shobber wrote: $ \odot O_1$ and $ \odot O_2$ meet at points $ P$ and $ Q$. The circle through $ P$, $ O_1$ and $ O_2$ meets $ \odot O_1$ and $ \odot O_2$ at points $ A$ and $ B$. Prove that the distance from $ Q$ to the lines $ PA$, $ PB$ and $ AB$ are equal. (Prove the following three cases: $ O_1$ and $ O_2$ are in the common space of $ \odot O_1$ and $ \odot O_2$; $ O_1$ and $ O_2$ are out of the common space of $ \odot O_1$ and $ \odot O_2$; $ O_1$ is in the common space of $ \odot O_1$ and $ \odot O_2$, $ O_2$ is out of the common space of $ \odot O_1$ and $ \odot O_2$. This one is quite easy compared to the other China TST problems. Note that $\triangle O_1PO_2\cong\triangle O_1QO_2\implies \angle O_2O_1P=\angle O_2O_1B.$ Also, $\angle PO_1O_2=\angle PAO_2,$ and $\angle QO_1O_2=\angle BAO_2,$ leading to $\angle PAQ=\angle BAQ.$ So $AQ$ bisects $\angle PAB.$ Similarly $BQ$ bisects $\angle PBA,$ so $Q$ is the incentre of $\triangle PAB.$ So it is equidistant from $PA,PB$ and $AB.\Box$

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Dear Mathlinkers, remark that the third circle is a Morley's circle. http://perso.orange.fr/jl.ayme vol. 2 Les cercles de Morley... Sincerely Jean-Louis

Potla wrote: shobber wrote: $ \odot O_1$ and $ \odot O_2$ meet at points $ P$ and $ Q$. The circle through $ P$, $ O_1$ and $ O_2$ meets $ \odot O_1$ and $ \odot O_2$ at points $ A$ and $ B$. Prove that the distance from $ Q$ to the lines $ PA$, $ PB$ and $ AB$ are equal. (Prove the following three cases: $ O_1$ and $ O_2$ are in the common space of $ \odot O_1$ and $ \odot O_2$; $ O_1$ and $ O_2$ are out of the common space of $ \odot O_1$ and $ \odot O_2$; $ O_1$ is in the common space of $ \odot O_1$ and $ \odot O_2$, $ O_2$ is out of the common space of $ \odot O_1$ and $ \odot O_2$. This one is quite easy compared to the other China TST problems. Note that $\triangle O_1PO_2\cong\triangle O_1QO_2\implies \angle O_2O_1P=\angle O_2O_1B.$ Also, $\angle PO_1O_2=\angle PAO_2,$ and $\angle QO_1O_2=\angle BAO_2,$ leading to $\angle PAQ=\angle BAQ.$ So $AQ$ bisects $\angle PAB.$ Similarly $BQ$ bisects $\angle PBA,$ so $Q$ is the incentre of $\triangle PAB.$ So it is equidistant from $PA,PB$ and $AB.\Box$ I'm missing something in the solution, why is $ O_1QB $ a straight line?

@above O_1PO_2B is a cyclic quad.

One line proof: It is well known that $A,Q,C_2$ and $C_1,QB$ are collinear and thus, $C_1C_2BA$ is cyclic. Thus, $2\angle APQ=\angle AC_1Q=\angle QC_2B=2\angle QPB$ and $\angle QAB=\angle BC_1C_2=\angle PC_1C_2=\angle PAQ$ completing the proof.