp B_1C_1$. This follows from $\overline{AA_1}, \overline{AT}$ isogonal in $\angle B_1AC_1$. $\square$ We are now done, since $OD^2 = OA^2 = OB_1 \cdot OC_1$ and so $\overline{OD}$ is tangent to $(DB_1C_1)$. $\blacksquare$ [asy][asy] size(12cm); defaultpen(fontsize(10pt)); pair A = dir(105.3); pair B = dir(193.3); pair C = dir(346.7); pair I = incenter(A,B,C); pair A1 = B+C-foot(I,B,C); pair B1 = A+C-foot(I,A,C); pair C1 = A+B-foot(I,A,B); pair D = extension(I,I+(A-I)*dir(90), B, C); pair Ma = intersectionpoints(circumcircle(A1,B1,C1),circumcircle(A,B,C))[0]; pair Ic = extension(A,Ma,A1,C1); pair Ib = extension(A,Ma,A1,B1); pair E = foot(A,B,C); pair O = (A+D)/2; pair X = 2O-I; pair T = intersectionpoints(Ma--3I-2Ma, circumcircle(A,B,C))[0]; filldraw(A--B--C--cycle,orange+white+white,orange); filldraw(anglemark(I,E,A),orange,orange); filldraw(anglemark(I,A,A1),orange,orange); filldraw(anglemark(I,X,A),orange,orange); draw(circumcircle(A,B,C), red); draw(circumcircle(A1,B1,C1), red); draw(circle((A+D)/2,abs(A-D)/2), red); draw(X--Ib, red+white); draw(D--B, orange); draw(X--B1, red+white+dashed); draw(D--I--A,red); draw(Ma--T, red+white); draw(Ic--A1--Ib, red+white+dashed); draw(A--A1, red+white); draw(A--T, red+white); draw(A--E, red+white); dot("$A$", A, dir(135)); dot("$B$", B, dir(225)); dot("$C$", C, dir(-45)); dot("$I$", I, dir(45)); dot("$A_1$", A1, dir(270)); dot("$B_1$", B1, dir(0)); dot("$C_1$", C1, dir(180)); dot("$D$", D, dir(180)); dot("$M_A$", Ma, dir(90)); dot("$I_B$", Ib, dir(90)); dot("$I_C$", Ic, dir(90)); dot("$E$", E, dir(-90)); dot("$O$", O, dir(90)); dot("$X$", X, dir(135)); dot("$T$", T, dir(-90)); [/asy][/asy]

Claim: [USA EGMO TST 2019/5] Draw the excenters $I_A,I_B,I_C$ and let $V$ be the circumcircle of $\triangle I_AI_BI_C$. Note that $\overline{A_1I_A},\overline{B_1I_B},\overline{C_1I_C}$ concur at $V$. Now $$\angle C_1A_1B = 90^{\circ} - \angle C_1BA_1, \angle B_1A_1C = 90^{\circ} - \angle A_1CB_1, \angle C_1A_1B_1 = 180^{\circ} - \angle C_1AB_1,$$so $V$ lies on $\overline{BC}$, and thus $V = A_1.$ Now $\overline{A_1B_1I_B}$ and $\overline{A_1C_1I_C}$ are collinear, so by Pappus $\overline{C_1IB_1}$ is collinear. Let $\overline{AP}$ be an altitude in $\triangle ABC,$ and let the tangents of $(A_1B_1C_1)$ at $A$ and $P$ intersect at $X.$ Note that $X$ lies on $\overline{B_1C_1}$ as $-1 = (\overline{I_BI_C} \cap \overline{BC},A; I_B,I_C) \stackrel{A_1}{=} (P,A;B_1,C_1),$ so $\overline{AP}$ is a symmedian in $\triangle AC_1B_1.$ This implies that $(AIP)$ is the Apollonian circle of $\overline{B_1C_1}$ passing through $A$. Thus $X$ is the midpoint of $\overline{AD},$ and we conclude that $XD^2 = XA^2 = XB_1 \cdot XC_1.$ Remarks: This was a nice problem (and there seems to be a trend with Ankan geometry diagrams that are hard to draw; USAMO 2019/2, TSTST 2019/9, and TST 2019/6). I wasn't able to find the last step of using Apollonian circles, unfortunately.

USA 2019 TST P6 wrote: Let $ABC$ be a triangle with incenter $I$, and let $D$ be a point on line $BC$ satisfying $\angle AID=90^{\circ}$. Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to $\overline{BC}$ at $A_1$. Define points $B_1$ on $\overline{CA}$ and $C_1$ on $\overline{AB}$ analogously, using the excircles opposite $B$ and $C$, respectively. Prove that if quadrilateral $AB_1A_1C_1$ is cyclic, then $\overline{AD}$ is tangent to the circumcircle of $\triangle DB_1C_1$. Ankan Bhattacharya Probably the longest solution in the thread Anyway, I think this solution is quite different from the others. Let $I_A, I_B$ and $I_C$ be the three excenters and let $M$ be the midpoint of $I_BI_C$. Since the circumcircle of $ABC$ is the Euler circle of $I_AI_BI_C$, point $M$ belongs to this circle and actually it is the midpoint of the arc $BAC$. Let $O \equiv I_CC_1 \cap I_BB_1$. We prove now several Claims: Claim 1: $O$ is the circumcenter of triangle $I_AI_BI_C$. Proof: Well known fact. By Nagel's theorem, the circumcenter of $I_AI_BI_C$, belongs to the perpendiculars from $I_C$ to $AB$ and from $I_B$ to $AC$, hence the result $\blacksquare$. Claim 2: $AC_1OB_1$ is cyclic. Proof: Obvious since $\angle AC_1O=\angle AB_1O=90^\circ$. Claim 3: $A_1 \equiv O$ Proof: We know that $A_1$ and $O$ both belong to the circumcircle of $AB_1C_1$. In addition, by Nagel's $I_AO \perp BC$ hence $I_A,A_1,O$ are collinear. Hence $O$ and $A_1$ must necessarily coincide. $\blacksquare$. Claim 4: $I \in B_1C_1$. Proof: Apply Pappus at $\overline{I_CAI_B}$ and $\overline{BA_1C}$ $\blacksquare$. Claim 5: Let $X$ be the midpoint of $B_1C_1$. Then $AX$ and $MA_1$ intersect on the antipode of $A$ on the circumcircle of $ABC$. Proof: Define $X \equiv AY \cap B_1C_1$ where $Y \equiv MA_1 \cap (A,B,C)$. We aim to prove two things: (i) $X$ is the midpoint of $B_1C_1$ (ii) $Y$ is the antipode of $A$. For (i), note that $$\frac{BA_1}{CA_1}=\dfrac{BM}{MC} \cdot \dfrac{\sin \angle BMY}{\sin \angle CMY}=\dfrac{\sin \angle BAY}{\sin \angle CAY}=\dfrac{AB_1}{AC_1} \cdot \dfrac{XC_1}{XB_1},$$and since $BA_1=AB_1$ and $CA_1=AC_1$, we obtain that $XC_1=XB_1$. For (ii), note that $$\angle BAY =\angle BMY=90^\circ-\angle I_CMB=90^\circ-\angle C,$$hence $\angle BAY+\angle AYB=90^\circ$, implying that $AY$ is the diameter, as desired $\blacksquare$. Now, let's attack the problem. Let $Z \equiv MI \cap BC$. We present the following angle-chase: $$\angle AA_1B-\angle AXC_1=\angle C+\angle CAA_1-\angle AB_1C_1-\angle XAB_1=\angle C+\angle CAA_1-\angle AB_1C_1-(90^\circ-\angle B)=90^\circ-\angle A+\angle CAA_1-\angle AB_1C_1=90^\circ-\angle C_1AA_1-\angle AB_1C_1=90^\circ-\angle C_1B_1A_1-\angle AB_1C_1=90^\circ-\angle AB_1A_1=0,$$therefore $\angle AA_1B=\angle AXC_1$. Note now that since $AIXM$ is cyclic to a circle with diameter $MI$, we have $$\angle AMZ=\angle AMI=\angle AXC_1=\angle AA_1B,$$therefore $AMA_1Z$ is cyclic. This combined with the fact that $\angle AMA_1=90^\circ$ readily implies that $AZ \perp BC$. Now, let's put point $D$ into the game. Since $\angle AID=\angle AZD=90^\circ$, $AIZD$ is cyclic. Now, it's time for a second angle-chase: $$\angle DAB=\angle DAI-\angle BAI=\angle IZC-\angle BAI=\angle IZC-\angle IAB_1=\angle MAA_1-\angle IAB_1=\angle MC_1A_1-\angle A/2=\angle I_CMC_1=\angle AB_1C_1,$$therefore $\angle DAB=\angle AB_1C_1$, which implies that $DA$ is tangent to the circumcircle of $AB_1C_1$. It's now time to end. Let $W \equiv B_1C_1 \cap AD$. Then, $$\angle AIW=\angle AB_1C_1+\angle A/2=\angle DAB+\angle A/2=\angle WAI,$$so $WA=WI$. Therefore, $WA=WD=WI$, thus $WD^2=WA^2=WC_1WB_1$, giving us that $WD$ is tangent to the circumcircle of $B_1C_1D$, as desired.

Here's my unnecessarily long solution. I totally missed the Appollonian circle The first two claims are from USA EGMO TST 2019 P5. Let $I_A,I_B$ and $I_C$ be the corresponding excenters. Let $M$ be the midpoint of $I_{C}I_{B}$, $N$ be the midpoint of $AD$, $P$ be the foot of perpendicular from $A$ to $BC$ and $Q$ be the intersection of $BC$ and $I_{B}I_{C}$. Also denote $(AB_{1}A_{1}C_{1})$ by $\omega$ for brevity. Claim 1: The center of $(I_{A}I_{B}I_{C})$ coincides with $A_1$. Proof. Assume that this is not the case, and let $O$ denote the circumcenter. Then since $\angle OC_1A=\angle OB_1A=90^\circ$, it follows that $O$ lies on $\omega$. But this means that \[\measuredangle B_1OC_1=\measuredangle B_1A_1C_1=\measuredangle B_1OC_1-(\measuredangle OBC+\measuredangle BCO)=\measuredangle B_1OC_1+\measuredangle BOC \Rightarrow \measuredangle BOC=0.\]Therefore, $O$ lies on $BC$. But as both $I_{A}O$ and $I_{A}A_{1}$ are perpendicular to $BC$, it follows that $O\equiv A_1$. Claim 2: $I$ lies on $C_1B_1$. Proof. This follows by Pappus on $\overline{I_{C}AI_{B}}$ and $\overline{CBA_1}$. Claim 3: $P$, $I$, $M$ are collinear. Proof. Obviously, $P$ lies on $\omega$. Therefore, projecting from line $I_{B}I_{C}$ gives us \[ (A,P;B_{1},C_{1}) \stackrel{A_1}{=} (A,Q,I_{B},I_{C})=-1\]Now notice that $M$ is the midpoint of arc $BAC$. Since $BC_1 = B_1C$, by spiral similarity, it follows that $M$ is also the midpoint of arc $B_{1}AC_{1}$ in $\omega$. Let $R$ be the midpoint of arc $A_{1}C_{1}$, and let $MI$ meet $\omega$ again at $P'$. Then \[(A,P';B_{1},C_{1})\stackrel{I}{=}(R,M;B_{1},C_{1})=-1\]which implies that $P=P'$ as desired. Claim 4: $AD$ is tangent to $\omega$. Proof. Due to right angles, $A_1MAP$ and $IADP$ are concyclic. Therefore by spiral similarity, $\triangle AMI\sim\triangle A_1AD$. In particular, $AD\perp AA_1$, so $AD$ is tangent to $\omega$. Finally, notice that $B_1C_1$, and $AD$ are concurrent on the perpendicular bisector of $AI$ as $AI$ is the internal angle bisector of $\angle B_1AC_1$. But this point is $N$, so by power of a point, \[NB_1\cdot NC_1=NA^2=ND^2\]and it follows that $AD$ is tangent to $(DB_{1}C_{1})$.

Similar solution for storage; had to approximate the asy diagram oops. I also discovered the following property: Quote: $\overline{HI}$ bisects $\overline{BC}$, where $H$ is the orthocenter of $\triangle ABC$. [asy][asy] size(10cm); defaultpen(fontsize(10pt)+linewidth(0.4)); dotfactor *= 1.2; pair A = dir(116.8), B = dir(192.4), C = dir(347.6), I = incenter(A,B,C), Ia = 2*circumcenter(B,I,C)-I, Ib = 2*circumcenter(C,I,A)-I, Ic = 2*circumcenter(A,I,B)-I, A1 = foot(Ia,B,C), B1 = foot(Ib,C,A), C1 = foot(Ic,A,B), L = dir(90), F = foot(A,B,C), D = extension(B,C,I,I+dir(A-L)), M = (A+D)/2, T = F+dir(L--F)*abs(B-F)*abs(C-F)/abs(L-F), S = A1+dir(A--A1)*abs(B-A1)*abs(C-A1)/abs(A-A1), I1 = 2M-I; draw(A--B--C--A, linewidth(0.8)+heavyred); draw(unitcircle, heavyred); draw(A--I--D--B^^F--A--D, heavyred); draw(circumcircle(A1,B1,C1), purple); draw(Ia--Ib--Ic--Ia, lightgreen); draw(B--Ib--A1--Ic--C, lightgreen); draw(circumcircle(A,D,F), heavyblue); draw(B1--M, orange+linewidth(0.8)); draw(M--extension(A,L,B1,C1)--A, orange+dashed); draw(L--T--S--A, magenta); draw(arc(circumcenter(D,B1,C1),circumradius(D,B1,C1),50,135), cyan); dot("$A$", A, dir(80)); dot("$B$", B, dir(225)); dot("$C$", C, dir(330)); dot("$A_1$", A1, dir(270)); dot("$B_1$", B1, dir(10)); dot("$C_1$", C1, dir(130)); dot("$D$", D, dir(175)); dot("$F$", F, dir(300)); dot("$L$", L, dir(90)); dot("$I_A$", Ia, dir(270)); dot("$I_B$", Ib, dir(45)); dot("$I_C$", Ic, dir(105)); dot("$M$", M, dir(115)); dot("$I$", I, dir(310)); dot("$T$", T, dir(225)); dot("$S$", S, dir(315)); dot("$I'$", I1, dir(180)); [/asy][/asy] Claim: $(AB_1C_1)$ passes through $L$, the midpoint of arc $BAC$. Proof. Let $L = (AB_1C_1) \cap (ABC)$. Then by spiral sim $\triangle LBC_1 \sim \triangle LCB_1$, but we know that $BC_1 = CB_1$, so in fact $\triangle LBC_1 \cong \triangle LCB_1$ and $LB = LC$ as desired. $\square$ Claim: If $(AB_1A_1C_1)$ meets $\overline{BC}$ again at $F$, then $L$, $I$, $F$ are collinear. Proof. Let $S = \overline{AA_1} \cap (ABC)$ and $T = \overline{LF} \cap \overline{ABC}$. By Reim's $\overline{TS} \parallel \overline{BC}$. But since $\overline{AA_1}$ is isogonal to the $A$-mixtilinear chord, $T$ must be the $A$-mixtilinear intouch point and $I \in \overline{LFT}$ as desired. $\square$ Claim: $(AB_1A_1C_1)$ has diameter $\overline{AA_1}$. Proof. Let $\triangle I_AI_BI_C$ be the excentral triangle and let $Be = \overline{I_BB_1} \cap \overline{I_CC_1}$ be the Bevan point (center of $(I_AI_BI_C)$). Then since $\measuredangle AB_1Be = \measuredangle AC_1Be = 90^\circ$, $Be$ lies on $(AB_1C_1)$. But by definition $Be$ also lies on $\overline{I_AA_1}$, so $A_1 = Be$, and it is the $A$-antipode on $(AB_1C_1)$. $\square$ This also implies $F$ is the foot of the altitude from $A$ to $\overline{BC}$. Claim: $B_1$, $C_1$, $I$ are collinear. Proof. Pappus on $\overline{I_CAI_B}$ and $\overline{BA_1C}$. $\square$ Claim: $\overline{AD}$ is tangent to $(AB_1A_1C_1)$. Proof. Let $M$ be the midpoint of $\overline{AD}$. Note that $AIFD$ is cyclic with diameter $\overline{AD}$. The key idea is to notice that $(AIFD)$ is the $A$-Apollonian circle in $\triangle AC_1B_1$, which is true because both $\overline{AI}$ and $\overline{FI}$ are angle bisectors. Therefore $M$ is also the midpoint of $\overline{II'}$ where $I' = \overline{AL} \cap \overline{C_1B_1}$ and the result follows. $\square$ We are done because $MD^2 = MA^2 = MC_1 \cdot MB_1$. $\blacksquare$

Ouch. Solution with judicious helpings of help from awang11. Let the arc midpoint of $\widehat{BAC}$ be $M$. If $A\ne M'=(AB_1C_1)\cap (ABC)$, then $M'B=M'C$ so config cases imply $M'=M$ as in IMO 2013/3. Note that since $\triangle ABC$ is the orthic triangle of its excentral triangle $\triangle I_AI_BI_C$, the circumcenter $V$ of $\triangle I_AI_BI_C$ lies on $I_BB_1,I_CC_1,I_AA_1$. Then $V$ unconditionally lies on $(AB_1C_1)$ as $\angle VB_1A=\angle VC_1A=90^\circ$. Note $V,A_1,B_1$ must lie on both $(AB_1C_1)$ and $(CA_1B_1)$, but $V,A_1,C_1$ must lie on both $(AB_1C_1)$ and $(BC_1A_1)$ so the three circles cannot have a common chord and so $V=A_1$. Next, note that by Pappus on $BA_1C$ and $I_CAI_B$, $I$ lies on $B_1C_1$. Now, let $(AB_1C_1A_1)$ intersect $BC$ again at $E$ so $\angle AEA_1=90^\circ$. Then $E\in (AID)$. Claim: $AD$ is tangent to $(AB_1C_1A_1E)$. Solution: It is enough to check $\angle DIE=\angle DAE=\angle AME$, so we need to check $EIM$ collinear. Let $K$ denote the reflection of $I_A$ over $A_1$. Then by well-known geo configs and linearity, it is sufficient to show $IMK$ collinear. But this is clear: wrt $(I_AI_BI_C)$, $K$ is the $I_A$-antipode so standard results on the orthocenter configuration imply this. $\fbox{}$ Now it is sufficient to show $B_1C_1$ bisects $AD$ (which has midpoint $S$) by radical axis. Let the $I$-antipode wrt $(AD)$ be $I'$ so by rectangles $I'$ lies on $I_BI_C$. It would be enough to demonstrate that $I'$ lies on $B_1C_1$. Now angle chase: if $I''$ is $B_1C_1\cap I_BI_C$, then \[\measuredangle B_1I''M = \measuredangle C_1B_1M+\measuredangle B_1MA=\measuredangle MC_1B_1+\measuredangle B_1C_1A=\measuredangle MC_1A, \]\[\measuredangle II'A=\measuredangle IEA = \measuredangle MEA=\measuredangle MC_1A.\]Thus $I''=I'$, done.

Define $I_A, I_B, I_C$ excentral and $A'$ to be foot from $A$ to $BC$, and $M$ to be midpoint of arc $BAC$ in $(ABC)$. We claim that $A_1$ is the bevan point $K$. Suppose otherwise, it is well known that $AB_1KC_1$ is cyclic with diameter $AK$. If $A_1$ is on $(AB_1C_1)$, then $\angle AA_1K = 90^{\circ}$ which means $A \in BC$, impossible. So $(AB_1C_1)$ has diameter $AA_1$. It follows that foot $A'$ lies on $(AB_1C_1)$. This circle is getting a lot of points so just call it $\omega$ from now on. Next, note that by Pappus on triangles $I_BA_1I_C$ and $BAC$, we have that $B_1, I, C_1$ are collinear. Furthermore, note that $AIA'D$ cyclic with diameter $AD$. Next, we push to prove that $(AIA'D)$ and $\omega$ are orthogonal. It is also well known that $M$ is center of spiral similarity sending $B_1C_1$ to $CB$, hence it turns out $M$ is also the midpoint of arc $B_1AC_1$ in $\omega$. We will show $A', I, M$ collinear, which would show by angle bisection that\[\frac{AB_1}{AC_1} = \frac{IB_1}{IC_1} = \frac{A'B_1}{A'C_1}\]and thus $AB_1A'C_1$ is harmonic. Let $AA_1$ and $MI$ hit $(ABC)$ again at $X$ and $T$. It is known that $T$ is the mixtillinear touchpoint, and that $AA_1, AT$ are isogonal. It follows that\[\angle MA'A = \angle MAA_1 = \angle MAX = \angle MTX\]but since $AT, AX$ isogonal it follows that $TX \parallel BC$ hence $MAA' = \angle MTX$ means that $A' \in \overline{MIT}$ as desired. It follows that $AIA'D$ is in fact the Appolonius circle to both $AB_1C_1$ and $A'B_1C_1$, and is indeed therefore orthogonal to $\omega$. It follows that center of $AIA'D$, the midpoint of $AD$ denoted as $O$, satisfies the property that $OA, OA'$ are both tangent to $\omega$ and thus $O \in \overline{B_1IC_1}$. It follows that $OD^2 = OA^2 = OA'^2 = OB_1 \cdot OC_1$ so indeed $(DB_1C_1)$ is tangent to $AD$.

Define $M$ as the midpoint of arc $BC$, not containing $A$. Let $I_A,I_B,I_C$ be the excenters of $\triangle ABC$ wrt $A,B,C$ respectively. Let $K$ be the intersection of $AM$ and $BC$. Note that the given condition of cyclic quadrilateral $AB_1A_1C_1$ yields that $A_1$ is the Bevan point of $\triangle ABC$. Note that $\measuredangle DII_A=90^\circ=\measuredangle DA_1I_A$, we get that $DIA_1I_A$ is cyclic quadrilateral. Since $BICI_A$ is cyclic, we get that $$KD\cdot KA_1=KI\cdot KI_A=KB\cdot KC=KA\cdot KM,$$hence $A_1MDA$ is cyclic quadrilateral. Also, $I_AM\cdot I_AA$ is the power of $I_A$ wrt $(ABC)$ and because $I_AA_1$ is tangent to $(A_1B_1C)$ and $(ABC)$ is the nine-point circle of $\triangle I_AI_BI_C$, we get that $I_AM\cdot I_AA=I_AA_1^2$, hence $I_AA_1$ is tangent to $(A_1MDA)$. This means that $\measuredangle DAA_1=\measuredangle DA_1I_A=90^\circ$. Hence $DA$ is tangent to $(AB_1A_1C_1)$. By Pappus theorem on triplets $I_C-A-I_b$ and $B-A_1-C$, we get that $I$ lies on $B_1C_1$. Now, easy angle chase implies $\measuredangle DAI=\measuredangle AIC_1\implies B_1C_1$ passes through the midpoint of $AD$. Hence, by PoP argument, we get that $AD$ is tangent to $(DB_1C_1)$. [asy][asy]import geometry; size(10cm);defaultpen(fontsize(10pt)); pair O,A,B,C,I,A1,B1,C1,Ia,Ib,Ic,D,M,K; O=(0,0);A=dir(110);B=dir(193.3);C=dir(347.2);I=incenter(A,B,C);A1=2midpoint(B--C)-foot(I,B,C); B1=2midpoint(A--C)-foot(I,A,C);C1=2midpoint(A--B)-foot(I,A,B);Ia=intersectionpoint(line(A,I),perpendicular(A1,line(A1,B))); Ib=intersectionpoint(line(B,I),perpendicular(B1,line(B1,C))); Ic=intersectionpoint(line(C,I),perpendicular(C1,line(C1,A))); path w=circumcircle(A,B,C);path o=circumcircle(A1,B1,C1);D=intersectionpoint(perpendicular(I,line(A,I)),line(B,C)); M=intersectionpoints(w,A--Ia)[1];K=extension(B,C,A,I); draw(A--B--C--cycle,red+1);draw(w,heavyblue+1);draw(o,cyan+1);draw(Ia--Ib--Ic--cycle,orange); draw(Ia--A1,heavygreen);draw(Ib--A1,heavygreen);draw(Ic--A1,heavygreen); draw(B--Ib,magenta);draw(A--Ia,magenta);draw(C--Ic,magenta); draw(D--B,red+0.3);draw(A--D,red+0.3);draw(I--D,red+0.3); draw(B1--C1,dashed+red+0.3); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$A_1$",A1,dir(A1)); dot("$B_1$",B1,dir(B1)); dot("$C_1$",C1,dir(C1)); dot("$I$",I,dir(I)); dot("$I_A$",Ia,dir(Ia)); dot("$I_B$",Ib,dir(Ib)); dot("$I_C$",Ic,dir(Ic)); dot("$D$",D,dir(D)); dot("$M$",M,dir(M)); dot("$K$",K,dir(K)); [/asy][/asy]

Oranges are yummy Let $N$ be the midpoint of major arc $\overarc{BAC}$ and $M$ the midpoint of minor arc $\overarc{BC}$. By spiral sim, $N$ also lies on $(AC_1 B_1 A_1)$. Now, observe by PoP, we have \[CB_1 \left(\frac{CA}{CA_1} + \frac{BA}{BA_1}\right) = BC\]\[(s-a)\cdot \frac{s(b+c) - 2bc}{(s+b)(s-c)} = a \Rightarrow s^2 (b+c-a) - s(2bc) + abc = 0\]\[s \left(\frac{(b+c)^2 - a^2}{2} - 2bc\right) + abc = 0 \Rightarrow s(a-b+c)(a+b-c) = 2abc\]\[\frac{K^2}{s-a} = \frac{abc}{2} = K\cdot r_a \Rightarrow r_a = \frac{a}{\sin A} = 2R\]Therefore, $r_a = 2R$. Letting $I_A$ be the A-excenter, we have $I_a A_1 = MN$, so $A_1N || AI$ so $\angle ANA_1 = 90$. This means that $AA_1$ is the diameter of $(AB_1C_1)$, so the foot from $A$ to $BC$, which we denote as $X$, must also lie on this circle. Let $T$ be the A-mixtilinear in-touch point. We first have $N,I,T$ collinear. Next, \[\angle ANX = \angle AA_1 B = 180 - \angle BAA_1 -\angle B = 180 - \angle B - \angle TAC= 180 - \angle TAAN = \angle ANT\]Therefore, $N,X,T$ are collinear, so $N,I,X,T$ are collinear. Observe that $\angle AXD = \angle AID = 90$, so $(AIXD)$ is cyclic. We can also angle chase to get \[\angle ADB = \angle AIN = 90 - \angle ANI = 90 - \angle AA_1 X = \angle XAA_1\]This means $AD\perp AA_1$, and since $AA_1$ is the diameter, we have $AD$ is tangent to $(AB_1C_1)$. Now, observe that by pascal on $C_1B_1 A_1 XNA$, we have $B, C_1B_1\cap XN, A_1B_1\cap AN$ are collinear. However, $A_1B_1\perp AC$ (since $AA_1$ is diameter, and $AN$ is external bisector of $\angle BAC$, so $AN\cap A_1B_1 = I_B$, or the B-excenter. Since $I\in XN$, and $B,I,I_A$ collinear, this means $C_1, I, B_1$ are also collinear. Finally, let $O$ be the midpoint of $AD$. If we show that $O,C_1,B_1$ are collinear, then we're done, since $O$ then $OA^2 = OC_1 OB_1 = OD\cdot OD'$, where $D'$ is the second intersection of $OD$ with $(DC_1B_1)$, but since $OD = OA$, we have $OD' = OA = OD$ so $D = D'$ and $OD$ is tangent to $(DC_1B_1)$. It remains to prove this collinearity. Since $O$ is the center of $(ADXI)$, we have \[\angle OIA = 180 - \angle OAI - \angle AOI = 180 - \frac{1}{2}A - \angle AB_1C_1 - 2\angle AXN\]\[ = 180 - \frac{1}{2}A - \angle AB_1C_1 - \angle AC_1B_1 + \angle AB_1C_1 = 180 - \frac{1}{2}A - \angle AC_1B_1 = \angle C_1IA\]Therefore, $O, I, C_1$ are coillinear, so $O, C_1, B_1$ are collinear (since $I\in C_1B_1$). Therefore, $AD$ is tangent to $(DC_1B_1)$

Solved with Awesome_360! Hopefully this relatively unique solution works :smile:

Proof. Assume WLOG that $C,P$ lie on the same side on the angle bisector of $\angle BAC$ (with the other case following similarly). Let $X=AQ\cap FE$ and $Y=AQ\cap FP$. Note that $\triangle YFX\sim\triangle YAF$, so \[\measuredangle QAF=\measuredangle EFP=\measuredangle EAP\]so $AQ,AP$ are isogonal. By symmetry, $BQ,BP$ are also isogonal and $CQ,CP$ are too. Let $A_1'=I_BB_1\cap I_CC_1$. Note that since $ABC$ is the pedal triangle of the orthocenter in $\triangle I_AI_BI_C$, $I_AA_1'\perp BC$. As $A_1'$ also lies on $(AB_1C_1)$, it follows that $A_1=A_1'$. Let $K$ be the $A$-sharky devil point and $P$ the foot from $I$ to $BC$; $A,K,B$ are collinear by radical axis on $(AFE),(ABC),$ and $(BIC)$. Claim. $\triangle KFE\sim\triangle AC_1B_1$. Proof. First, we prove that $KT$ bisects $\angle BAC$; letting $M$ to be the midpoint of minor arc $BC$, it suffices to show that $K$ and $P$ swap under inversion in $(BIC)$. However, $A$ inverts to $Q=AI\cap BC$, so $K$ inverts to $(IQ)\cap BC=P$. Then, \[\frac{KF}{KE}=\frac{KB}{KC}=\frac{BP}{CP}=\frac{AC_1}{AB_1}\]by angle bisector theorem, and from $\measuredangle FKE=\measuredangle C_1AB_1$ the conclusion follows. Then note that $B_1,I,C_1$ is collinear by Pascal on $A_1I_BBACI_C$. Let $N=B_1IC\cap AD$. Claim. $NA=NI$. Proof. Note that \[\measuredangle NAI=\measuredangle KAM=\measuredangle KBM=\measuredangle KBC+\measuredangle CBM=\measuredangle AC_1I+\measuredangle IAC_1=\measuredangle AIN.\] Since $\measuredangle AID=90^\circ$, it follows that $N$ is the midpoint of $AD$. Since $\measuredangle AB_1C_1=\measuredangle KEF=\measuredangle KAF$, $ND$ is tangent to $(AA_1B_1C_1)$; thus, $ND^2=NA^2=NC_1\cdot NB_1$ which implies the conclusion.

Very nice one Ankan! Let $I_A,I_B,I_C,I$ be the $A,B,C$ excenters and the incenter respectivily, let $O$ the center of $(I_AI_BI_C)$ and let the projection of $A$ to $BC$ be $H$ and let $I_BI_C \cap B_1C_1=G$ Claim 1: $H$ lies on $(AB_1A_1C_1)$. Proof: Let $I_A,I_B,I_C$ be the $A,B,C$ excenters respectivily and let $O$ be the center of $(I_AI_BI_C)$, note that since $\angle BI_AC=90-\frac{\angle BAC}{2}$ we get $\angle I_BOI_C=180-\angle BAC$ hence $O$ lies on $(AB_1A_1C_1)$ but using orthic duality we have that $O-A_1-I_A$ hence $O \equiv A_1$ and since by more orthic duality we have $I_B-B_1-O$ and $I_C-C_1-O$ we get $\angle AB_1A_1=90=\angle AC_1A_1$ hence $AA_1$ is diameter of $(AB_1A_1C_1)$ as desired. Claim 2: $I$ lies on $B_1C_1$ Proof: We use Pascal on the degenrate conic $\mathcal C$ passing through $A,B.C,I_B,I_C,A_1$ to get the desired result. Claim 3: $AGDHI$ cyclic. Proof: We do my favorite thing in geometry (we must take in count Claim 2) $$(G, I; B_1, C_1) \overset{A}{=} (I_BI_C \cap BC, AI \cap BC; B, C)=-1=(I_BI_C \cap BC, A; I_B, I_C) \overset{A_1}{=} (H, A; B_1, C_1) \implies \frac{B_1H}{HC_1}=\frac{B_1A}{AC_1}=\frac{B_1I}{IC_1}=\frac{MB_1}{C_1M}$$Hence $HIAG$ is cyclic becuase its the apolonian circle of $\triangle AB_1C_1$ and since $\angle AHD=90$ we have $AGDHI$ cylcic wih diameters $GI$ and $AD$ (we let $K$ to be center of this circle) Final steps: First letting $HI \cap (AB_1A_1C_1)=J \ne H$ we get by the previous ratios that $J$ is the midpoint of the arc $\widehat{B_1AC_1}$ hence we get $AI \parallel JA_1$ and now since $\angle ADH=\angle AIJ=\angle HJA_1=\angle HAA_1$ we get $AA_1$ tangent to $(AGDHI)$ (Claim 3 goes brrr) and becuase $\angle DAA_1=90$ we can get $KA$ tangent to $(AB_1A_1C_1)$ and now using tangent rule $KD^2=KA^2=KB_1 \cdot KC_1$ thus we are done

I'm not usually a fan of config geo, but this is too good. Let $I_A$, $I_B$, and $I_C$ be the $A$, $B$, and $C$-excenters of $\triangle ABC$, respectively. Let $\Gamma$ and $O$ be the circumcircle and circumcenter of $\triangle ABC$, respectively. Let $\omega$ be the circumcircle of $AB_1A_1C_1$. Let $N$ be the midpoint of $\widehat{BAC}$ of $\omega$, let $K$ be the foot of the $A$-altitude, let $T$ be the $A$-mixtilinear intouch point of $\triangle ABC$, and let $M$ be the midpoint of $\overline{AD}$. Claim 1 (USA EGMO TST 2019/5 + IMO 2013/3): $ANB_1A_1KC_1$ is cyclic. Proof: Notice that since $NB=NC$, $NB_1=NC_1$, and $\measuredangle NBC_1=\measuredangle NCB_1$, $\triangle NBC_1 \cong \triangle NCB_1$. Thus, \[\measuredangle B_1AC_1=\measuredangle CNB=\measuredangle B_1NC_1,\]so $ANB_1A_1C_1$ is cyclic. Let $I'$ be the reflection of $I$ over $O$. Notice that lines $I_AA'$, $I_BB'$, and $I_CC'$ are concurrent. Also notice that $I'$ is the antipode of $A$ in $\omega$. Thus, either $\overline{AA_1} \parallel \overline{BC}$, or $A_1=I'$. The former is impossible, so $A_1$ is the antipode of $A$ in $\omega$. Since $\angle AKA_1=90^\circ$, $AB_1A_1KC$ is cyclic and the claim is proven. Claim 2: $N$, $I$, $K$, and $T$ are collinear. Proof: It's well-known that $N$, $I$, and $T$ are collinear. To prove that $N$, $K$, and $T$ are collinear, do a direct angle chase with $\measuredangle TAI=\measuredangle IAA_1$ to prove that $\measuredangle ANK=\measuredangle ANT$. Claim 3: The circumcircle of $\triangle AKT$ is tangent to $\overline{AI}$. Proof: Do a direct angle chase to prove that $\measuredangle ATI=\measuredangle IAK$. Claim 4: $\measuredangle DAA_1=90^\circ$. Proof: Do a direct angle chase with $\measuredangle AKI=\measuredangle IAT=\measuredangle A_1AI$. By the previous claim, $\overline{AD}$ is tangent to $\omega$. Claim 5: $I$, $B_1$, and $C_1$ are collinear. Proof: Use Pappus on $\overline{BA_1C}$ and $\overline{I_BAI_C}$. Claim 6: $\overline{B_1C_1} \parallel \overline{IM}$. Proof: Do a direct angle chase to prove that $\measuredangle(\overline{AN},\overline{B_1C_1})=\measuredangle(\overline{AN},\overline{IM})$. Hence, $I$, $B_1$, $C_1$, and $M$ are collinear. Therefore, we have \[MB_1 \cdot MC_1=MA^2=MD^2,\]so $\overline{AD}$ is tangent to the circumcircle of $\triangle DB_1C_1$.

Let $A_2 = (AA_1B_1C_1) \cap BC$. Then $a = BC = BA_2 + A_2C = \frac{(s-a)c}{s-c} + \frac{(s - a)b}{s-b}$. Expanding gives $a^3 + a^2c + a^2b - ab^2 - 2abc - ac^2 - c^3 + c^2b + cb^2 - b^3 = 0$ I claim $A_2$ is the altitude from $A$. To confirm this, notice that expanding $\frac{\frac{(s-a)c}{s-c}}{\frac{(s-a)b}{s-b}} = \frac{c\cos B}{b \cos C}$ using LoC yields the claim is equivalent to $(b-c)(a^3 + a^2c + a^2b - ab^2 - 2abc - ac^2 - c^3 + c^2b + cb^2 - b^3) = 0$ as desired. Now, notice that using barycentric coordinates $\text{det} \begin{vmatrix} a & b & c\\ s-a & s-b & 0\\ s-a & 0 & s-c \end{vmatrix} = 0 \Longleftrightarrow a^3 + a^2c + a^2b - ab^2 - 2abc - ac^2 - c^3 + c^2b + cb^2 - b^3 = 0$ so $B_1IC_1$ is collinear. Thus by angle bisector theorem $\frac{s - b}{s - c} = \frac{AC_1}{AB_1} = \frac{IC_1}{IB_1}$. Then notice by the lemma on my blog that if $T = AA_2 \cap (ABC)$ and $M$ is the midpoint of arc $BAC$ then $\triangle B_1C_1A_2$ is spiral similar to $\triangle CBT$ through $M$ so $\frac{A_2C_1}{A_2B_1} = \frac{TB}{TC} = \frac{s-b}{s-c}$ by ratio lemma in $\triangle ABC$. Notice $(AIA_2D)$ is concyclic and thus an Appolonius circle of $C_1B_1$. Hence $\frac{DB_1}{DC_1} = \frac{s - c}{s - b}$. Now notice that computing the $D$ through barycentric coordinates, that $A_2, D$ are harmonic conjugates wrt $B, C$ via cross ratio, so if $D_1 = B_1C_1 \cap AD$ then $\frac{B_1D_1}{C_1D_1} = \frac{s-c}{s-b} \cdot \frac{s-c}{s-b}$ by ratio lemma in $\triangle AB_1C_1$ so it follows that since $\left( \frac{D_1B_1}{D_1C_1} \right)^2 = \frac{DB_1}{DC_1}$, we have $AD = D_1D$ is tangent to $(B_1C_1D)$ as desired. EDIT: WOW are you kidding me? I did bary because it was way too simple but I came here hoping to see some nice synthetic solution given all the nice observations made in the solution... but it was just a Bevan point configuration knowledge check? SMH.

nooooooooooo. Angle chase to get that $AA_1$ is a diameter, so draw $A_1B_1\cap AB=V$ and $A_1C_1\cap AC=U$ and note $A_1$ is orthocenter of $\triangle AUV$. Let $K=AA_1\cap B_1C_1$ and let $Z$ be the harmonic conjugate of $B_1C_1$ with respect to $K$. So now $Z$ lies on $UV$ and $BC$. Let $E$ be the foot of the altitude from $A$ to $BC$, note $AB_1EC_1$ harmonic. Now we need to show that $I$ lies on $B_1C_1$. This is simple by Menelaus bash: let $AI\cap BC=J$ and Menelaus on $\triangle CAJ$ to show $ZB_1I$ collinear. Now $(AEID)$ is obviously an Apollonian circle. From here we can let $O$ be the midpoint of $AD$, notice that $(AEID)$ and $(B_1C_1D)$ orthogonal and we're done. rip bevan point oh i never drew the excenters that might have been the issue LOL!

USA TST 2019 Nice problem. Here is a sketch of the solution. Firstly we show that $B_1IC_1$ are collinear by pappus theorem. Let $(AB_1A_1C_1) \cap BC=H’$ then we can easily how that $H’$ is the $A$ feet of altitude. The main idea is showing that $AH’B_1C_1$ is harmonic. While showing that the tangents at $A$ and $H’$ intersect at $M$ which the intersection of $AD$ with $B_1C_1$. We how that $M$ is the Center of $AIH’D$ hence by power of point we get that. $$MD^2=MA^2=MB_1 \cdot MC_1$$Hence we are done.

Let’s AK this. Parts of this solution will be similar to other solutions above. Step 1: Proving as much as possible without the sticky condition. We prove the usual claim that $(AB_1C_1)$ is tangent to $\overline{AD}$, and $\overline{IM}\parallel\overline{B_1C_1T}$ where $M$ is the midpoint of $AD$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10) + blue; defaultpen(dps); /* default pen style */ pen dotstyle = blue; /* point style */ real xmin = -21, xmax = 18, ymin = -11, ymax = 11; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccqqqq = rgb(0.8,0,0); pen ffvvqq = rgb(1,0.3333333333333333,0); draw((-5.00130207163223,8.65950215591446)--(-9.152407121756008,-4.029074816583814)--(9.101968524336803,-4.141759165135286)--cycle, linewidth(0.5) + ccqqqq); /* draw figures */ draw(circle((0,0), 10), linewidth(0.5) + qqwuqq); draw((-5.00130207163223,8.65950215591446)--(-9.152407121756008,-4.029074816583814), linewidth(0.5) + ccqqqq); draw((-9.152407121756008,-4.029074816583814)--(9.101968524336803,-4.141759165135286), linewidth(0.5) + ccqqqq); draw((9.101968524336803,-4.141759165135286)--(-5.00130207163223,8.65950215591446), linewidth(0.5) + ccqqqq); draw((-19.759779416499267,-3.9635954564539646)--(-2.845029399842456,0.5141498919484042), linewidth(0.5) + ffvvqq); draw((-19.759779416499267,-3.9635954564539646)--(-9.152407121756008,-4.029074816583814), linewidth(0.5) + ffvvqq); draw((-19.759779416499267,-3.9635954564539646)--(-5.00130207163223,8.65950215591446), linewidth(0.5) + ffvvqq); draw(circle((-3.9231657357373417,4.586826023931432), 4.212964376160934), linewidth(0.5) + qqwuqq); draw(circle((-1.0781363358948877,4.072676131983027), 6.035743729142097), linewidth(0.5) + qqwuqq); draw(circle((-0.061728871928146724,-9.999809475503547), 10.876125385867832), linewidth(0.5) + qqwuqq); draw(circle((-3.1536888609567755,-2.271612354266112), 6.2508635466551326), linewidth(0.5) + qqwuqq); draw(circle((5.97349896208963,-2.3279545285418446), 3.616242395309652), linewidth(0.5) + qqwuqq); draw((-11.056367069694842,3.4805345203159774)--(3.8660813226853814,0.6107526625138946), linewidth(0.5) + ffvvqq); draw((-6.953736677062792,2.6915450756578996)--(2.8228754214215614,-4.10299829324825), linewidth(0.5) + ffvvqq); draw((2.8228754214215614,-4.10299829324825)--(3.8660813226853814,0.6107526625138946), linewidth(0.5) + ffvvqq); draw((-5.00130207163223,8.65950215591446)--(-2.845029399842456,0.5141498919484042), linewidth(0.5) + ffvvqq); draw((-7.77946021700683,6.283311128060428)--(-7.199972516325446,1.9388822636727459), linewidth(0.5) + ffvvqq); draw((-7.199972516325446,1.9388822636727459)--(0.23458513001919146,3.906990328265279), linewidth(0.5) + ffvvqq); draw((0.23458513001919146,3.906990328265279)--(-7.77946021700683,6.283311128060428), linewidth(0.5) + ffvvqq); draw((-7.77946021700683,6.283311128060428)--(-2.845029399842456,0.5141498919484042), linewidth(0.5) + ffvvqq); /* dots and labels */ dot((0,0),dotstyle); label("$O$", (0.12817675840733747,0.2926007430266706), NE * labelscalefactor); dot((-5.00130207163223,8.65950215591446),dotstyle); label("$A$", (-4.89166511339295,8.939495990405241), NE * labelscalefactor); dot((-9.152407121756008,-4.029074816583814),dotstyle); label("$B$", (-9.041014175054459,-3.7406826106163535), NE * labelscalefactor); dot((9.101968524336803,-4.141759165135286),dotstyle); label("$C$", (9.210318410855255,-3.856748318634858), NE * labelscalefactor); dot((-2.845029399842456,0.5141498919484042),dotstyle); label("$I$", (-2.7154330880460043,0.81489642910994), NE * labelscalefactor); dot((2.8228754214215614,-4.10299829324825),dotstyle); label("$A_1$", (2.9427701778560533,-3.798715464625606), NE * labelscalefactor); dot((3.8660813226853814,0.6107526625138946),dotstyle); label("$B_1$", (3.987361550022587,0.9019457101238182), NE * labelscalefactor); dot((-6.953736677062792,2.6915450756578996),dotstyle); label("$C_1$", (-6.835765722702887,2.9911284544568955), NE * labelscalefactor); dot((-19.759779416499267,-3.9635954564539646),dotstyle); label("$D$", (-19.573977177733674,-3.6826497566071015), NE * labelscalefactor); dot((-7.199972516325446,1.9388822636727459),dotstyle); label("$C_2$", (-7.09691356574452,2.2367013523366177), NE * labelscalefactor); dot((0.23458513001919146,3.906990328265279),dotstyle); label("$B_2$", (0.360308174444345,4.209818388651191), NE * labelscalefactor); dot((-2.873314018840767,-4.067835688470851),dotstyle); label("$A_2$", (-2.74444951505063,-3.76969903762098), NE * labelscalefactor); dot((-7.77946021700683,6.283311128060428),dotstyle); label("$S$", (-7.677242105837039,6.560148976025903), NE * labelscalefactor); dot((-0.06172887192814658,-9.999809475503545),dotstyle); label("$M_a$", (0.04112747739345968,-9.718066573569326), NE * labelscalefactor); dot((-11.056367069694842,3.4805345203159774),dotstyle); label("$T$", (-10.927081930355143,3.7745719835817995), NE * labelscalefactor); dot((2.845029399842456,-0.5141498919484042),dotstyle); label("$V$", (2.971786604860679,-0.2296949430565987), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] I claim that if $S$ is the $A$-sharkydevil point then $\triangle SB_2C_2\stackrel{+}{\sim}\triangle AB_1C_1$. We however have angles $\measuredangle C_2SB_2=\measuredangle C_2AB_2=\measuredangle C_1AB_1$, and ratios $\frac{SC_2}{SB_2}=\frac{BC_2}{CB_2}=\frac{AC_1}{AB_1}$, as desired. So $\measuredangle DAC_1=\measuredangle SAC_2=\measuredangle SB_2C_2=\measuredangle AB_1C_1$, as desired. Now further notice that $\measuredangle (\overline{B_1C_1},\overline{B_2C_2})=\measuredangle (\overline{AC_1},\overline{SC_2})=\measuredangle AC_2S=\measuredangle AIS=\measuredangle IDA=\measuredangle MID=\measuredangle(\overline{MI},\overline{B_2C_2})$, as desired. Step 2: Interpreting the condition Introduce the Bevan point $V$; we claim that $V=A_1$. However if $AA_1B_1C_1$ cyclic and $V\neq A_1$ then as $AV$ is the diameter of $(AA_1B_1C_1)$ then $\overline{AA_1}\perp\overline{A_1V}$. But then again $\overline{A_1V}\perp\overline{BC}$ by definition of the Bevan point, and hence $\overline{AA_1}\parallel\overline{BC}$, which is a contradiction. So we have that $V=A_1$. Now $I_aA_1=r_a$, the $A$-exradius, whilst $I_aV=2R$ due to the excentre-orthocentre duality so $r_a=2R$, where $I_a$ is the $A$-excentre. Since I'm actually very dumb, let's draw in $M_b$ and $M_c$ being defined as per the diagram below: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10) + blue; defaultpen(dps); /* default pen style */ pen dotstyle = blue; /* point style */ real xmin = -21.65110060692449, xmax = 20.977842173753757, ymin = -11.055305559931867, ymax = 12.390612969441175; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccqqqq = rgb(0.8,0,0); pen ffvvqq = rgb(1,0.3333333333333333,0); draw((-5.476119010842156,8.367324577132951)--(-9.78267371561353,-2.0734741314624983)--(9.786194275204696,-2.056794012036894)--cycle, linewidth(0.5) + ccqqqq); /* draw figures */ draw(circle((0,0), 10), linewidth(0.5) + qqwuqq); draw((-5.476119010842156,8.367324577132951)--(-9.78267371561353,-2.0734741314624983), linewidth(0.5) + ccqqqq); draw((-9.78267371561353,-2.0734741314624983)--(9.786194275204696,-2.056794012036894), linewidth(0.5) + ccqqqq); draw((9.786194275204696,-2.056794012036894)--(-5.476119010842156,8.367324577132951), linewidth(0.5) + ccqqqq); draw((-3.595934733432582,2.0708437955383086)--(-3.5924066956343808,-2.0681976692155213), linewidth(0.5) + ffvvqq); draw((-5.476119010842156,8.367324577132951)--(-3.595934733432582,2.0708437955383086), linewidth(0.5) + ffvvqq); draw((-9.244471768878217,3.813101324960256)--(9.786194275204696,-2.056794012036894), linewidth(0.5) + ffvvqq); draw((-9.78267371561353,-2.0734741314624983)--(5.640013234593206,8.257738837818346), linewidth(0.5) + ffvvqq); draw((5.640013234593206,8.257738837818346)--(5.486618265075023,-8.411225757322281), linewidth(0.5) + ffvvqq); draw((-9.244471768878217,3.813101324960256)--(5.486618265075023,-8.411225757322281), linewidth(0.5) + ffvvqq); draw((-9.244471768878217,3.813101324960256)--(-5.476119010842156,8.367324577132951), linewidth(0.5) + ffvvqq); draw((-5.476119010842156,8.367324577132951)--(5.486618265075023,-8.411225757322281), linewidth(0.5) + ffvvqq); draw((-7.836531405897288,2.6447476274567396)--(3.595927255225547,-2.0620704742838716), linewidth(0.5) + ffvvqq); draw((3.595927255225547,-2.0620704742838716)--(5.57158428578019,0.8217731824280499), linewidth(0.5) + ffvvqq); /* dots and labels */ dot((0,0),dotstyle); label("$O$", (0.12962484507830163,0.33461508928060524), NE * labelscalefactor); dot((-5.476119010842156,8.367324577132951),dotstyle); label("$A$", (-5.3322084486960994,8.693884337679233), NE * labelscalefactor); dot((-9.78267371561353,-2.0734741314624983),dotstyle); label("$B$", (-9.661710449858735,-1.7302243266584978), NE * labelscalefactor); dot((9.786194275204696,-2.056794012036894),dotstyle); label("$C$", (9.920960140015337,-1.7302243266584978), NE * labelscalefactor); dot((-3.595934733432582,2.0708437955383086),dotstyle); label("$I$", (-3.4671922020414256,2.3994545052197083), NE * labelscalefactor); dot((3.595927255225547,-2.0620704742838716),dotstyle); label("$A_1=V$", (3.726441892198029,-1.7302243266584978), NE * labelscalefactor); dot((5.57158428578019,0.8217731824280499),dotstyle); label("$B_1$", (5.691369723494917,1.1672116279657274), NE * labelscalefactor); dot((-7.836531405897288,2.6447476274567396),dotstyle); label("$C_1$", (-7.696782618561846,2.9656201515255916), NE * labelscalefactor); dot((-3.5924066956343808,-2.0681976692155213),dotstyle); label("$A_2$", (-3.4671922020414256,-1.7302243266584978), NE * labelscalefactor); dot((0.008523800852425989,-9.999996367240291),dotstyle); label("$M_a$", (0.12962484507830163,-9.656543374940862), NE * labelscalefactor); dot((5.640013234593206,8.257738837818346),dotstyle); label("$M_b$", (5.757977446589726,8.593972753037018), NE * labelscalefactor); dot((-9.244471768878217,3.813101324960256),dotstyle); label("$M_c$", (-9.095544803552851,4.131255305684762), NE * labelscalefactor); dot((5.486618265075023,-8.411225757322281),dotstyle); label("$U$", (5.624762000400107,-8.091261882212832), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We are going to force an application of Pascal to get $B_1$, $I$, $C_1$ collinear. To do that I claim the hexagon $AM_cBUCM_b$ works where $U$ is the antipode of $A$ on $(ABC)$. All I actually need to show is that $B_1=\overline{AC}\cap\overline{M_bU}$. I will do this using the ratio lemma. Let $B_1'=\overline{AC}\cap\overline{M_bU}$; then: \begin{align*} \frac{AB_1}{B_1C}&=\frac{AM_b}{M_bC}\frac{AU}{UC}\\ &=\frac{1}{\cos(\angle B)}\\ &=\frac{2ac}{a^2+c^2-b^2} \end{align*}I claim that with the condition $2R=r_a$, $\frac{2ac}{a^2+c^2-b^2}=\frac{AB_1}{B_1C}=\frac{s-c}{s-a}$. Notice however $2R=\frac{abc}{2rs}=\frac{s}{s-a}r$, so $2(rs)^2=abc(s-a)$. Now $\frac{2ac}{a^2+2ac+c^2-b^2}=\frac{2ac}{(a+c-b)(a+b+c)}=\frac{ac}{2s(s-b)}$, whilst $\frac{s-c}{2s-a-c}=\frac{s-c}{b}$, so it suffices to show that $abc=2s(s-b)(s-c)$, i.e. $\frac{2(rs)^2}{s-a}=2s(s-b)(s-c)$, or $(rs)^2=s(s-a)(s-b)(s-c)$, which is just Heron’s formula squared. So the Pascal works. Part 3: The finish [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10) + blue; defaultpen(dps); /* default pen style */ pen dotstyle = blue; /* point style */ real xmin = -23.259339754245026, xmax = 19.541271812197504, ymin = -10.447898799466639, ymax = 13.092437562076874; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccqqqq = rgb(0.8,0,0); pen ffvvqq = rgb(1,0.3333333333333333,0); draw((-5.529490044046149,8.332150973955915)--(-9.78267371561353,-2.0734741314624983)--(9.786194275204696,-2.056794012036894)--cycle, linewidth(0.5) + ccqqqq); /* draw figures */ draw(circle((0,0), 10), linewidth(0.5) + qqwuqq); draw((-5.529490044046149,8.332150973955915)--(-9.78267371561353,-2.0734741314624983), linewidth(0.5) + ccqqqq); draw((-9.78267371561353,-2.0734741314624983)--(9.786194275204696,-2.056794012036894), linewidth(0.5) + ccqqqq); draw((9.786194275204696,-2.056794012036894)--(-5.529490044046149,8.332150973955915), linewidth(0.5) + ccqqqq); draw((-17.336151248875133,-2.0799125676251085)--(-3.6344941786712726,2.0592624641139796), linewidth(0.5) + ffvvqq); draw((-17.336151248875133,-2.0799125676251085)--(-9.78267371561353,-2.0734741314624983), linewidth(0.5) + ffvvqq); draw((-17.336151248875133,-2.0799125676251085)--(-5.529490044046149,8.332150973955915), linewidth(0.5) + ffvvqq); draw(circle((-4.58199211135871,5.1957067190349475), 3.2764363104863987), linewidth(0.5) + qqwuqq); draw(circle((-0.9474979326874364,3.1364442549209683), 6.927482949728449), linewidth(0.5) + qqwuqq); draw(circle((0.008523800852427224,-9.999996367240291), 12.597511800380618), linewidth(0.5) + qqwuqq); draw(circle((-3.074089768471129,-2.0663682977880526), 6.708587710444645), linewidth(0.5) + qqwuqq); draw(circle((6.710344226938108,-2.058028238075522), 3.0758502958916925), linewidth(0.5) + qqwuqq); draw((-11.432001519352662,3.12684157533726)--(5.574190950974938,0.8002948325823711), linewidth(0.5) + ffvvqq); draw((-7.857008451664908,2.6377615833571366)--(3.634496544141483,-2.0620375985781543), linewidth(0.5) + ffvvqq); draw((3.634496544141483,-2.0620375985781543)--(5.574190950974938,0.8002948325823711), linewidth(0.5) + ffvvqq); draw((-5.529490044046149,8.332150973955915)--(-3.6344941786712726,2.0592624641139796), linewidth(0.5) + ffvvqq); draw((-7.575344228330761,6.527952192096369)--(-7.455155307994771,3.62091525913628), linewidth(0.5) + ffvvqq); draw((-7.455155307994771,3.62091525913628)--(-1.3174867198163902,5.47506212933665), linewidth(0.5) + ffvvqq); draw((-1.3174867198163902,5.47506212933665)--(-7.575344228330761,6.527952192096369), linewidth(0.5) + ffvvqq); draw((-7.575344228330761,6.527952192096369)--(-3.6344941786712726,2.0592624641139796), linewidth(0.5) + ffvvqq); draw((-5.5257089064128975,4.203787559149821)--(-3.6309759845503167,-2.0682305449212386), linewidth(0.5) + ffvvqq); draw((-3.6344941786712726,2.0592624641139796)--(0.008523800852428653,-9.999996367240293), linewidth(0.5) + ffvvqq); draw((-7.575344228330761,6.527952192096369)--(0.008523800852428653,-9.999996367240293), linewidth(0.5) + ffvvqq); draw((-3.6344941786712726,2.0592624641139796)--(-7.455155307994771,3.62091525913628), linewidth(0.5) + ffvvqq); draw((-3.6344941786712726,2.0592624641139796)--(-1.3174867198163902,5.47506212933665), linewidth(0.5) + ffvvqq); draw((-3.6344941786712726,2.0592624641139796)--(3.634496544141483,-2.0620375985781543), linewidth(0.5) + ffvvqq); /* dots and labels */ dot((0,0),dotstyle); label("$O$", (0.1472446961532296,0.3191300477166152), NE * labelscalefactor); dot((-5.529490044046149,8.332150973955915),dotstyle); label("$A$", (-5.403459616369786,8.678624494287465), NE * labelscalefactor); dot((-9.78267371561353,-2.0734741314624983),dotstyle); label("$B$", (-9.650082795227755,-1.7540245750329557), NE * labelscalefactor); dot((9.786194275204696,-2.056794012036894),dotstyle); label("$C$", (9.911134209747932,-1.7205865972466723), NE * labelscalefactor); dot((-3.6344941786712726,2.0592624641139796),dotstyle); label("$I$", (-3.497494882551642,2.392284670466186), NE * labelscalefactor); dot((3.634496544141483,-2.0620375985781543),dotstyle); label("$A_1=V$", (3.758546297071818,-1.7205865972466723), NE * labelscalefactor); dot((5.574190950974938,0.8002948325823711),dotstyle); label("$B_1$", (5.697949008676245,1.1216415145874168), NE * labelscalefactor); dot((-7.857008451664908,2.6377615833571366),dotstyle); label("$C_1$", (-7.710680083623329,2.960730292833004), NE * labelscalefactor); dot((-17.336151248875133,-2.0799125676251085),dotstyle); label("$D$", (-17.207065774927766,-1.7540245750329557), NE * labelscalefactor); dot((-7.455155307994771,3.62091525913628),dotstyle); label("$C_2$", (-7.30942435018793,3.9638696264215056), NE * labelscalefactor); dot((-1.3174867198163902,5.47506212933665),dotstyle); label("$B_2$", (-1.1902744152980995,5.802958404667093), NE * labelscalefactor); dot((-3.6309759845503167,-2.0682305449212386),dotstyle); label("$A_2$", (-3.497494882551642,-1.7205865972466723), NE * labelscalefactor); dot((-7.575344228330761,6.527952192096369),dotstyle); label("$S$", (-7.443176261333063,6.872973693828162), NE * labelscalefactor); dot((0.008523800852428653,-9.999996367240293),dotstyle); label("$M_a$", (0.1472446961532296,-9.678825310382122), NE * labelscalefactor); dot((-11.432001519352662,3.12684157533726),dotstyle); label("$T$", (-11.288543706755634,3.462299959627255), NE * labelscalefactor); dot((3.6344941786712726,-2.0592624641139796),dotstyle); dot((-5.5257089064128975,4.203787559149821),dotstyle); label("$P$", (-5.403459616369786,4.532315248788324), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] So what happens now is that $\overline{IM}$ is the same as $\overline{C_1B_1}$, so $DM^2=AM^2=MC_1\cdot MB_1$, thus done by PoP!