Easy problem matinyousefi wrote: A circle $\omega$ is strictly inside triangle $ABC$. The tangents from $A$ to $\omega$ intersect $BC$ in $A_1,A_2$ define $B_1,B_2,C_1,C_2$ similarly. Prove that if five of six points $A_1,A_2,B_1,B_2,C_1,C_2$ lie on a circle the sixth one lie on the circle too. Let $O$ be center of $\omega$.Let $$\angle{OAB}=x \; \angle{OBC}=y \; \angle{OCA}=z$$Claim : $A_1,A_2,B_1,B_2,C_1,C_2$ lies on a Conic Proof : Let $\theta = \text{angle between tangent from } A \text{to} \; \omega \text{and} \; OA$ .Note that \[\left(\frac{BA_1}{A_1C}\cdot \frac{BA_2}{A_2C}\right)=\left(\frac{\sin{(x-\theta})}{\sin{(A-x+\theta})}\cdot \frac{\sin{(x+\theta})}{\sin{(A-x-\theta})}\right)=\left(\frac{\sin^2{x}-\sin^2{\theta}}{\sin^2{A-x}-\sin^2{\theta}}\right)=\frac{\left(\frac{OB^2OA^2\sin^2{\angle{AOB}}}{AB^2}\right)-r^2}{\left(\frac{OC^2OA^2\sin^2{\angle{AOC}}}{AC^2}\right)-r^2}\]which obviously is symmetric.Hence we have \[\prod_{cyc}\left(\frac{BA_1}{A_1C}\cdot \frac{BA_2}{A_2C}\right)=1\]Thus by Carnot's Conic Theorm we have $A_1,A_2,B_1,B_2,C_1,C_2$ lie on a conic.$\square$ Back to the main problem by our lemma we have $A_1,A_2,B_1,B_2,C_1,C_2$ lie on a conic but since $5$ points uniquely determine a Conic whence the circle through $5$ points is the Conic which gives the sixth point also lies on that Conic a.k.a circle so we are done.$\blacksquare$

Nice solution. Just for the record there is a solution without conics.

First, we will prove that it is enough to show that \[ \frac{|AB_1| \cdot |AB_2| \cdot |BC_1| \cdot |BC_2| \cdot |CA_1| \cdot |CA_2|}{|AC_1| \cdot |AC_2| \cdot |BA_1| \cdot |BA_2| \cdot |CB_1| \cdot |CB_2|} = 1. \]Notice that if all six points lie on a circle, then the equality is trivial by applying the power of a point theorem three times. Furthermore, as any one of the six points varies along its side, the value of the expression is clearly injective. Thus, any five of these points lying on a circle would force the sixth to be the other intersection of that circle with the respective side. Therefore, all six points must lie on a circle. (This is essentially the proof of Carnot's Theorem.) Proof: Let $BB_1$ and $CC_1$ meet at $P_1$, and let $BB_2$ and $CC_2$ meet at $P_2$. Let $AP_1$ and $AP_2$ meet side $BC$ at $Q_1$ and $Q_2$. Applying the dual of Desargues' Involution Theorem to the conic $w$, the quadrilateral $BP_1CP_2$, and the point $A$, we find that $(AP_1, AP_2)$, $(AB, AC)$, and $(AA_1, AA_2)$ are reciprocal pairs of some involution of pencils. (For further details, refer to this Handout on Desargues' Involution Theorem.) Projecting onto $BC$, we see that there is an involution on line $BC$ swapping pairs $(B, C)$, $(Q_1, Q_2)$, and $(A_1, A_2)$. Thus, $$(A_1, Q_1; C, B) = (A_2, Q_2; B, C).$$Expanding this relation, we have \[ \frac{|CA_1| \cdot |BQ_1| \cdot |CA_2| \cdot |BQ_2|}{|BA_1| \cdot |CQ_1| \cdot |BA_2| \cdot |CQ_2|} = 1. \]Finally, applying Ceva's theorem to the cevians concurrent at $P_1$ and $P_2$ simplifies this to the desired result.

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