EDIT: MY 100th Post!! Pretty nice problem and soo classic flavour! Sharygin P22 wrote: Chords $A_1 A_2 , A_3 A_4 , A_5 A_6$ of a circle $\Omega$ concur at point $O.$ Let $B_i$ be the second common point of $\Omega$ and the circle with diameter $OA_i$ . Prove that chords $B_1 B_2 , B_3 B_4 , B_5 B_6$ concur. Given $A_1A_2,A_3A_4,A_5A_6$ concur at $O.$ Let $A$ denote the centre of $\Omega.$ We present a stronger claim. Claim: We claim that $AO,B_1B_2,B_3B_4,B_5B_6$ concur. Note that proving this claim, we will be done. So, to prove the stronger claim, Enough to show $AO,B_1B_2,B_3B_4$ concur and by symmetry, we are done. So let's remove $B_5,B_6,A_5,A_6.$ Claim: $A_1B_2,A_2B_1,A_3B_4,B_3A_4$ concur Proof: Since all of the points lie on a circle, so it's enough to show that $$(A_1,A_2;A_3,A_4)=(B_2,B_1;B_4,B_3)$$But $$(B_2,B_1;B_4,B_3)=\frac{B_2B_4}{B_1B_4}\frac{B_3B_1}{B_3B_2}=(B_1,B_2;B_3,B_4).$$Now, let $L,M,N,P$ be the antipodes of $A_1,A_2,A_3,A_4$ wrt $\Omega.$ Then, note that $$(A_1,A_2;A_3,A_4)\stackrel{A}{=}(L,M;N,P)\stackrel{O}{=}(B_1,B_2;B_3,B_4)=(B_2,B_1;B_4,B_3)$$ Let the concurrency point of $A_1B_2,A_2B_1,A_3B_4,B_3A_4$ be $H.$ Let $\ell$ be the polar of $H$ wrt $\Omega.$ Define $G=A_1B_1\cap B_2A_2,H=B_1B_2\cap A_1A_2 B_1B_2\cap A_1A_2=J.$ Note that, by brokard's theorem, $GJ$ is the polar of $H\implies GJ\in \ell.$ Similarly, $I\in \ell.$ Note that $AH\cap \ell=F$ is the miquel point of the quadrilateral $A_1B_1B_2A_2,A_3B_4B_3A_4.$ ( As $\angle GFA=90$ and miquel point lies on polar. ) By the definition of miquel point, we get $F\in (GB_1B_2).$ But $$\angle OB_1G=\angle OB_2G=90\implies O\in(GB_1B_2)\implies \angle GFO=90\implies F-H-O-A. $$Similarly, we get $F,O\in (IB_3B_4).$ Claim: $B_1B_2,B_4B_3,OF$ concur Proof: We use radical axis theorem on $(OB_1B_2),(OB_3B_4),\Omega.$ But $OF=OA.$ Hence $AO,B_1B_2,B_3B_4$ concur. And we are done!

Fix a point $O$ inside the circle and say $A_1$ is a point on the circle, let $A_2 = A_1O \cap \Omega$, define $B_1, B_2$ as in the problem, it suffices to show $B_1B_2$ goes through a fixed point as $A_1$ varies. Let $O'$ be the center of the circle and say $C_1, C_2$ are antipodes of $A_1, A_2$ so $B_1, O, C_1$ and $A_1, O', C_1$ are collinear. Say $X = OO' \cap B_1B_2$, I claim $X$ is the fixed point. But this follows by ping pong lemma using $$A_1 = (((((O'A_1 \cap \Omega)O \cap \Omega)X \cap \Omega)O \cap \Omega)O' \cap \Omega)O \cap \Omega$$so we are done. $\blacksquare$

Pretty trivial by Involution---

Just invert with center $O$ and radius $\sqrt{Pow(\Omega, O)}$ and then reflect over $O$. Then $B_i*$ is the reflection of $A_i$ in the center of $\Omega$. All this means is that $B_1*B_2*$ is the reflection of $A_1A_2$ across the center of $\Omega$. Now use the Forgotten Coaxiality Lemma to prove the circles $OB_1*B_2*$, $OB_3*B_4*$ and $OB_5*B_6*$ are coaxial. (Here $X$ is the original point and $X*$ is its image after the inversion and the reflection)

Master_of_Aops wrote: Just invert with center $O$ and radius $\sqrt{Pow(\Omega, O)}$ and then reflect over $O$. Then $B_i*$ is the reflection of $A_i$ in the center of $\Omega$. All this means is that $B_1*B_2*$ is the reflection of $A_1A_2$ across the center of $\Omega$. Now use the Forgotten Coaxiality Lemma to prove the circles $OB_1*B_2*$, $OB_3*B_4*$ and $OB_5*B_6*$ are coaxial. (Here $X$ is the original point and $X*$ is its image after the inversion and the reflection) Yes. My solution is same as yours but does it require that unique radius?

No it does not, but it is convenient if the images are already defined in the problem and you don't have to go about defining all of them again.

Master_of_Aops wrote: No it does not, but it is convenient if the images are already defined in the problem and you don't have to go about defining all of them again. Oh i see. Clever way!

Kinda strange that no one has posted a complex bash yet. Well, here it is: We will use complex numbers. WLOG let $\Omega$ be the unit circle and let $a,b,c,x$ denote the complex numbers of $A_{1},A_{3},A_{5},O$ respectively and denote by $a_{2},a_{4},a_{6},b_{i}$ the complex numbers of $A_{2},A_{4},A_{6},B_{i}$ respectively. Then: $$\overline{a}=\frac{1}{a},\quad \overline{b}=\frac{1}{b},\quad \overline{c}=\frac{1}{c},\quad \overline{a_{i}}=\frac{1}{a_{i}},\quad \overline{b_{i}}=\frac{1}{b_{i}}$$We will calculate $a_{2}$ (we can multiply by $a$ and $a_{2}$ because they lie on the unit circle, so they are nonzero and divide by $a-a_{2}$, because $A_{1}\not\equiv A_{2}$): $$A_{2}\in OA_{1}\Longrightarrow (a-x)\overline{a_{2}}-(\overline{a}-\overline{x})a_{2}-a\overline{x}+\overline{a}x=0$$$$\Longrightarrow (a-x)\frac{1}{a_{2}}-(\frac{1}{a}-\overline{x})a_{2}-a\overline{x}+\frac{x}{a}=0$$$$\Longrightarrow a^2-ax-(1-a\overline{x})a_{2}^2-a_{2}(a^2\overline{x}-x)=0$$$$\Longrightarrow (a-a_{2})(a+a_{2}-x-\overline{x}aa_{2})=0$$$$\Longrightarrow a_{2}=\frac{x-a}{1-\overline{x}a}$$Analogically, we get that $a_{4}=\frac{x-b}{1-\overline{x}b}$ and $a_{6}=\frac{x-c}{1-\overline{x}c}$. Now we will calculate the complex numbers $b_{i}$. For uniformity, let $Z\in\Omega$ and we'll express the complex number of the point $Y$, which is the intersection of the circle with diameter $OZ$ and $\Omega$ and then substitute $a_{i}$ instead of $z$ into the formula for $y$ (we can multiply by $y$ and $z$ since they lie on the unit circle, therefore are not zero and divide/multiply by $y-z$ as $Y\not\equiv Z$): $$Y\in\Omega\Longrightarrow \overline{y}=\frac{1}{y}$$$$YZ\perp OY\Longrightarrow \frac{x-y}{y-z}=-\overline{\left(\frac{x-y}{y-z}\right)}=-\left(\frac{\overline{x}-\frac{1}{y}}{\frac{1}{y}-\frac{1}{z}}\right)=\frac{\overline{x}yz-z}{y-z}$$$$\Longrightarrow x-y=\overline{x}yz-z\Longrightarrow y=\frac{x+z}{1+\overline{x}z}$$$$\Longrightarrow b_{1}=\frac{x+a}{1+\overline{x}a},\quad b_{2}=\frac{2x-ax\overline{x}-a}{1-2a\overline{x}+x\overline{x}}$$$$\Longrightarrow b_{3}=\frac{x+b}{1+\overline{x}b},\quad b_{4}=\frac{2x-bx\overline{x}-b}{1-2b\overline{x}+x\overline{x}}$$$$\Longrightarrow b_{5}=\frac{x+c}{1+\overline{x}c},\quad b_{6}=\frac{2x-cx\overline{x}-c}{1-2c\overline{x}+x\overline{x}}$$We will prove that the point $T$ lies on $B_{1}B_{2}$, $B_{3}B_{4}$, $B_{5}B_{6}$, where $t=\frac{x(x\overline{x}+3)}{3x\overline{x}+1}$ (this seems random, but in the process of solving I've just computed this point as the intersection of the lines and it's easier to show that it lies on the three lines for this solution). Firstly, $t$ is a complex number, because $3x\overline{x}+1\neq 0$, because $x\overline{x}$ is equal to the distance from $O$ to the center of $\Omega$ (i.e. the origin of the complex plane) squared, so $3x\overline{x}+1\geq 1>0$, thus the denominator is not zero and therefore such a point exists. Now: $$T\in B_{1}B_{2}\iff (b_{1}-b_{2})\overline{t}-(\overline{b_{1}}-\overline{b_{2}})t-b_{1}\overline{b_{2}}+\overline{b_{1}}b_{2}=0$$$$\iff \left(\frac{x+a}{1+\overline{x}a}-\frac{2x-ax\overline{x}-a}{1-2a\overline{x}+x\overline{x}}\right)\cdot\frac{\overline{x}(x\overline{x}+3)}{3x\overline{x}+1}-\left(\frac{1+\overline{x}a}{x+a}-\frac{1-2a\overline{x}+x\overline{x}}{2x-ax\overline{x}-a}\right)\cdot\frac{x(x\overline{x}+3)}{3x\overline{x}+1}-$$$$-\frac{x+a}{1+\overline{x}a}\cdot \frac{1-2a\overline{x}+x\overline{x}}{2x-ax\overline{x}-a}+\frac{1+\overline{x}a}{x+a}\cdot \frac{2x-ax\overline{x}-a}{1-2a\overline{x}+x\overline{x}}=0$$$$\iff \Big((x+a)(1-2a\overline{x}+x\overline{x})-(1+\overline{x}a)(2x-ax\overline{x}-a)\Big)\overline{x}(x\overline{x}+3)(x+a)(2x-ax\overline{x}-a)$$$$-((1+\overline{x}a)(2x-ax\overline{x}-a)-(x+a)(1-2a\overline{x}+x\overline{x}))x(x\overline{x}+3)(1+\overline{x}a)(1-2a\overline{x}+x\overline{x})$$$$-(x+a)^2(1-2a\overline{x}+x\overline{x})^2(3x\overline{x}+1)$$$$+(1+\overline{x}a)^2(2x-ax\overline{x}-a)^2(3x\overline{x}+1)=0\quad(\star)$$We will do this by parts as the computation is getting long. Firstly, we can factor(by factoring out $3x\overline{x}+1$ and then noting that $\overline{x}=\frac{1}{x}$ would make the expression $0$, so it must be a factor and then also factoring $x\overline{x}+3$ which we know will come up, because it appears in the other two terms in $\star$ and then finishing the factorization by considering what's left as a quadratic in $x$): $$-(x+a)^2(1-2a\overline{x}+x\overline{x})^2(3x\overline{x}+1)+(1+\overline{x}a)^2(2x-ax\overline{x}-a)^2(3x\overline{x}+1)=$$$$=(3 x \overline{x} + 1) (x \overline{x} - 1) (x \overline{x} + 3) (-a^2\overline{x} + 2 a - x) (x - a^2\overline{x})$$We will do a similar thing for the sum of the other two terms. We have that: $$\Big((x+a)(1-2a\overline{x}+x\overline{x})-(1+\overline{x}a)(2x-ax\overline{x}-a)\Big)\overline{x}(x\overline{x}+3)(x+a)(2x-ax\overline{x}-a)$$$$-((1+\overline{x}a)(2x-ax\overline{x}-a)-(x+a)(1-2a\overline{x}+x\overline{x}))x(x\overline{x}+3)(1+\overline{x}a)(1-2a\overline{x}+x\overline{x})=$$$$=(x\overline{x}+3)(x \overline{x} - 1) (3 x \overline{x} + 1) (x - a^2 \overline{x}) (a^2 \overline{x} -2a+ x)$$Therefore we can finish $(\star)$: $$(\star)\iff (x\overline{x}+3)(x \overline{x} - 1) (3 x \overline{x} + 1) (x - a^2 \overline{x}) (a^2 \overline{x} -2a+ x)+$$$$+(3 x \overline{x} + 1) (x \overline{x} - 1) (x \overline{x} + 3) (-a^2\overline{x} + 2 a - x) (x - a^2\overline{x})=0$$$$\iff (x\overline{x}+3)(x \overline{x} - 1) (3 x \overline{x} + 1) (x - a^2 \overline{x})(a^2 \overline{x} -2a+ x-a^2\overline{x} + 2 a - x)=0$$But $a^2 \overline{x} -2a+ x-a^2\overline{x} + 2 a - x=0$, so going back the equivalences, we have shown that $T\in B_{1}B_{2}$. Now note that the calculations for $T\in B_{3}B_{4}$ will be exactly the same, but $a$ replaced by $b$, so they are analogues. The same can be said about $T\in B_{5}B_{6}$: we would have the same calculations, but with $c$ instead of $a$, so therefore $T$ lies on $B_{1}B_{2},B_{3}B_{4},B_{5}B_{6}$, so the three lines concur, which was what we wanted to prove.

Consider an inversion with arbitrary radius centered at $O$. Denote the image of $X$ with $X'$. The points $A_i'$ will still lie on a circle $\Omega'$. However, since \[ 90^\circ= \measuredangle OB_1A_1 \ = -\measuredangle OA_1'B_1' = -\measuredangle A_2'A_1'B_1', \]$B_1$ is the antipode of $A_2$ in $\Omega'$. The images of the rest of the $B_i$ can be found similarly. Thus, the problem transforms into the following (now we drop the prime notation for clarity): Claim. Let $A_i$ be points on a circle $\Omega$, $i=1,\ldots,6$ in such a way that lines $A_1A_2$, $A_3A_4$, and $A_5A_6$ concur at $O$. Let $B_2$, $B_1$, $B_4$, $B_3$, $B_6$, and $B_5$ be the antipodes of $A_1$, $A_2$, $A_3$, $A_4$, $A_5$, and $A_6$ respectively. Then the circles $(OB_1B_2)$, $(OB_3B_4)$, and $(OB_5B_6)$ intersect at a point other than $O$. Proof. Let $X$ be the center of $\Omega$. Since $B_1,B_2$ are the reflections of $A_2,A_1$ in $X$, the line $B_1B_2$ is the reflection of the line $A_1A_2$ in $X$. It follows that lines $B_1B_2$, $B_3B_4$, and $B_5B_6$ concur at $P$, the reflection of $O$ in $X$. It is now clear that the three circles are coaxial: $P$ has equal power with respect to all three circles, namely the power of $P$ with respect to $\Omega$. Combined with the fact that $O$ lies on all three circles implies the claim. Inverting back, the problem follows.

Let $X'$ be reflection of $X$ over the center of circle. Thus, $\overline{A_1'A_2'},\overline{A_3'A_4'},\overline{A_5'A_6'}$ are concurrent at $O'$. Observe that $A_i',O,B_i$ are collinear. By inversion at $O$ with radius $\sqrt{-OA_1\cdot OA_2}$, we get that $(OB_1B_2),(OB_3B_4),(OB_5B_6)$ are coaxial. Finally, radical axis theorem on $(OB_1B_2),(OB_3B_4),(OB_5B_6),(B_1B_2B_3B_4B_5B_6)$ yields desired concurrency.

Let $X\neq a(X)=OX\cap\Omega$ and $X\neq h(X)=NX\cap\Omega$ where $N$ is the center of $\Omega$. It is clear by angle chasing that $B_i=(a\circ h(A_i))$ where $a\circ h$ is the composition of the two involutions. Therefore $B_2=((a\circ h)\circ a\circ( h\circ a))(B_1)=l(B_1)$. Since $$l\circ l$$$$=a\circ h\circ a\circ h\circ a\circ a\circ h\circ a\circ h\circ a=$$$$=a\circ h\circ a\circ h\circ h\circ a\circ h\circ a=$$$$=a\circ h\circ a\circ a\circ h\circ a=$$$$=a\circ h\circ h\circ a=$$$$=a\circ a=$$$$id$$$l$ is an involution, and therefore there exists a point $M$ such that $X\neq l(X)=MX\cap\Omega$, as wanted.

Woah thats a very nice problem! I might have used geogebra to draw the diagram, but thats irrelevant since sharygin allows everything Let the centre of the circle be $P$ and also remember that $i$ is odd throughout this proof Invert about $O$ with radius -$\sqrt{OA_1 \cdot OA_2}$, the point $A_i$ are mapped to $A_i+1$ and $B_i$ gets mapped such that $A_I'B_i' \perp OA_i'$. This basically means that $B_i'$ is the reflection of $A_i$ wrt to $P$. This means that $B_i'B_{i+1}'$ also concur at a point that is the reflection of $O$ about $P$ (O_1) . Now we have to prove $X \in \odot(B_i'B_{i+1}'O)$. Let $O'O \cap \odot(B_1B_2O) = X$, notice that $$O_1X \cdot O_1O = O_1B_1' \cdot O_1B_2' = O_1B_3' \cdot O_1B_4' \implies X \in \odot(B_3'B_4'O)$$, similar such results imply $X \in \odot(B_i'B_{i+1}'O)$ thus finishing the problem $\blacksquare$