The circles $\gamma_1$ and $\gamma_2$ intersect at the points $Q$ and $R$ and internally touch a circle $\gamma$ at $A_1$ and $A_2$ respectively. Let $P$ be an arbitrary point on $\gamma$. Segments $PA_1$ and $PA_2$ meet $\gamma_1$ and $\gamma_2$ again at $B_1$ and $B_2$ respectively. a) Prove that the tangent to $\gamma_{1}$ at $B_{1}$ and the tangent to $\gamma_{2}$ at $B_{2}$ are parallel. b) Prove that $B_{1}B_{2}$ is the common tangent to $\gamma_{1}$ and $\gamma_{2}$ iff $P$ lies on $QR$.
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Tags: geometry, power of a point, radical axis, geometry unsolved
shobber
28.05.2006 06:16
Let the centres of the circles be $O_1$, $O_2$ and $O$ respectively. Then: (a). Connect $O_1B_1$, $O_2B_2$ and $PO$. Then $O_1B_1 || PO || O_2B_2$, which means the tangents at $B_1$ and $B_2$ are also paralel. (b). Since P lies on QR, so $PB_1 \cdot PA_1 = PB_2 \cdot PA_2$. Thus $A_1, A_2, B_2, B_1$ lies on the same circle. Therefore $\angle{PB_2B_1}+\angle{O_2B_2A_2}=\angle{B_1A_1A_2}+\angle{O_2A_2B_2}=$ $=\angle{O_2A_2B_2}+\angle{OA_2A_1}+\angle{OA_1B_1}=\angle{PB_1B_2}+\angle{O_1B_1A_1}$ Thus $\angle{B_1B_2O_2}=\angle{B_2B_1O_1}$. But by the conclusion of the part (a) we have that those two angles have the sum of $180^o$. So there must be $\angle{B_1B_2O_2}=\angle{B_2B_1O_1}=90^o$, which yields $B_1B_2$ is the common tangent of $\gamma_1$ and $\gamma_2$.
Mashimaru
03.08.2009 05:23
Mr shobber, part (b) of the problem require an iff Here is my solution: Consider the inversion $ \mathcal{H}$ through pole $ P$ with product $ k = \overline{PA_1}.\overline{PB_1}$ which takes $ A_1\mapsto B_1$ and $ A_2\mapsto A_2'$ then $ \mathcal{H}$ takes $ (O)\mapsto (B_1A_2')$, but since $ (O)$ is tangent to $ (O_1), (O_2)$, so is $ B_1A_2'$. Then $ B_1B_2$ is the external tangent of $ (O_1),(O_2)$ iff $ B_2\equiv A_2'$ or $ \overline{PA_2}.\overline{PB_2} = k = P_{P/(O_1)}$, which implies that $ P$ lies on the radical axis of $ (O_1),(O_2)$, which is $ QR$.