Solution: Let's begin with a lemma. Lemma: Let $A,B,C,D $ be points on a line $\ell $ such that $(A,C;B,D)=(-1) $. Let $M $ be the midpoint of $BD $. Then, $(B,D)$ and $(C,M) $ are reciprocal pairs of an involution on $\ell$ with focus $A $. Proof: Length chasing is enough! Here's a sketch: (All lengths are directed) $AB\cdot CD=AD\cdot BC $. So, $AB\cdot AD=AB\cdot AC+AD\cdot BC$ $\implies 2\cdot AB\cdot AD=AC\cdot (AB+AD) $ $\implies AB\cdot AD=AC\cdot AM $. Main problem: Let the concurrency point of $AA_1$, $BB_1$, and $CC_1$ be $P $ and let $M $ be the midpoint of $A_2C_2$. Also, let $AC\cap A_2C_2=K $, $AP\cap B_1C_1=Q $, and $BQ\cap AC=R $. We have, $(-1)=(A,B_1;R,C)\stackrel{B}\doublebarwedge (A,P;Q,A_1)\stackrel {B_1}\doublebarwedge (K,BB_1\cap A_2C_2;C_2,A_2)$. Applying the above lemma, we get $KL\cdot KM=KC_2\cdot KA_2=KA\cdot KC$ $\implies A$, $C $, $M $, and $BB_1\cap A_2C_2$ are concyclic.

Projective freebie TheDarkPrince wrote: Let $C_1, A_1, B_1$ be points on sides $AB, BC, CA$ of triangle $ABC$, such that $AA_1, BB_1, CC_1$ concur. The rays $B_1A_1$ and $B_1C_1$ meet the circumcircle of the triangle at points $A_2$ and $C_2$ respectively. Prove that $A, C$, the common point of $A_2C_2$ and $BB_1$ and the midpoint of $A_2C_2$ are concyclic. Let $X=A_2C_2 \cap AC$ and $Y=A_1C_1 \cap AC$; let $P=BB_1 \cap A_2C_2$ and $R=BB_1 \cap AC$ and let $Q$ be the midpoint of $A_2C_2$. Now project $-1=(AC, B_1Y) \overset{B}{=} (A_1C_1, RY) \overset{B_1}{=} (A_2C_2, PX)$ proving $XA\cdot XB=XA_2 \cdot XC_2=XP \cdot XQ$ and hence $APQC$ is cyclic.

Looking at the above two solutions I feel kinda bad, since I was able to get that \[(K,BB_1\cap A_2C_2;C_2,A_2)=-1\]in @2above's answer's terms

Extend $AC$ and $C_2A_2$ to meet at a point $X$. Let $AA_1 \cap B_1C_1=R, \ \ B_1A_2 \cap C_1C=S, \ \ BB_1 \cap C_2A_2=K$. Let $M$ be the midpoint of $A_2C_2$. Then $B_1(A,P,R,A_1)=-1 \ \Longrightarrow B_1(C,P,C_1,S)=-1 \ \Longrightarrow (X,K,C_2,A_2)=-1$. By power of a point and Mclaren's identity $XC \cdot AX= XA_2 \cdot XC_2=XK \cdot XM$ implying $ACKM$ is cyclic.

The solution that I submitted during the exam

Attachments:
Solution_Q 18_1.pdf (430kb)

Tumon2001 wrote: Solution: Let's begin with a lemma. Lemma: Let $A,B,C,D $ be points on a line $\ell $ such that $(A,C;B,D)=(-1) $. Let $M $ be the midpoint of $BD $. Then, $(B,D)$ and $(C,M) $ are reciprocal pairs of an involution on $\ell$ with focus $A $. A synthetic proof of Lemma ,invert with center $A$ and power $\sqrt{AB.AD}$ ,now $-1=(A,C;B,D)=(\infty_{\overline{ABCD}},C^*;D,B)$ ,$\implies C^*=M$ so done $\blacksquare$