We use barycentric coordinates. Let $A=(1,0,0)$, $B=(0,1,0)$, and $C=(0,0,1)$. Denote $a=BC$, $b=CA$, and $c=AB$. We claim that the common point is \[ K = \left( a^2-b^2+c^2 : b^2-a^2+c^2 : -c^2 \right). \]Let $C_1 = (u:v:0)$. Let $A_0$ be the intersection of $\overline{C_1B_1}$ and $\overline{BC}$, and observe that $AC_1A_1C$ is cyclic. Define $B_0$ analogously. [asy][asy] size(8cm); /* Initialize Objects */ pair A = (-3.0, -1.0); pair B = (0.0, -1.0); pair C = (-2.0105, 0.7088); pair C_1 = (-1.9336, -1.0); pair B_1 = 2*(foot(circumcenter(C,B,C_1),C,A))-C; pair A_1 = 2*(foot(circumcenter(C,A,C_1),C,B))-C; pair C_2 = extension(A, A_1, B, B_1); /* Draw objects */ draw(A--B--C--cycle, rgb(0.0,0.6,1.0)); draw(circumcircle(C,C_1,B), rgb(0.0,0.8,0.4)); draw(circumcircle(C,C_1,A), rgb(0.0,0.8,0.4)); draw(C_1--C_2, rgb(0.0,0.6,0.6)); draw(A--A_1, rgb(0.0,0.6,0.6)); draw(B--B_1, rgb(0.0,0.6,0.6)); draw(A_1--C_1--B_1); /* Label points */ dot("$A$", A, dir(222)); dot("$B$", B, dir(-42)); dot("$C$", C, dir(100)); dot("$C_1$", C_1, dir(-95)*2); // dot("$B_0$", B_0, dir(-10)); // dot("$A_0$", A_0, dir(120)); dot("$A_1$", A_1, dir(30)); dot("$B_1$", B_1, dir(170)); dot("$C_2$", C_2, dir(105)); [/asy][/asy] By power of a point, we observe that $BA_1 = \frac{uc^2}{a}$. Therefore, we obtain that \[ A_1 = \left( 0:a-\frac{uc^2}{a}:uc \right) = \left( 0:a^{2}-uc^2 : uc^2 \right). \]Similarly, \[ B_1 = \left( b^2-vc^2 : 0 : vc^2 \right). \]Therefore, \[ C_2 = \left( u(b^2-vc^2) : v(a^2-uc^2) : uvc^2 \right). \]Now we show that $C_1$, $C_2$, and $K$ are collinear. Expand \begin{align*} \left\lvert \begin{array}{ccc} u (b^2-vc^2) & v (a^2-uc^2) & uv c^2 \\ u & v & 0 \\ a^2-b^2+c^2 & b^2-a^2+c^2 & -c^2 \\ \end{array} \right\rvert &= uvc^2 \left\lvert \begin{array}{ccc} b^2-vc^2 & a^2-uc^2 & uv \\ 1 & 1 & 0 \\ \frac{a^2-b^2+c^2}{u} & \frac{b^2-a^2+c^2}{v} & -1 \\ \end{array} \right\rvert \\ &= uvc^2 \Big[ (a^2-uc^2)-(b^2-vc^2) \\ &\qquad+ u(b^2-a^2+c^2) - v(a^2-b^2+c^2) \Big] \\ &= uvc^2 (b^2-a^2)(u+v-1) = 0 \end{align*}which implies that $C_1$, $C_2$, and $K$ are collinear, as desired.

Wow this problem turned out really nicely with bary bashing [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 = -5.500000000000006, xmax = 18.00000000000000, ymin = -8.000000000000000, ymax = 3.000000000000000; /* image dimensions */ /* draw figures */ draw((7.440000000000008,1.800000000000002)--(4.680000000000005,-6.080000000000007)); draw((6.473206427087516,-0.9602657081704480)--(8.947583452694158,0.5114671344494492)); draw((6.473206427087516,-0.9602657081704480)--(8.416659765892497,-6.116996631345478)); draw((7.440000000000008,1.800000000000002)--(16.80000000000002,-6.200000000000007)); draw((16.80000000000002,-6.200000000000007)--(4.680000000000005,-6.080000000000007)); draw((-4.866111544640798,1.388638053706691)--(7.814796184733804,-1.238171772405358)); draw((7.440000000000008,1.800000000000002)--(8.416659765892497,-6.116996631345478)); draw((8.947583452694158,0.5114671344494492)--(4.680000000000005,-6.080000000000007)); /* dots and labels */ dot((7.440000000000008,1.800000000000002),dotstyle); label("$A$", (7.552618979049386,1.954501745273012), NE * labelscalefactor); dot((4.680000000000005,-6.080000000000007),dotstyle); label("$B$", (4.163363714359152,-6.870553287845752), NE * labelscalefactor); dot((16.80000000000002,-6.200000000000007),dotstyle); label("$C$", (16.89874713319518,-6.047224538400270), NE * labelscalefactor); dot((6.473206427087516,-0.9602657081704480),dotstyle); label("$C_1$", (5.486372227942535,-0.5926715733239493), NE * labelscalefactor); dot((8.947583452694158,0.5114671344494492),dotstyle); label("$B_1$", (9.041837201413278,0.6680505742644456), NE * labelscalefactor); dot((8.416659765892497,-6.116996631345478),dotstyle); label("$A_1$", (8.220199561488371,-7.024927428366780), NE * labelscalefactor); dot((7.814796184733804,-1.238171772405358),dotstyle); label("$C_2$", (7.860733094021226,-1.776206650651830), NE * labelscalefactor); dot((-4.866111544640798,1.388638053706691),dotstyle); label("$P$", (-4.771945619824194,1.542837370550270), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We use barycentric coordinates. Let $BC = a, CA = b, AB = c$, and let $A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1)$. Let $AC_1 = br$ for some $r$. We claim that all lines $C_1C_2$ pass through $P = (a^2-b^2+c^2: -a^2+b^2+c^2: -c^2)$, which is independent of $r$. It follows that $C_1 = (c-br : br : 0), A_1 = (0: a^2 - (c-br)c : (c-br)c), B_1 = (b-cr, 0, cr)$. The equations of lines $AA_1$ and $BB_1$ are: \begin{align*} AA_1&: (c-br)cy - (a^2-(c-br)c)z = 0,\\ BB_1&: crx - (b-cr)z = 0. \end{align*} Thus their intersection $C_2$ has coordinates $\left((b-cr)(c-br):(a^2-(c-br)c)r : cr(c-br)\right)$. The equation of line $C_1C_2$ is \[C_1C_2: (bcr)x+(br-c)cy + (-b^2+(a^2-c^2)+2bcr)z = 0.\] Substituting $(a^2-b^2+c^2: -a^2+b^2+c^2: -c^2)$ yields \begin{align*} (bcr)(a^2-b^2+c^2)+(br-c)c(-a^2+b^2+c^2)+(-b^2+(a^2-c^2)+2bcr)(-c^2) &= 0,\end{align*} which shows that $C_1C_2$ always passes through $P$, as desired.

robinpark wrote: Wow this problem turned out really nicely with bary bashing Yeah, intersect a lne with a side, and then two cevians. I won't be surprised if the official solution turns out to be Menelaus-based, as Menelaus tends to just be bary in disguise.

See attachment for a syn. solution.

v_Enhance wrote: We claim that the common point is \[ K = \left( a^2-b^2+c^2 : b^2-a^2+c^2 : -c^2 \right). \] Could you explain how you computed this point? I tried picking some values of $u,v$ that seemed to be nice, but I don't know a nice way to intersect arbitrary lines in bary and so the computation soon became unbearable. v_Enhance wrote: [asy][asy] /* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(7cm); real lsf=0.7000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair A = (-3.0, -1.0); pair B = (0.0, -1.0); pair C = (-2.01050760446784, 0.7088170680463545); pair C_1 = (-1.933608876813893, -1.0); pair B_0 = (2)*(foot(circumcenter(C,B,C_1),C,A))-C; pair A_0 = (2)*(foot(circumcenter(C,A,C_1),C,B))-C; pair A_1 = IntersectionPoint(Line(B_0,C_1,lisf),Line(B,C,lisf)); pair B_1 = IntersectionPoint(Line(C_1,A_0,lisf),Line(A,C,lisf)); pair C_2 = IntersectionPoint(Line(A,A_1,lisf),Line(B,B_1,lisf)); /* Draw objects */ draw(A--B, rgb(0.0,0.6,1.0)); draw(B--C, rgb(0.0,0.6,1.0)); draw(C--A, rgb(0.0,0.6,1.0)); draw(circumcircle(C,C_1,B), rgb(0.0,0.8,0.4)); draw(circumcircle(C,C_1,A), rgb(0.0,0.8,0.4)); draw(C_1--C_2, rgb(0.0,0.6,0.6)); draw(A--C_2, rgb(0.0,0.6,0.6)); draw(B--C_2, rgb(0.0,0.6,0.6)); draw(C--A_1, rgb(0.0,0.2,0.8)); draw(C--B_1, rgb(0.0,0.2,0.8)); draw(C_1--A_1); draw(C_1--B_1); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(C_1); dot(B_0); dot(A_0); dot(A_1); dot(B_1); dot(C_2); /* Label points */ label("$A$", A, lsf * dir(222)); label("$B$", B, lsf * dir(-42)); label("$C$", C, lsf * dir(100)*2.7); label("$C_1$", C_1, lsf * dir(-95)*2); label("$B_0$", B_0, lsf * dir(-10)); label("$A_0$", A_0, lsf * dir(10)); label("$A_1$", A_1, lsf * dir(150)); label("$B_1$", B_1, lsf * dir(30)); label("$C_2$", C_2, lsf * dir(45)); [/asy][/asy] I'm also a bit confused on the locations of $A_1,B_1$. Shouldn't $B_1$ be where $B_0$ is located, or am I misreading?

v_Enhance wrote: We will use barycentric coordinates. Let $A=(1,0,0)$, $B=(0,1,0)$ and $C=(0,0,1)$. Denote $a=BC$, $b=CA$, and $c=AB$. We claim that the common point is \[ K = \left( a^2-b^2+c^2 : b^2-a^2+c^2 : -c^2 \right). \] Let $C_1 = (u:v:0)$. Let $A_0 = C_1B_1 \cap BC$ and observe that $AC_1A_0C$ is cyclic. Define $B_0$ analogously. By Power of a Point, we observe that $BA_0 = \frac{uc}{a}$. Therefore, we obtain that \[ A_0 = \left( 0:a-\frac{uc}{a}:uc \right) = \left( 0:a^{2}-uc : uc \right). \]Combining with $C_1=(u:v:0)$ we therefore observe that \[ B_1 = AC \cap C_1A_0 = (a^2-uc: 0 : -vc) \]Similarly, \[ A_1 = \left( 0 : b^2-vc : -uc \right). \]Therefore, \[ C_2 = \left( u(a^2-uc) : v(b^2-vc) : -uvc \right). \] Now we compute \begin{align*} \frac{-1}{c} \cdot \det \left<C_1,C_2,K\right> &= \frac{-1}{c} \cdot \det \left[ \begin{array}{ccc} u(a^2-uc) & v(b^2-vc) & -uvc \\ u & v & 0 \\ a^2-b^2+c^2 & b^2-a^2+c^2 & -c^2 \\ \end{array} \right] \\ &= \det \left[ \begin{array}{ccc} a^2-uc & b^2-vc & 1 \\ 1 & 1 & 0 \\ v(a^2-b^2+c^2) & u(b^2-a^2+c^2) & c \\ \end{array} \right] \\ &= \left( u(b^2-a^2+c^2) -v(a^2-b^2+c^2) \right) + c\left( (a^2-uc)-(b^2-vc) \right) \\ &= (u+v)(b^2-a^2) + (u-v)c^2 + c(a^2-b^2) - (u-v)c^2 \\ &= 0 \end{align*}which implies that $C_1$, $C_2$, and $K$ are collinear, as desired. [asy][asy] /* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(11cm); real lsf=0.7000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair A = (-3.0, -1.0); pair B = (0.0, -1.0); pair C = (-2.01050760446784, 0.7088170680463545); pair C_1 = (-1.933608876813893, -1.0); pair B_0 = (2)*(foot(circumcenter(C,B,C_1),C,A))-C; pair A_0 = (2)*(foot(circumcenter(C,A,C_1),C,B))-C; pair A_1 = IntersectionPoint(Line(B_0,C_1,lisf),Line(B,C,lisf)); pair B_1 = IntersectionPoint(Line(C_1,A_0,lisf),Line(A,C,lisf)); pair C_2 = IntersectionPoint(Line(A,A_1,lisf),Line(B,B_1,lisf)); /* Draw objects */ draw(A--B, rgb(0.0,0.6,1.0)); draw(B--C, rgb(0.0,0.6,1.0)); draw(C--A, rgb(0.0,0.6,1.0)); draw(circumcircle(C,C_1,B), rgb(0.0,0.8,0.4)); draw(circumcircle(C,C_1,A), rgb(0.0,0.8,0.4)); draw(C_1--C_2, rgb(0.0,0.6,0.6)); draw(A--C_2, rgb(0.0,0.6,0.6)); draw(B--C_2, rgb(0.0,0.6,0.6)); draw(C--A_1, rgb(0.0,0.2,0.8)); draw(C--B_1, rgb(0.0,0.2,0.8)); draw(C_1--A_1); draw(C_1--B_1); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(C_1); dot(B_0); dot(A_0); dot(A_1); dot(B_1); dot(C_2); /* Label points */ label("$A$", A, lsf * dir(222)); label("$B$", B, lsf * dir(-42)); label("$C$", C, lsf * dir(100)*2.7); label("$C_1$", C_1, lsf * dir(-95)*2); label("$B_0$", B_0, lsf * dir(-10)); label("$A_0$", A_0, lsf * dir(10)); label("$A_1$", A_1, lsf * dir(150)); label("$B_1$", B_1, lsf * dir(30)); label("$C_2$", C_2, lsf * dir(45)); [/asy][/asy] angle BC1A1 is not angle C in your picture

similar to above solutions, but shows how one can find the fixed point We use barycentrics wrt $ABC.$ Let $C_1=(k:1-k:0).$ From $ABC\sim AB_1C_1,$ $$AB_1=\frac{AB\cdot AC_1}{AC}=\frac{c^2(1-k)}{b}.$$Then we have $$B_1=\left(1-\frac{c^2(1-k)}{b^2},0,\frac{c^2(1-k)}{b^2}\right)=\left(\frac{b^2}{1-k}-c^2:0:c^2\right).$$Similarly, $A_1=\left(0:\frac{a^2}{k}-c^2:c^2\right),$ so $$C_2=\left(\frac{b^2}{1-k}-c^2:\frac{a^2}{k}-c^2:c^2\right).$$We wish to find a fixed point $P=(x:y:z)$ on $C_1C_2,$ such a point satisfies \begin{align*} \begin{vmatrix} x & y & z\\ k & 1-k & 0\\ \frac{b^2}{1-k}-c^2 & \frac{a^2}{k}-c^2 & c^2 \end{vmatrix} &=c^2(1-k)x-c^2ky+z(a^2-c^2k-b^2+c^2(1-k))\\ &=c^2x+z(a^2-b^2+c^2)-c^2k(x+y+2z)\\ &=0.\\ \end{align*}Then $x+y+2z=0$ gives an equation for $x,z$ independent of $k,$ so there must be a fixed point $P$ on $C_1C_2,$ as desired.

The fixed point is the reflection of $C$ through $AB$.

By DDIT, $(C_1A,C_1B),(C_1A_1,C_1B_1),(C_1C,C_1C_2)$ are pairs of involution. Obviously, this involution is the reflection over $AB$, so $C_1C_2$ passes through point symmetric to $C$ wrt $AB$.

This also can be done by isogonal theorem: since $C_1B_1$ and $C_1A_1$ are isogonals in $\angle AC_1B$, the lines $CC_1$ and $CC_2$ are isogonals as well, thus they are symmetric to each other wrt. $AB$.

Define $A_0$ and $B_0$ same as #2. Similarly $BC_1B_0C$ and $AC_1A_0C$ are cyclic. Let $X=AA_0 \cap BB_0$. By Pappus on $AB_0B_1$ and $BA_0A_1$ it follows that $C_1$, $X$, $C_2$ are collinear. Now we have to prove that $C_1X$ passes trough a fixed point. Perform an inversion with radius $ \sqrt{CA*CB}$ followed by a reflection over bisector of $\angle ACB$. This inversion swaps $A$ and $B$, $C_1$ is a point on $(ABC)$, $A_0 = AC_1 \cap BC$, $B_0= BC_1 \cap AC$ and finally $X$ is the Miquel point of $CBC_1A$. We have to prove that $(XC_1C)$ passes trough a fixed point but it's well known that $O$ lies on that circle where $O$ is the circumcenter of $(ABC)$.

Here's a linearity solution. [asy][asy] import geometry; defaultpen(fontsize(9pt)); size(6cm); real r = 0.5; pair C = dir(110), A = dir(180+30), B = dir(-30), C1 = r*A+(1-r)*B; guide c1 = circumcircle(A, C, C1), c2 = circumcircle(B, C, C1); pair A1 = intersectionpoints(line(B, C), c1)[1], B1 = intersectionpoints(line(A, C), c2)[1], C2 = extension(A, A1, B, B1); draw(C--A--B--cycle, blue+linewidth(0.8)); draw(A--A1^^B--B1, deepcyan); draw(circumcircle(B, A1, C1)^^circumcircle(A, B1, C1), purple); draw(c1^^c2, royalblue+dotted); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$A_1$", A1, dir(50)); dot("$B_1$", B1, dir(B1)); dot("$C_1$", C1, dir(C1)); dot("$C_2$", C2, dir(C2)); [/asy][/asy] Since $\measuredangle C_1B_1C_2 = \measuredangle C_1CB = \measuredangle C_1AA_1$, $AB_1C_2C_1$ is cyclic; similarly, so is $BA_1C_2C_1$. Denote by $\omega_A, \omega_B$ their circumcircles, respectively. Define $f:\mathbb{R}^2 \to \mathbb{R}$ as \[ f(X) = \mathrm{Pow}(X, \omega_A) - \mathrm{Pow}(X, \omega_B), \]which is linear. We claim that $f(C^\prime) = 0$, where $C^\prime$ is the reflection of $C$ in $\overline{AB}$. Let $D$ be the foot of $C$ onto $AB$. Then \begin{align*} f(C) &= CB_1 \cdot AC - CA_1 \cdot BC \\ &= AC (AC - AB_1) - BC (BC - BA_1) \\ &= AC^2 - BC^2 + AB(BC_1-AC_1) \\ &= AD^2 - DB^2 + (AD+DB)(BC_1-AC_1) \\ &= DB(BC_1-DB) - AD(AC_1-AD) + AD \cdot BC_1 - DB \cdot AC_1 \\ &= - DB \cdot DC_1 - AD \cdot DB_1 + f(D) \\ &= 2f(D). \end{align*}Since $f(D) = \frac12 f(C) + \frac12 f(C^\prime)$, we are done.

The fixed point (call this point $D$) is the reflection of $C$ across $AB$.