orl wrote: An acute triangle $ ABC$ is given. Points $ A_1$ and $ A_2$ are taken on the side $ BC$ (with $ A_2$ between $ A_1$ and $ C$), $ B_1$ and $ B_2$ on the side $ AC$ (with $ B_2$ between $ B_1$ and $ A$), and $ C_1$ and $ C_2$ on the side $ AB$ (with $ C_2$ between $ C_1$ and $ B$) so that \[ \angle AA_1A_2 = \angle AA_2A_1 = \angle BB_1B_2 = \angle BB_2B_1 = \angle CC_1C_2 = \angle CC_2C_1. \] The lines $ AA_1,BB_1,$ and $ CC_1$ bound a triangle, and the lines $ AA_2,BB_2,$ and $ CC_2$ bound a second triangle. Prove that all six vertices of these two triangles lie on a single circle. We assume that, \[ \angle AA_1A_2 = \angle AA_2A_1 = \angle BB_1B_2 = \angle BB_2B_1 = \angle CC_1C_2 = \angle CC_2C_1 = \theta \] At first we notice that for any given point $ A_1$ the constructions of the other points are unique. To construct $ A_2$ we have to take point symmetric to the foot of perpendicular from $ A$. And analogously we have to construct other points by rotating the altitudes by angle $ \frac \pi2 - \theta$. Now we chase the angles. We get easily that, $ \angle P_1Q_1R_1 = \pi - (\pi - \theta) - (\frac \pi2 - B - (\frac \pi2 - \theta)) = B$. Analogous angle chasing assures that $ \triangle ABC \sim \triangle P_1Q_1R_1 \sim \triangle P_2Q_2R_2$ Now We shall prove that $ P_1R_2 \parallel AC$. We have $ BB_1 = BB_2$. So it is enough to prove that $ BP_1 = BR_2$. We have $ \angle R_2AB = \angle CP_1B = \pi - B - \theta = \phi$. So, Sine law in $ \triangle R_2AB,\triangle CP_1B$ gives, \[ \frac {BR_2}{\sin \phi} = \frac {AB}{\sin C} \text{ and } \; \frac {BP_1}{\sin \phi} = \frac {BC}{\sin A} \] Dividing the first equation by the second one we get, $ BZ_2 = BP_1$ Now we shall prove $ R_1P_2 \parallel CA$ i.e. $ BR_1 = BP_2$. Sine law in $ \triangle BP_2C_2, \triangle BR_1A_1$ gives, \[ \frac {P_2B}{\sin \theta} = \frac {BC_2}{\sin A} \; \text{ and } \frac {BR_1}{\sin \theta} = \frac {A_1B}{\sin C} \] Let, $ \angle C_2CB = \angle AA_1B = \theta - B = \alpha$ Again by sine law, \[ \frac {P_2B}{\sin \alpha} = \frac {BC}{\sin \theta}\; \text{ and } \frac {A_1B}{\sin \alpha} = \frac {AB}{\sin \theta} \] From these four equations we get, $ BR_1 = BP_2 \iff R_1P_2 \parallel AC \parallel P_1R_2$. And also $ R_2Q_1 \parallel BC$. Now straight forward angle chasing gives, $ \angle P_1R_2Q_1 = \pi - (\theta - A) - (\pi - \theta - B) = A + B = \pi - C$. Which proves that $ R_2$ lies on the circumcircle of $ \triangle P_1Q_1R_1$. Analogously, $ Q_2,R_2$ lies on the circle proving the claim. EDIT: Fixed typo. I hope that I haven't made any more typo

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This is my solution We have $ \angle BC_1J = \angle AA_1A_2 \Rightarrow BC_1JA_1$ is cyclic $ \Rightarrow AJ.AA_1 = AC_1.AB$ Similarly, $ AK.AA_2 = AB_2.AC$ But $ \angle BC_1C = \angle BB_2C \Rightarrow BC_1B_2C$ is cyclic. $ \Rightarrow AC_1.AB = AB_2.AC$ $ \Rightarrow AJ.AA_1 = AK.AA_2$ Similarly, $ AN.AA_1 = AM.AA_2$ We get $ AJ.AN = AK.AM \Rightarrow NJKM$ is cyclic. But $ JA_1A_2K$ and $ LC_1C_2M$ are cyclic then $ \angle JKM = \angle AA_1A_2 = \angle CC_2C_1 = \angle JLM$ $ \Rightarrow JKLM$ is cyclic. Similarly, $ JPNM$ is cyclic. $ \rightarrow$ QED

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@livetolove212: Nice solution, but hard for me to think of. BTW there is another solution, faster than ours. For this we need to show that $ H$ is the center of the circle. That can be done by calculating the radius using trigonometry.

Moonmathpi496 wrote: BTW there is another solution, faster than ours. For this we need to show that $ H$ is the center of the circle. That can be done by calculating the radius using trigonometry. Can you post this solution? Thanks a lot

Actually this solution is from IMO compendium. This is surprisingly short for an ISL 04. Triangles $ AHB, BHC, CHA, ABC$ have the same circumradius $ R$. $ BCP_1H$ is concyclic and therefore $ HP_1 =2R \cos \theta$. The same holds for the other radii and the conclusion follows.

I’m going to use the second post’s picture and write my solution using it. First it’s easy to see that $AR_2HR_1B$ , $CQ_2HQ_1A$ and $CP_1HP_2B$ lie on one circle.Then we know that $AH$ and $BH$are the angle bisector of $\angle A_1AA_2$ and $\angle B_1BB_2$ so $R_2H=HR_1$ ( using the same way $P_1H=HP_2$ & $Q_1H=HQ_2$ ) $(1)$ By angle chasing we find that $\angle AB_1C_2=\angle AQ_2C_2=\angle B$ and also $\angle B_1R_1A= \angle B_1C_2A= \angle C$ thus $B_1,Q_2,R_1,C_2,A$ lie on one circle. Then we have $B_1Q_2R_1C_2$ and also $B_1C_2BC$ are concyclic using this it’s easy to see that: $Q_2R_1 \parallel CB$ Thus $AH$ is the perpendicular bisector of $Q_2R_1 \Rightarrow HQ_2=HR_1$ (with the same way: $HQ_1=HP_2$ & $HP_1=HR_2$) combining with $(1)$ $\Rightarrow$ the points $P_1,P_2,Q_1,Q_2,R_1,R_2$ lie on a circle which $H$ is it’s center.