5-page bary bash incoming a) [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(400); 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 = -9.500000000000010, xmax = 12.50000000000001, ymin = -5.000000000000000, ymax = 4.000000000000000; /* image dimensions */ /* draw figures */ draw((3.000000000000000,3.000000000000000)--(2.000000000000000,-4.000000000000000)); draw((2.000000000000000,-4.000000000000000)--(12.00000000000000,-4.000000000000000)); draw((3.000000000000000,3.000000000000000)--(12.00000000000000,-4.000000000000000)); draw(circle((4.834656780437054,-1.541515208943204), 2.458484791056800)); draw((3.000000000000000,3.000000000000000)--(5.827822185373194,-4.000000000000000)); draw((3.213033967190907,-0.9851238539119858)--(12.00000000000000,-4.000000000000000)); draw((7.449444998619339,0.7262785890927330)--(2.000000000000000,-4.000000000000000)); draw((3.213033967190907,-0.9851238539119858)--(-8.518800293636604,-3.999999999999962)); draw((7.449444998619339,0.7262785890927330)--(-8.518800293636604,-3.999999999999962)); draw((3.213033967190907,-0.9851238539119858)--(7.449444998619339,0.7262785890927330)); draw((-8.518800293636604,-3.999999999999962)--(2.000000000000000,-4.000000000000000)); /* dots and labels */ dot((3.000000000000000,3.000000000000000),dotstyle); label("$A$", (3.075845136715499,3.116962329614384), NE * labelscalefactor); dot((2.000000000000000,-4.000000000000000),dotstyle); label("$B$", (1.897830819781713,-4.576943677859385), NE * labelscalefactor); dot((12.00000000000000,-4.000000000000000),dotstyle); label("$C$", (12.07661077703770,-3.895904150882041), NE * labelscalefactor); dot((2.400881006356696,-1.193832955503153),dotstyle); label("$C_0$", (2.376399136036064,-1.687127306631199), NE * labelscalefactor); dot((6.344020441375566,0.3990952122634552),dotstyle); label("$B_0$", (6.168132718666686,0.6136819061300954), NE * labelscalefactor); dot((5.827822185373194,-4.000000000000000),dotstyle); label("$L$", (5.800003244624878,-4.632163098965656), NE * labelscalefactor); dot((3.213033967190907,-0.9851238539119858),dotstyle); label("$P$", (3.149471031523860,-1.558281990716566), NE * labelscalefactor); dot((7.449444998619339,0.7262785890927330),dotstyle); label("$Q$", (7.530211772621375,0.8345595905551796), NE * labelscalefactor); dot((-8.518800293636604,-3.999999999999962),dotstyle); label("$R$", (-8.925175717047441,-4.521724256753114), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We use barycentric coordinates. Let $A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1)$, let $a = BC, b = CA, c = AB$, and let $s$ be the semiperimeter of $\triangle ABC$. It is known that $C_0 = (s-b : s-a : 0), B_0 = (s-c : 0 : s-a)$. Since $P$ and $Q$ lie on the angle bisectors of $\angle C$ and $\angle B$ respectively, let $P = (ap, bp, 1-p(a+b)), Q = (aq, 1-q(a+c), cq)$ for some $p$ and $q$. It is well-known that $L = \left(0, \frac{b}{b+c}, \frac{c}{b+c}\right)$. Therefore the midpoint $M$ of $AL$ has coordinates $M = \left(\frac{1}{2}, \frac{b}{2(b+c)}, \frac{c}{2(b+c)}\right)$. Thus the displacement vectors $\overrightarrow{AL}, \overrightarrow{PM}, \overrightarrow{QM}$ are: \begin{align*} \overrightarrow{AL}&: \left(1, -\frac{b}{b+c}, -\frac{c}{b+c}\right)\\ \overrightarrow{PM}&: \left(ap-\frac{1}{2}, bp-\frac{b}{2(b+c)}, 1-p(a+b)-\frac{c}{2(b+c)}\right)\\ \overrightarrow{QM}&: \left(aq-\frac{1}{2}, 1-q(a+c)-\frac{b}{2(b+c)}, cq-\frac{c}{2(b+c)}\right)\end{align*} From the conditions $\overrightarrow{AL} \perp \overrightarrow{PM}$ and $\overrightarrow{AL} \perp \overrightarrow{QM}$, we have \begin{align*} &\overrightarrow{AL} \perp \overrightarrow{PM} \\ \Rightarrow \ 0 &= a^2\left(\left(bp-\frac{b}{2(b+c)}\right)\left(-\frac{c}{b+c}\right)+\left(-\frac{b}{b+c}\right)\left(1-p(a+b)-\frac{c}{2(b+c)}\right)\right)\\ &+b^2\left(\left(-\frac{c}{b+c}\right)\left(ap-\frac{1}{2}\right)+\left(1-p(a+b)-\frac{c}{2(b+c)}\right)\right) \\ &+c^2\left(\left(ap-\frac{1}{2}\right)\left(-\frac{b}{b+c}\right)+\left(bp-\frac{b}{2(b+c)}\right)\right) \\ \Rightarrow \ p &= \frac{b}{(a+b-c)(b+c)}\end{align*} and \begin{align*} &\overrightarrow{AL} \perp \overrightarrow{QM} \\ \Rightarrow \ 0 &= a^2\left(\left(1-q(a+c)-\frac{b}{2(b+c)}\right)\left(-\frac{c}{b+c}\right)+\left(-\frac{b}{b+c}\right)\left(cq-\frac{c}{2(b+c)}\right)\right)\\ &+b^2\left(\left(-\frac{c}{b+c}\right)\left(aq-\frac{1}{2}\right)+\left(cq- \frac{c}{2(b+c)}\right)\right) \\ &+c^2\left(\left(aq-\frac{1}{2}\right)\left(-\frac{b}{b+c}\right)+\left(1-q(a+c)-\frac{b}{2(b+c)}\right)\right) \\ \Rightarrow \ q &= \frac{c}{(a-b+c)(b+c)}.\end{align*} After substitution, we see that the lines $PC_0, QB_0, BC$ have equations \begin{align*} PC_0 &: c(c-a)(-a+b+c)x+c(c-a)(-a+b-c)y+b(-a^2+b^2-bc+ac)z = 0\\ QB_0 &: b(b-a)(-a+b+c)x+c(-a^2+ab-bc+c^2)y+b(b-a)(-a-b+c)z = 0 \\ BC&: x + 0y + 0z = 0. \end{align*} Therefore it remains to show that \begin{align*} \det \left| \begin{array}{ccc} c(c-a)(-a+b+c) & c(c-a)(-a+b-c) & b(-a^2+b^2-bc+ac) \\ b(b-a)(-a+b+c) & c(-a^2+ab-bc+c^2) & b(b-a)(-a-b+c) \\ 1 & 0 & 0 \end{array} \right| = 0.\end{align*} But this is not difficult, since it is just straightforward computation: \begin{align*}&c(c-a)(-a+b-c)b(b-a)(-a-b+c)- b(-a^2+b^2-bc+ac)c(-a^2+ab-bc+c^2)\\&=(a^4bc-a^3b^2c-a^2b^3c+ab^4c-a^3bc^2-ab^3c^2-b^4c^2-a^2bc^3-ab^2c^3+2b^3c^3+abc^4-b^2c^4+3a^2b^2c^2)\\&-(a^4bc-a^3b^2c-a^2b^3c+ab^4c-a^3bc^2-ab^3c^2-b^4c^2-a^2bc^3-ab^2c^3+2b^3c^3+abc^4-b^2c^4+3a^2b^2c^2) = 0. \end{align*} b) [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(400); 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 = -9.500000000000010, xmax = 15.50000000000002, ymin = -5.500000000000006, ymax = 8.000000000000000; /* image dimensions */ /* draw figures */ draw((2.997845573765339,3.003596319012727)--(1.997845573765337,-3.996403680987302)); draw((2.997845573765339,3.003596319012727)--(11.99784557376532,-3.996403680987302)); draw((2.997845573765339,3.003596319012727)--(5.825667759138527,-3.996403680987302)); draw(circle((3.911756666451933,-0.6983909799425150), 3.813127777423874)); draw(circle((8.911756666451927,1.321482009609752), 6.148483793757588)); draw((3.051084486883713,-0.9267006134126927)--(11.99784557376532,-3.996403680987302)); draw((7.704952354951876,0.9533437873185249)--(1.997845573765337,-3.996403680987302)); draw((7.704952354951876,0.9533437873185249)--(11.99784557376532,-3.996403680987302)); draw((1.997845573765337,-3.996403680987302)--(3.051084486883713,-0.9267006134126927)); draw((-8.520954719871137,-3.996403680987341)--(11.99784557376532,-3.996403680987302)); draw((-8.520954719871137,-3.996403680987341)--(3.911756666451933,-0.6983909799425153)); draw((-8.520954719871137,-3.996403680987341)--(8.911756666451927,1.321482009609752)); /* dots and labels */ dot((2.997845573765339,3.003596319012727),dotstyle); label("$A$", (2.542186356907239,3.072476274997438), NE * labelscalefactor); dot((1.997845573765337,-3.996403680987302),dotstyle); label("$B$", (1.698373636065485,-4.698450177405645), NE * labelscalefactor); dot((11.99784557376532,-3.996403680987302),dotstyle); label("$C$", (11.84374983781402,-3.756519698326484), NE * labelscalefactor); dot((5.825667759138527,-3.996403680987302),dotstyle); label("$L$", (5.603460413914536,-4.678826625758162), NE * labelscalefactor); dot((3.911756666451933,-0.6983909799425153),dotstyle); label("$O_1$", (3.994329178820957,-0.5775043314343133), NE * labelscalefactor); dot((8.911756666451927,1.321482009609752),dotstyle); label("$O_2$", (8.998334848929039,1.443721488256388), NE * labelscalefactor); dot((3.051084486883713,-0.9267006134126927),dotstyle); label("$C_1$", (3.130892906331720,-0.8129869512041038), NE * labelscalefactor); dot((7.704952354951876,0.9533437873185249),dotstyle); label("$B_1$", (7.408827165482942,1.070874006954219), NE * labelscalefactor); dot((-8.520954719871137,-3.996403680987341),dotstyle); label("$S$", (-9.016085563460052,-4.619955970815716), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We use barycentric coordinates. We preserve the notation from part a). Let $C_1 = (ar, br, 1-r(a+b)), B_1 = (as, 1-s(a+c), cs)$ since $C_1$ and $B_1$ lie on the angle bisectors of $\angle C$ and $\angle B$ respectively. Therefore the displacement vectors $\overrightarrow{BC_1}, \overrightarrow{CC_1}, \overrightarrow{CB_1}, \overrightarrow{BB_1}$ are: \begin{align*}\overrightarrow{BC_1}&: \left(ar, br-1, 1-r(a+b)\right) \\ \overrightarrow{CC_1}&: \left(ar, br, -r(a+b)\right) \\ \overrightarrow{CB_1}&: \left(as, 1-s(a+c), cs-1\right) \\ \overrightarrow{BB_1} &: \left(as, -s(a+c), cs\right)\end{align*} From the conditions $\overrightarrow{BC_1} \perp \overrightarrow{CC_1}$ and $\overrightarrow{CB_1} \perp \overrightarrow{BB_1}$ we have \begin{align*} &\overrightarrow{BC_1} \perp \overrightarrow{CC_1} \\ \Rightarrow \ 0 &= a^2\left((br-1)(-r(a+b))+(1-r(a+b))(br)\right)\\&+b^2\left((ar)(-r(a+b))+(1-r(a+b))(ar)\right)\\&+c^2\left((ar)(br)+(br-1)(ar)\right)\\ \Rightarrow r &= \frac{1}{2b}\end{align*} and \begin{align*} &\overrightarrow{CB_1} \perp \overrightarrow{BB_1} \\ \Rightarrow \ 0 &= a^2\left((1-s(a+c))(cs)+(cs-1)(-s(a+c))\right)\\&+b^2\left((as)(cs)+(cs-1)(as)\right)\\&+c^2\left((as)(-s(a+c))+(1-s(a+c))(as)\right)\\ \Rightarrow s &= \frac{1}{2c}.\end{align*} We now compute the coordinates of $O_1$ and $O_2$. Denote by $\mathcal{P}(XY)$ the perpendicular bisector of segment $XY$. It is evident that $O_1 = \mathcal{P}(AB) \cap \mathcal{P}(AL)$ and $O_2 = \mathcal{P}(AC) \cap \mathcal{P}(AL)$. It is well-known that the equations of $\mathcal{P}(AB)$ and $\mathcal{P}(AC)$ are: \begin{align*} \mathcal{P}(AB)&: c^2(y-x)+z(b^2-a^2)=0, \\ \mathcal{P}(AC)&: b^2(x-z)+y(a^2-c^2)=0.\end{align*} Incidentally, we have almost calculated the equation of $\mathcal{P}(AL)$ in part a): It is the line connecting points $P$ and $Q$, as defined from the previous problem. From this, we can deduce the equation of $\mathcal{P}(AL)$: \[\mathcal{P}(AL): bcx-c^2y-b^2z = 0.\] Now performing the desired intersections (by both adding and subtracting equations) we obtain the coordinates of $O_1$ and $O_2$: \begin{align*} O_1&: \left(a^2c:a^2b+b^2c-b^3:bc^2-c^3\right) \\ O_2&: \left(a^2b:b^2c-b^3:a^2c+bc^2-c^3\right)\end{align*} Finally, we can find the equations of lines $O_1C_1$ and $O_2B_1$: \begin{align*} O_1C_1&: x(a^3b-a^2b^2-ab^3+b^4+ab^2c-b^3c+b^2c^2-bc^3)\\&+y(-a^3c+a^2bc-abc^2+ac^3)\\&+z(a^3b-ab^3-a^2bc+ab^2c) = 0, \\ O_2B_1&: x(a^3c-b^3c-a^2c^2+abc^2+b^2c^2-ac^3-bc^3+c^4)\\&+y(a^3c-a^2bc+abc^2-ac^3)\\&+z(-a^3b+ab^3+a^2bc-ab^2c) = 0.\end{align*} Adding the two equations immediately yields $x = 0$, implying that the intersection of the two lines lie on line $BC$. c) We now explicitly find the coordinates of the intersections. From a) we find the intersection $R$ of $PC_0$ and $QB_0$, and from b) we find the intersection $S$ of $O_1C_1$ and $O_2B_1$. We have proven that the $x$-coordinates of $R$ and $S$ are $0$. It is not difficult to see that, after substituting $x = 0$ into the equations of $PC_0$ and $O_1C_1$, that \begin{align*}R&:\left(0:b(-a^2+b^2-bc+ac):c(c-a)(-a+b-c)\right), \\ S&: \left(0:a^2b-b^3-abc+b^2c:-a^2c+abc-bc^2+c^3\right).\end{align*} Expanding the coordinates of $R$ immediately shows that $R = -S$, so after normalizing, we see that $R$ and $S$ coincide, as desired.

Welcome to the dark side Here's a synthetic solution: First, we claim that $B_1$ and $C_1$ lie on line $B_0C_0$. To see this, note that $\angle IB_1C=\angle IB_0C = \frac{\pi}{2}$ which implies that $IB_1B_0C$ is a cyclic quadrilateral. Yet it is easy to show that $\angle B_1IC + \angle C_0B_0C = \pi$, implying the conclusion. Next, remark that $Q$ lies on $(ABL)$, since $BI$ is an internal angle bisector and we know that $QA = QL$. Similarly, $P \in (ACL)$. We claim that $\triangle A_0B_0C_0$ and $LQP$ are homothetic. Since $B_0C_0$ and $PQ$ are both perpendicular to $AL$, $B_0C_0 \parallel PQ$. Also, $\angle C_0A_0B = \frac{\pi-B}{2}$, and $\angle PLB = \angle PAC = \angle PAL + \angle LAC = %Error. "half" is a bad command. C + %Error. "half" is a bad command. A = \frac{\pi - B}{2}$, which shows that $C_0A_0 \parallel PL$. Similarly, $B_0A_0 \parallel LQ$. Hence $\triangle A_0B_0C_0$ and $LQP$ are homethetic, say with homethety $h$. Let $K$ be the center of homethety; remark that $K \in LA_0 = BC$ hence solving part (a). Let $O_1'$ be the intersection of $PQ$ and $C_1K$. Then $O_1'$ is the image of $C_1'$ under $h$. Since $B_0C_1 = A_0C_1$, it follows that $QO_1' = LO_1'$. But $PQ$ happens to be the perpendicular bisector of $AL$, so in fact $O_1'A = O_1'Q = O_1'L$. Hence $O_1'$ is the circumcenter of $(ABL)$; that is, $O_1 = O_1'$. Similarly $O_2 = O_2'$ and the proof is complete. [asy][asy] /* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(14cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair A = (-4.5, 4.0); pair B = (-5.5, -0.5); pair C = (-1.5, -0.5); path a = B--C; pair I = incenter(A,B,C); pair B_0 = foot(I,C,A); pair C_0 = foot(I,A,B); pair L = IntersectionPoint(Line(A,bisectorpoint(B,A,C),lisf),a); pair Q = IntersectionPoint(Line(midpoint(A--L),bisectorpoint(A,L),lisf),Line(B,I,lisf)); pair P = IntersectionPoint(Line(midpoint(A--L),bisectorpoint(A,L),lisf),Line(C,I,lisf)); pair A_0 = foot(I,B,C); pair O_1 = circumcenter(A,L,B); pair O_2 = circumcenter(A,L,C); pair C_1 = foot(B,C,I); pair B_1 = foot(C,B,I); pair K = IntersectionPoint(Line(P,C_0,lisf),Line(B,C,lisf)); pair O = circumcenter(A,B,C); /* Draw objects */ draw(A--B, rgb(0.6,0.2,0.0)); draw(a, rgb(0.6,0.2,0.0)); draw(C--A, rgb(0.6,0.2,0.0)); draw(incircle(A,B,C), rgb(1.0,0.0,0.0) + linewidth(1.0) + dotted); draw(A--L); draw(C--P); draw(B--Q); draw(P--Q); draw(circumcircle(A,L,B), rgb(0.0,0.0,1.0)); draw(circumcircle(A,C,L), rgb(0.0,0.0,1.0)); draw(K--P, rgb(0.0,1.0,0.0) + linewidth(1.0) + linetype("4 4")); draw(K--O_1, rgb(0.0,1.0,0.0) + linewidth(1.0) + linetype("4 4")); draw(K--Q, rgb(0.0,1.0,0.0) + linewidth(1.0) + linetype("4 4")); draw(K--O_2, rgb(0.0,1.0,0.0) + linewidth(1.0) + linetype("4 4")); draw(circumcircle(B,C,P), rgb(0.0,0.6,0.0)); draw(C_0--B_0, rgb(0.0,0.4,0.0) + linetype("4 4")); draw(circumcircle(L,P,Q), rgb(1.0,0.0,0.0) + linewidth(1.0) + linetype("4 4")); draw(circumcircle(A,B,C)); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(I); dot(B_0); dot(C_0); dot(L); dot(Q); dot(P); dot(A_0); dot(O_1); dot(O_2); dot(C_1); dot(B_1); dot(K); /* Label points */ label("$A$", A, lsf * dir(75)); label("$B$", B, lsf * dir(225)); label("$C$", C, lsf * dir(-45)); label("$I$", I, lsf * dir(45)); label("$B_0$", B_0, lsf * dir(45)); label("$C_0$", C_0, lsf * dir(45)); label("$L$", L, lsf * dir(70) * 2); label("$Q$", Q, lsf * dir(45)); label("$P$", P, lsf * dir(190)); label("$A_0$", A_0, lsf * dir(90)); label("$O_1$", O_1, lsf * dir(45)); label("$O_2$", O_2, lsf * dir(45)); label("$C_1$", C_1, lsf * dir(45)); label("$B_1$", B_1, lsf * dir(90)); label("$K$", K, lsf * dir(-90)); [/asy][/asy] BTW, Conjecture: the concurrence point is $IO \cap BC$, where $I$ and $O$ are the incenter. Anyone care to try and prove this? I haven't had much success, although I suppose can be done with $I=(a:b:c)$, $O=(a^2S_A, b^2S_B, c^2S_C)$ and the work above.

v_Enhance wrote: BTW, Conjecture: the concurrence point is $IO \cap BC$, where $I$ and $O$ are the incenter. Anyone care to try and prove this? I haven't had much success, although I suppose can be done with $I=(a:b:c)$, $O=(a^2S_A, b^2S_B, c^2S_C)$ and the work above. Well, the equation of $IO$ is $bc(b - c)(s - a)x + ac(c - a)(s - b)y + ab(a - b)(s - c)z = 0$, so intersecting this with $BC: x = 0$ gives us $T = (0:ab(a-b)(s-c):-ac(c-a)(s-b))$, where $T = IO \cap BC$. Now substituting $s = \frac{a+b+c}{2}$ gives us \begin{align*}ab(a-b)(s-c) &= ab(a-b)\left(\frac{a+b-c}{2}\right) = \frac{1}{2}(a^3b-a^2bc-ab^3+ab^2c), \\ -ac(c-a)(s-b) &= -ac(c-a)\left(\frac{a-b+c}{2}\right) = \frac{1}{2}(a^3c-a^2bc+abc^2-ac^3),\end{align*} so the coordinates of $T$ reduce to the coordinates of $R$ after scaling with a factor of $\frac{2}{a}$, proving your conjecture true.

My synthetic solution is mostly the same as v_Enhance's solution, so I won't bore you with all the details. I'll just mention some key facts, which show that this is a very interesting configuration: Let $A_0$ be the projection of $I$ onto $BC$, $M$ the midpoint of $BC$, and $B_2$ and $C_2$ the intersections of $BI$ and $CI$ with circle $ABC$. Let $D$ and $E$ be the intersections of $PQ$ and $B_0C_0$ with $BC$. The following polygons are cyclic: $BPQC$, $BC_1B_1C$, $PAQI$, $BC_0C_1IA_0$, $CB_0B_1IA_0$, $B_1C_1A_0M$, $AO_1OO_2$. Furthermore, circles $ABC$ and $APQ$ are tangent to $DA$, so circle $LPQ$ is tangent to $BC$ by reflection over $PQ$. By angle chasing, figures $LDPO_1O_2Q$ and $A_0EC_0C_1B_1B_0$ are similar with same orientation, so $DE = BC$, $PC_0$, $O_1C_1$, $O_2B_1$, and $QB_0$ concur at a point $X$. If $O'$ is the center of circle $PQL$, then $O'I$ also passes through $X$. Because $B_2C_2$ and $PQ$ are parallel, and $\angle QO'P = \angle B_2OC_2$, a homothety about $I$ takes circle $PQL$ to circle $ABC$, and hence takes $O'$ to $O$. Because $I$, $O$, and $O'$ are collinear, $OI$ passes through $X$.

Hi! Here is my solution. We see that $B_0C_0 \parallel PQ$ and also that $\triangle A_0B_0C_0 \sim \triangle LPQ$ (This is angle chasing) Also, $LP \parallel A_0B_0$ and $LQ \parallel A_0C_0$ (can be seen easily by chasing angles) and so these triangles are in fact homothetic. Let $T$ be the centre of Homothety. Then $\frac{TA_0}{TL}$ is the ratio of Homothety. Thus, $PC_0,QB_0,LA_0=BC$ are concurrent at $T$. Proves Part a.) Also, we see that $\triangle LO_1O_2 \sim \triangle ABC$ and also, by the right angles on the intouch chord Lemma, we have $\triangle A_0B_1C_1 \sim \triangle ABC$ and so we see that $\triangle LO_1O_2 \sim \triangle A_0B_1C_1$ and so since $O_1O_2 \parallel B_1C_1$ we get that these triangles are infact homothetic. Let $T'$ be the centre of homothety. Thus, the ratio of Homothety is $\frac{T'A_0}{T'L}$ and so $O_1B_1,O_2C_1,LA_0=BC$ are concurrent at point $T'$. Proves part b.) For part c.) we get that the Homothety about $T$ and that about $T'$ both send $A_0$ to $L$ and line $B_0C_0=B_1C_1$ to $PQ=O_1O_2$ thus, $T=T'$. $\blacksquare$

Cute problem! I'll also provide a synthetic proof of Evan's conjecture about the point in (c) being $IO \cap BC$. robinpark wrote: a) The incircle of a triangle $ABC$ touches $AC$ and $AB$ at points $B_0$ and $C_0$ respectively. The bisectors of angles $B$ and $C$ meet the perpendicular bisector to the bisector $AL$ in points $Q$ and $P$ respectively. Prove that the lines $PC_0, QB_0$ and $BC$ concur. b) Let $AL$ be the bisector of a triangle $ABC$. Points $O_1$ and $O_2$ are the circumcenters of triangles $ABL$ and $ACL$ respectively. Points $B_1$ and $C_1$ are the projections of $C$ and $B$ to the bisectors of angles $B$ and $C$ respectively. Prove that the lines $O_1C_1, O_2B_1,$ and $BC$ concur. c) Prove that the two points obtained in pp. a) and b) coincide. Let $A'$ be the intouch point on $BC$. Let $\ell_1$ be the $A$-intouch chord, $\ell_2$ be the perpendicular bisector of $AL$. Let $R, S$ be the projections of $A', L$ on $\ell_1, \ell_2$, respectively. Notice the direct similarities $\triangle A'B_0C_0 \sim \triangle LPQ$ and $\triangle A'B_1C_1 \sim \triangle LO_1O_2$. Moreover, $B_0C_0 \parallel PQ$ and $B_1C_1 \parallel O_1O_2$, hence they are homothetic. Part (a) It is clear that $X=PC_0 \cap QB_0$ is the homothety center for $\triangle A'B_0C_0 \mapsto \triangle LPQ$. Thus, $X=RS \cap LA'$ lies on line $BC$. Part (b) It is clear that $Y=O_1B_1 \cap O_2C_1$ is the homothety center for $\triangle A'B_1C_1 \mapsto LO_1O_2$. Thus, $Y=RS \cap BC$ also lies on $BC$. Part (c) Consequently, $X=Y$ is the intersection point $Z=RS \cap BC$. Part (d) Observe that the orthocenter $H$ of $\triangle A'B_0C_0$ lies on line $IO$, and $I$ is the orthocenter of $\triangle LPQ$. Hence, $X=IO \cap BC$, as desired.

We show the following more general problem. Quote: Let $ABC$ be a triangle with circumcenter $O$ and incenter $I$. The incircle is tangent to sides $BC$, $CA$, $AB$ at $A_0$, $B_0$, $C_0$. Point $L$ lies on $BC$ so that $\angle BAL=\angle CAL$. The perpendicular bisector of $AL$ meets $BI$ and $CI$ at $Q$ and $P$ respectively. Let $C_1$ and $B_1$ denote the projections of $B$ and $C$ onto lines $CI$ and $BI$. Let $O_1$ and $O_2$ denote the circumcenters of triangles $ABL$ and $ACL$. Prove that the six lines $BC$, $PC_0$, $QB_0$, $C_1O_1$, $B_1O_2$, and $OI$ are concurrent. Let $M$ be the midpoint of $AL$, let $\ell$ denote the perpendicular bisector of $AL$, and let $m$ denote the line $B_0C_0$. Let $S=\ell\cap BC$ and $T=m\cap BC$, and let $I_A$ denote the $A$-excenter. Let $H$ be the unique point on $BC$ such that $HT/HS=HA_0/HL$ (directed ratios). This point exists and is unique because it is the image of $\infty$ under the unique projective involution swapping $\{S,A_0\}$ and $\{T,L\}$. We claim that $H$ is the desired concurrency point. Let $\phi$ be the homothety centered at $H$ sending $\ell$ to $m$. By definition of $H$, it also sends $A_0$ to $L$. Note that $B_1$ and $C_1$ are on $B_0C_0$ by the Iran lemma, so it suffices to show (ignoring the part about $OI$ for now) that the homothety sends \begin{align*} Q &\mapsto B_0 \\ P &\mapsto C_0 \\ O_2 &\mapsto B_1 \\ O_1 &\mapsto C_1. \end{align*}Given any point $X\in\ell$ and $Y\in m$, we see that \[\phi(X)=Y\iff XL\parallel YA_0\]since $\phi(L)=A_0$ and $L,A_0$ are not on the lines $\ell$ and $m$. Thus, it suffices to show that \begin{align*} QL &\parallel B_0A_0 \\ PL &\parallel C_0A_0 \\ O_2L &\parallel B_1A_0 \\ O_1L &\parallel C_1A_0. \end{align*}To show these, we need an important claim. Claim: The quadrilaterals $APLC$ and $AQLB$ are cyclic. Proof: Note that $(AL;II_A)=-1$. Inverting this at $I$ with power $IA\cdot IL$, we see that $(LA;\infty I_A')=-1$, so we see that \[IM\cdot II_A=IA\cdot IL.\]Note that $\angle I_AMQ=\angle I_ABI=\pi/2$, so $MBI_AQ$ is cyclic. Thus, \[IA\cdot IL=IM\cdot II_A=IB\cdot IQ,\]so $AQLB$ is cyclic. We similarly derive that $APLC$ is cyclic. $\blacksquare$ We'll now show that $PL\parallel C_0A_0$, or that $PL\perp BI$. We'll be using directed angles mod $\pi$. Note that \[\angle LPC+\angle PIB = \angle LAC+\angle CIB = \pi/2,\]so $PL\perp BI$. Similarly, we get $QL\perp CI$, so we have $PL\parallel C_0A_0$ and $QL\parallel B_0A_0$. Note that a corollary of these perpendicularities is that $I$ is the orthocenter of $LPQ$, which is a fact we'll be needing later. We'll now show $O_1L\parallel C_1A_0$. Note that \[\angle LO_1M = \angle LBA\]as $O_1$ is the center of $(ABL)$. Note that $IC_1C_0BA_0$ is cyclic with diameter $BI$, so \[\angle A_0C_1C_0 = A_0BC_0,\]so $\angle LO_1M=\angle A_0C_1C_0$, showing that $O_1L\parallel C_1A_0$. Similarly, we have $O_2L\parallel B_1A_0$. Now, it suffices to show that $H\in OI$. Let $J=\phi(I)$. The main claim is that $J$ is the orthocenter of $A_0B_0C_0$. Indeed, we saw before that $I$ is the orthocenter of $LPQ$, so applying the homothety immediately implies the claim. Thus, $H$ is on the Euler line of $A_0B_0C_0$, which is $OI$ (this is well known, and perhaps the simplest way to see it is by incircle inversion). This completes the proof.

We also solve the above formulation. (OTIS noise.) First note that lines $\overline{B_1C_0B_0C_1}$ and $\overline{O_1PQO_2}$ are parallel. Moreover, $\angle BB_1I = \angle BA_0I = 90^\circ$, so $B,B_1,I,A_0$ lie on a circle, and so $\angle B_1A_0B = \angle B_1IB = 90-\tfrac{1}{2}\angle A$. In addition, $\angle O_1LB = 90 - \tfrac{1}{2} \angle BO_1L = 90 -\tfrac{1}{2} \angle A$. Thus, $BB_1 \parallel O_1Q$ and $B_1A_0 \parallel O_1L$ and so $\triangle B_1B_0A_0$ and $\triangle O_1QL$ are homothetic. Similarly, $\triangle C_1C_0A_0$ and $\triangle O_2PL$ are homothetic. The parallelisms also imply $\triangle PQL$ and $\triangle C_0B_0A_0$ are homothetic, yielding that $B_1O_1$, $C_0P$, $B_0Q$, $C_1O_2$, and $BC$ are concurrent. The final homothety sends the orthocenter $I$ of $\triangle PQL$ to the orthocenter of $\triangle A_0B_0C_0$, which is well-known to lie on $\overline{OI}$ as desired. [asy][asy] defaultpen(fontsize(11pt)); size(15cm); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.1861458613535216, xmax = 3.379103065824605, ymin = -1.3989944617865513, ymax = 1.5871786378402377; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((0.3304188560837271,1.033576029123633)--(-0.4404094829837761,-0.5056237797597282)--(2.27580920404537,-0.4880997882305078)--cycle, rvwvcq); /* draw figures */ draw((0.3304188560837271,1.033576029123633)--(-0.4404094829837761,-0.5056237797597282), rvwvcq); draw((-0.4404094829837761,-0.5056237797597282)--(2.27580920404537,-0.4880997882305078), rvwvcq); draw((2.27580920404537,-0.4880997882305078)--(0.3304188560837271,1.033576029123633), rvwvcq); draw(circle((0.5396161557709501,0.10400909160496959), 0.6032975698182648), wrwrwr); draw((xmin, 0.22504812858947407*xmin + 0.374115207646452)--(xmax, 0.22504812858947407*xmax + 0.374115207646452), wrwrwr); /* line */ draw((xmin, 0.2250481285894732*xmin + 0.15441929576946273)--(xmax, 0.2250481285894732*xmax + 0.15441929576946273), wrwrwr); /* line */ draw(circle((0.04960333639358699,-0.20080734407737932), 0.57708372236292), wrwrwr); /* dots and labels */ dot("$C$",(2.27580920404537,-0.4880997882305078),dir(0)); dot("$B$",(-0.4404094829837761,-0.5056237797597282),dir(180)); dot("$A$",(0.3304188560837271,1.033576029123633),dir(90)); dot("$I$",(0.5396161557709501,0.10400909160496959)); dot("$A_0$",(0.5435083171553992,-0.49927592298463663),dir(-135)); dot("$B_0$",(0.911311688329085,0.579204197666626),dir(90)); dot("$C_0$",(0.00018272360283666522,0.37415632925131953),dir(90)); dot("$L$",(0.6751931230579493,-0.498426343591717),dir(-45)); dot("$C_1$",(1.5258526882978543,0.7175054996511006),dir(90)); dot("$B_1$",(-0.15205472963377745,0.33989557529919195),dir(135)); dot("$O_1$",(0.11299263849332875,0.1798480776067732),dir(-90)); dot("$O_2$",(1.4691894171609752,0.48505762464499913),dir(-90)); dot("$Q$",(0.9724763592115533,0.3732732805075271),dir(-90)); dot("$P$",(0.2360411103372472,0.20753990592104155),dir(-90)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

$\textbf{Claim: }$ $\overline{PL}\parallel\overline{C_{0}A_{0}}$ and $\overline{QL}\parallel\overline{B_{0}A_{0}}.$ $\emph{Proof: }$ Since $\overline{BI}$ bisects $\angle ABC$ and $Q$ is the intersection point of $\overline{BI}$ and the perpendicular bisector of $\overline{AL},$ we know quadrilateral $QLBA$ is cyclic. This implies that $\angle QLA=\angle QBA=\frac{1}{2}\angle CBA.$ But we have $\angle(\overline{B_{0}A_{0}},\overline{AL})=90^\circ-\angle A_{0}B_{0}C_{0}=\frac{1}{2}\angle CBA$ as well, so $\overline{QL}\parallel\overline{B_{0}A_{0}}.$ Similarly, we can show that $\overline{PL}\parallel\overline{C_{0}A_{0}},$ so we are done. $\blacksquare$ $\textbf{Claim: }$ $\overline{LO_{1}}\parallel\overline{A_{0}C_{1}}$ and $\overline{LO_{2}}\parallel\overline{A_{0}B_{1}}.$ $\emph{Proof: }$ Since $O_2$ is the circumcenter of $\triangle ACL,$ we know $\angle O_{2}LC=90^\circ-\angle CAL=90^\circ-\frac{1}{2}\angle CAB.$ But since $\angle IA_{0}C=\angle IB_{0}C=\angle IB_{1}C=90^\circ,$ pentagon $IA_{0}B_{0}B_{1}C$ is cyclic. Hence, $\angle B_{1}A_{0}C=\angle B_{1}IC=180^\circ-\angle CIB=90^\circ-\frac{1}{2}\angle CAB,$ so $\overline{LO_{2}}\parallel\overline{A_{0}B_{1}}.$ Similarly, $\overline{LO_{1}}\parallel\overline{A_{0}C_{1}},$ as desired. $\blacksquare.$ $\textbf{Claim: }$ $\overline{B_{1}B_{0}}\parallel\overline{O_{2}Q},\overline{B_{0}C_{0}}\parallel\overline{QP},$ and $\overline{C_{0}C_{1}}\parallel\overline{PO_{1}}.$ $\emph{Proof: }$ By Iran Lemma, we know $B_1,B_0,C_0,C_1$ are collinear. Moreover, points $O_1,P,Q,O_2$ all lie on the perpendicular bisector of $\overline{AL}.$ Now the claim follows from the fact that $\overline{B_{0}C_{0}}\perp\overline{AL}.$ $\blacksquare$ Now, it is easy to see that $\triangle B_{1}B_{0}A_{0},\triangle A_{0}B_{0}C_{0},\triangle C_{1}C_{0}A_{0}$ are homothetic to $\triangle O_{2}QL,\triangle LQP,\triangle O_{1}PL$ respectively. This implies that $\overline{BC},\overline{PC_{0}},\overline{QB_{0}},\overline{C_{1}B_{1}},B_{1}O_{2}$ concur. To see that the concurrency point lies on $\overline{OI},$ just note that the orthocenter of $\triangle A_{0}B_{0}C_{0}$ lies on $\overline{OI}$ and $I$ is the orthocenter of $\triangle LQP.$

Firstly, note that $P,Q,C,B$ are cyclic because: \[\angle QPC = 90 - \angle AIP = 90 - (C/2 + A/2) = B/2 = \angle QBC \] Additionally, $B,L,Q,A$ are cyclic because Q is the fact 5 arc midpoint of AL. A similar result holds for $C,L,P,A$. \begin{claim} I is the orthocenter of LPQ \end{claim} \begin{proof} Let $S = PL\cap BI$. Clearly, $\angle SIL = \angle BIL = A/2+B/2 = 90 - C/2$. Then, \[\angle ILS = \angle PAL = \angle PCL = C/2\]Thus, $\angle SIL + \angle ILS =90$, from which it follows that $\angle ISL=90$ and $BI$ is perpendicular to $PL$ \end{proof} Note that $QL\parallel DE$ since they're both perpendicular to CI. Denote with $\mathcal{H}$ by the homothety taking $L$ to $D$. Since $PQ\parallel EF$ and $QL\parallel DE, PL\parallel DF$, then $\mathcal{H}$ takes $\triangle LPQ$ to $\triangle DFE$. Thus, $FP, EQ, DL=BC$ concur at smoe point $X$. $\blacksquare$. Now, note that $I, C, D, E, B_1$ are cyclic since they have a 90 degree angle with arc $IC$. \begin{claim} $C_1 , F, E, B_1$ are collinear \end{claim} \begin{proof} Firstly, $\angle FEI = \frac12 A$ and then \[\angle IEB_1 = 180 - \angle ICB_1 = 180 - (90 - \angle CIB_1)= 90 + \angle CI B_1 = 90+B/2 + C/2\]Thus, $\angle FEB_1 = 180$ and we're done. \end{proof} \begin{claim} $B_1 D \parallel L O_2$ \end{claim} \begin{proof} \[\angle O_2 L C = (180 - \angle L O2 C)/2 = 90 - \angle LAC = 90 - A/2\]Additionally, \[\angle B_1 DC = \angle B_1 I C = B/2 + C/2 = 90 - A/2\]Thus, $\angle O_2LC = B_1 DC$, so we're done. \end{proof} Now, since $O_1PQO_2\parallel C_1FEB_1$, the same homothety $\mathcal{H}$ takes $\triangle LO_2O_1$ to $\triangle DB_1C_1$, so the lines $O_1C_1,PC_0,QB_0,O_2B_1$ concur at $X$ on $BC$. We now attempt to get $O,I,X$ collinear. Note that under $\mathcal{H}$ we have $X,I, T$ collinear where $T$ is the orthocenter of $\triangle DEF$. Now, it suffices to note that $OI$ obviously passes through the nine point center of $DEF$ (by inversion at $I$). Thus, it follows that $O$ lies on the Euler line of $\triangle DEF$, so, $X\in OI$ and we're done. $\blacksquare$.

Very elegant length chase, involving almost no computations. Let $OI$ intersect $BC$ at $X$.Let $M$ be the second intersection of $CI$ with $(ABC)$ Part 1:Proving $X-P-C_0$. Let $H$ be the center of Homothety taking the incircle to the circumcirle. We have that $O-I-H$ and $H-C_0-M$. We would like to prove that $\frac{IP}{PM}\frac{MC_0}{C_0H}\frac{HX}{XI}=1$, because then Menelaus would imply the desired result. First of, $\frac{MC_0}{C_0H}=\frac{R-r}{r}$. We can also check that $P$ lies on $(LCA)$, and angle chasing gives us $MPA\sim BLA$, thus $$\frac{MP}{MI}=\frac{MP}{MA}=\frac{BL}{BA}=\frac{a}{c+b}$$, thus $\frac{IP}{PM}=\frac{c+b-a}{a}$. Let $N$ be the midpoint of $BC$ and $T$ the foot of the perpendicular from $H$ to $BC$. Let the lenght of $HT=h$. Since $OH;IA_0;HT$ are all perpendicular to $BC$ we can see that $\frac{r-OH}{OI}=\frac{h-r}{HI}$, which , since $\frac{HI}{IO}=\frac{r}{R-r}$, is equivalent to $\frac{h}{r}=\frac{R-OH}{R-r}$. This gives us $\frac{HX}{XI}=\frac{h}{r}=\frac{R-OH}{R-r}$. Now since $OH=r.cos(\angle BAC)=R\frac{b^2+c^2-a^2}{2bc}$ and by using the formulas $R=\frac{abc}{4S}$ and $S=p.r=\sqrt{p(p-a)(p-b)(p-c)}$, we actually get that $$\frac{IP}{PM}\frac{MC_0}{C_0H}\frac{HX}{XI}=\frac{c+b-a}{a}\frac{R-r}{r}\frac{R-OH}{R-r}=\frac{b+c-a}{a}\frac{R(1-cosA)}{r}=1$$, implying the desired result. I'll do the calculations for this at the end of the sol for part 2. Analogically, we can see that $X-B_0-Q$. Part 2:Proving $X-C_1-O_1$. Since $O-O_1-M$ we would like to show that $\frac{OO_1}{O_1M}\frac{MC_1}{C_1I}\frac{IX}{XO}=1$, because Menelaus would again imply the desired result. We know that $\frac{IX}{XO}=\frac{r}{OH}=\frac{r}{RcosA}$. Also since $\angle BAC_1=A$ and $\angle BC_1M=90$, we get that $\frac{MC_1}{MB}=\frac{MC_1}{MI}=cosA$, thus $\frac{MC_1}{C_1I}=\frac{cosA}{1-cosA}$. Angle chasing gives us $BO_1O\sim ACL$, thus $$\frac{OO_1}{OB}=\frac{OO_1}{OM}=\frac{LC}{AC}=\frac{a}{b+c}$$so $\frac{OO_1}{O_1M}=\frac{a}{b+c-a}$ Combining all of these together we can see that $$\frac{OO_1}{O_1M}\frac{MC_1}{C_1I}\frac{IX}{XO}=\frac{a}{b+c-a}\frac{cosA}{1-cosA}\frac{r}{RcosA}=\frac{a}{b+c-a}\frac{r}{R(1-cosA)}$$Also $1-cosA=1-\frac{b^2+c^2-a^2}{2bc}=\frac{(a+b-c)(a+c-b)}{2bc}$ so we can see that $$\frac{a}{b+c-a}\frac{r}{R(1-cosA)}=\frac{2abcr}{8(p-a)(p-b)(p-c)R}$$Since $R=\frac{abc}{4S}$ we get $$\frac{2abcr}{8(p-a)(p-b)(p-c)R}=\frac{8Sr}{8(p-a)(p-b)(p-c)}=\frac{S^2}{p(p-a)(p-b)(p-c)}=1$$and we're finally done with this case. Analogically $B_1-O_2-X$ so we got that all of these lines pass trough $X$.

Beautiful! We also solve the generalized problem, mentioned in #8. Obviously $Q \in (ABL)$ and $B_1,C_1 \in B_0C_0$. We have that $\angle (A_0B_0,AL)=\angle QLA = \frac{\beta}{2}$, so $QL \parallel A_0B_0$ and similarly $PL \parallel A_0C_0$, so triangles $LPQ$ and $A_0B_0C_0$ are homothetic (apparently $PQ \parallel B_0C_0$) and thus $PC_0,QB_0,LA_0$ are concurrent, say at $T$. Since the orthocenter of $\triangle LPQ$ is $I$ and the orthocenter of $\triangle A_0B_0C_0$ lies on $OI$ (provable by inversion wrt the incircle), we have that $IO$ also passes through $T$, again by the same homothety. We are left to prove that $LO_1 \parallel C_1A_0$ (then we will be done by the same homothety). Indeed, we have that $\angle LO_1B=90-\frac{\alpha}{2}=\angle C_1A_0B$, since $C_1 \in (BI)$.

Wow what is this The key is to notice that the triangles $LQP$ and $A_0B_0C_0$ are homothetic. To see this, notice that $\overline{B_0C_0}\parallel \overline{PQ}$ because they are both perpendicular to $\overline{AI}$. On the other hand, by Fact 5 the quadrilaterals $AQBL$ and $APCL$ are cyclic, so $\angle LPC = \frac A2,$ which is enough to show that all three angles of the triangles are correspondingly equal. Now, observe that $\overline{BC}, \overline{PC_0}, \overline{QB_0}$ concur as they comprise the lines connecting vertices of the homothetic triangles; $\angle IC_1A_0 = \frac B2$ because $C_1IA_0B$ is cyclic by Iran lemma, so in turn $\overline{C_1I}$ bisects $\angle B_0C_1A_0$ and thus $C_1B_0=C_1A_0$. But $O_1Q=O_1L$ too by our previous concyclicity, so $O_1$ and $C_1$ are corresponding parts of the homothety. As a result, $\overline{C_1O_1}$ and $\overline{B_1O_2}$ pass through the concurrency point too. Setting $O_3$ to be the circumcenter of $LPQ$, a homothety at $I$ takes $LPQ$ to the triangle formed by the minor arc midpoints in $(ABC)$. Therefore $O, O_3, I$ are collinear. But $I$ is the circumcenter of $A_0B_0C_0$, so $\overline{O_3I}$ passes through the concurrency point, and thus so does $\overline{OI}$.

Wow, great problem! Since $P$ lies on the bisector of $\angle ACL$ and lies on the perpendicular bisector of $AL$, therefore $P$ lies on $(ACL)$. Similarly $Q$ lies on $(ABL)$. Thus $\angle CPL + \angle BIP = \angle CAI + \angle CBI + \angle BCI = 90^{\circ}$ means $PL \perp BI$. Similarly $QL \perp CI$. Note that $A_0B_0 \perp CI$, thus $A_0B_0 \parallel LQ$ and similarly $A_0C_0 \parallel PL$. Since $B_0C_0 \parallel PQ$, hence triangles $A_0B_0C_0$ and $LQP$ are homothetic, which means $BC, B_0Q, C_0P$ are concurrent. Note that EGMO lemma 1.45 yields $C_1, C_0, B_0, B_1$ are collinear. (1). Let $R = BC \cap B_0Q \cap C_0P$. We'll prove that lines $B_1O_2, C_1O_1$ pass through $R$. Since $\angle IA_0C + \angle IB_1C = 90^{\circ} + 90^{\circ} = 180^{\circ}$, thus $IA_0CB_1$ is cyclic. Therefore $\angle B_1A_0C = B_1IC = 90^{\circ} - \frac{\angle BAC}{2}$, thus $O_2L \parallel B_1A_0$. Combining this and (1) yields triangles $B_0B_1A_0$ and $QO_2L$ are homothethic, which means $BO_2$ passes through $R$. Similarly $C_1O_1$ passes through $R$. Now it suffices to prove that $R, I, O$ are collinear. Let $I_A$ be the $A$-excenter of $\triangle ABC$ and define $I_B, I_C$ similarly. Then $I$ is the orthocenter of $\triangle I_AI_BI_C$ and $O$ is the nine-point-center of $\triangle I_AI_BI_C$. Hence $IO$ is the euler line of $I_AI_BI_C$. Now note that triangles $I_AI_BI_C, A_0B_0C_0$ are homothetic means $I_AA_0, I_BB_0, I_CC_0$ are concurrent. Let $T = I_AA_0 \cap I_BB_0$ and let $X$ be the circumcenter of $\triangle I_AI_BI_C$. Then $X, I, T$ are collinear and thus $T$ lies on $IO$. Let $H$ be the orthocenter of $\triangle A_0B_0C_0$. Then by homothethy, $H$ lies on $TI$, which means $H$ lies on $IO$. Note that $I$ is the orthocenter of $PQL$ and $H$ ies the orthocenter of $A_0B_0C_0$, hence $I, H, R$ are collinear and since $I ,H, O$ are collinear, so we're done. $\blacksquare$

Solution also contains proof to Evan's "conjecture". First we define some points. \begin{itemize} \ii Let $K$ be the center of $(LPQ)$. \ii Let $M_A$, $M_B$, $M_C$ be the respective $3$ minor arc midpoints of $\triangle ABC$. \ii Let $T= \ol{C_0P} \cap \ol{BC}$ . \ii Define $R_1=\ol{B_0C_0} \cap \ol{AI}$ and $R_2=\ol{PQ} \cap \ol{AI}$. \end{itemize} Also note some stuff. \begin{itemize} \ii By Iran Lemma we have $C_0,C_1,B_1,B_0$ collinear. \ii Obviously $P,O_1,Q,O_2$ collinear. \ii Both these lines are parallel. \end{itemize} Claim $(BC_0C_1IA_0)$ and $(A_0CB_1B_0I)$ are cyclic. Proof Just see that $\dang BA_0I=\dang BC_0I=\dang BC_1I=90\dg$ and analogous. $\blacksquare$ Claim $(ALPC)$ and $AQLB)$ are cyclic. Proof For the first circle, it is equivalent to prove that $IP \cdot IC=IA \cdot IL$. This is just a long boring trigonometry bash. See that RHS is easily tractable and for LHS, the only mysterious quantity is $IP$. See that the homothety at $I$ which sends $C_1$ to $P$ also sends $R_1$ to $R_2$. And so we get $IP=IC \cdot \frac{IR_1}{IR_2}$ and again the only unknown quantity is $IC_1$ which can be rewritten as $CC_1-CI$=$BC \cdot \cos \left(\frac{\angle C}{2} \right)-CI$, which is much more nicer. Similar for the other circle. Details left to the reader, I really could not care less. $\blacksquare$ Claim There is a homothety which maps $\triangle A_0C_0B_0$ to $\triangle LPQ$. Proof See that \[\dang LPQ=\half \dang LPA=90\dg+\half \dang LCA=90\dg+\half \dang BCA=\dang A_0C_0B_0\]and same for $\dang LQP$ gives us both the triangles are similar. But see that $\ol{B_0C_0} \parallel \ol{PQ}$ and they are in the same orientation and we are done. $\blacksquare$ Claim There is a homothety which maps $\triangle A_0C_1B_1$ to $\triangle LO_1O_2$. Proof Again just an angle chase \[\dang LO_1O_2=\half \dang LO_1A=\dang LBA=\dang CBA=\dang A_0BC_0=\dang A_0C_1C_0=\dang A_0C_1B_1\]and the same continuation as the proof of the previous claim gives us our desired.$\blacksquare$ And because of that fact that $\ol{C_0C_1B_0B_1}$ and $\ol{PO_1QO_2}$ are lines we get that the dilation points of both these homotheties coincide, namely it is $T$. So all we are left to prove is $T$, $O$, $I$ are collinear. See that $T$ send center of $(LPQ)$ to center of $(A_0C_0B_0)$ and so $T$, $K$, $I$ are collinear. Claim $K$, $I$, $O$ are collinear. Proof It is well known that $\triangle M_AM_BM_C$ and $\triangle A_0B_0C_0$ are homothetic and hence $\triangle LQP$ and $\triangle M_AM_BM_C$ are homothetic with dilation point $I$ which is obviously collinear with the the two circumcenters of those two triangles.$\blacksquare$ Now this will solve the problem. Bro, AOPS sadly does not have \ol and \dang functions and also not \itemize???? Anyways...., just take this.

The formulation I solved asked to prove that all the $5$ lines mentioned in the problem coincide with line $OI$, which might make it easier but not really (you probably end up constructing the intouch orthocenter either way, my formulation just forces you to prove that the Euler line of the intouch triangle passes through $O$). Claim 1: Euler line of the intouch triangle passes through $O$. Proof: Consider the external homothety taking the incircle to the circumcircle, this also takes the intouch triangle to the triangle formed by the arc midpoints, whose orthocenter is $I$. Letting $H$ be the orthocenter of the intouch triangle, we see that a homothety centered on $IO$ takes $IO$ to $HI$, so $H$ lies on $IO$, and we are done. We show that there is a unique homothety taking $LPQO_1O_2I$ to $A_0C_0B_0C_1B_1H$, which will finish. Note that $B_1,C_1$ are the Iran points who lie on the circle with diameter $BC$, and are collinear with $B_0, C_0$. Since both $B_1C_1$ and $PQ$ are perpendicular to the $A$ angle bisector, Reim's theorem forces that $PQBC$ is cyclic. Claim 2: $ALBQ, ALCP$ are both cyclic. It suffices to prove $O_1B = O_1Q$. Since $PQBC$ is cyclic, we know that $\angle O_1QB = \angle PQB = \angle PCB = \angle \frac C2$. By circumcenter properties, we know that $\angle O_1BQ = \angle O_1BC - \angle QBC = 90 - \angle \frac A2 - \angle \frac B2 = \angle \frac C2$. Since $O_1BQ$ is isosceles, we are done. The symmetrical arguments hold for $Q,C,O_2$. By angle chasing using the cyclicities above, we see that $\angle KAQ + \angle KAP = \angle \frac B2 + \angle \frac C2= \angle 90 - \frac A2$, thus $APQI$ is cyclic. Now since $L$ is the reflection of $A$ over $PQ$, and $AI$ is perpendicular to $PQ$, so we see that $LPQ$ is similar to $A_0C_0B_0$ by angles (all should be easily accessible using $APQI$ cyclicity and reflection), with $I$ as the orthocenter. Since we can explicitly confirm $PQ$ is parallel to $C_0B_0$ (consider angles to $BC$) and cyclic variants over the two triangles, we see there is a homothety taking $LPQ$ to $A_0C_0B_0$, said homothety also takes $I$ to $H$. By angle chasing (use circumcenter properties and then basically done), we see that $LO_1O_2$ is similar to $ABC$. Now observe $(BIC_0A_0C_1)$ is cyclic with diameter $BI$, so we can get angle $\angle B_1C_1A_0 = \angle C_0C_1A_0 = \angle C_0BA_0 = \angle B $, likewise we get $\angle C_1B_1A_0 = \angle C$. We can then once again manually check that $O_1O_2$ is parallel $C_1B_1$ and cyclic variants (consider angles to $BC$ again), so there exists a homothety taking $LO_1O_2$ to $A_0C_1B_1$ We now show the homothety taking $LO_1O_2$ to $A_0C_1B_1$ is the same as the other homothety. Consider the foot from $L$ to $O_1PQO_2$, and the foot from $A_0$ to $B_1B_0C_0C_1$, clearly both homotheties take the former point to the latter, and $L$ to $A_0$, and since homotheties are defined by two input/output pairs, we are done. Thus there is a unique homothety taking $LPQO_1O_2I$ to $A_0C_0B_0C_1B_1H$, so we are done.

Notice that since $P$ is the intersection of the angle bisector of $\angle ACL$ and the perpendicular bisector of $AL$, it is the arc midpoint of minor arc $AL$ in $(ACL)$. Similarly $Q$ lies on $(ABL)$ with analogous characterisation. Now by trivial angle chasing which I omit, $\Delta PQL$ and the intouch triangle of $\Delta ABC$ are homothetic so the corresponding vertices concur, which is $PC_0, QB_0$ and $BC$, at a point which we call $X$. Now by Iran Lemma, $\overline{B_0-B_1-C_1-C_0}$ are collinear and $C_1$ is in fact the intersection of the perpendicular bisector of one chord with another chord in the incircle and it is easy to see that $O_1$ has an analogous characterisation in $\Delta PQL$ so the prior homothety maps $C_1$ to $O_1$ and by similar logic, $B_1$ to $O_2$, so $C_1O_1$ and $B_1O_2$ also concur at $X$. We also prove a side fact which is given in the $EGMO$ book along with the problem, where we have to prove that $X$ also lies on $OI$. Seeing the line $OI$, we are motivated to construct the midpoints of arcs $AB,BC,CA$ not containing the opposite vertices. We remark that the triangle thus formed is also homothetic to $\Delta PQL$ and the intouch triangle, so the homothety sending any one triangle to any other is actually a composition of the other two homotheties, and so we get by the composition of homotheties lemma that $X$ lies on the line $\overline{O-I-E_X}$ where $E_X$ is the exsimilicenter of the incircle and circumcircle.

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