Three circles have a common point

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Three circles have a common point

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Source: Swiss Math Olympiad 2010 - final round, problem 9

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#1 h Mar 16, 2010, 8:56 PM • 2 Y

Y by Adventure10, Mango247

Let $k$ and $k'$ two concentric circles centered at $O$ , with $k'$ being larger than $k$ . A line through $O$ intersects $k$ at $A$ and $k'$ at $B$ such that $O$ seperates $A$ and $B$ . Another line through $O$ intersects $k$ at $E$ and $k'$ at $F$ such that $E$ separates $O$ and $F$ .
Show that the circumcircle of $\triangle{OAE}$ and the circles with diametres $AB$ and $EF$ have a common point.

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#2 h Mar 17, 2010, 12:12 AM • 2 Y

Y by Adventure10, Mango247

Moreover the circumcircle of $\triangle OBF$ also passes through such a common point. Consider the inversion through pole $O$ and power $\overline{OE} \cdot \overline{OF}.$ Circle with diameter $EF$ is double and the opposite rays of $OB,OA$ cut $k'$ and $k,$ respectively at the inverse images $A',B'$ of $A,B.$ Then $\odot(OBF) \mapsto EB',$ $\odot(OEA) \mapsto FA'$ and circle with diameter $AB$ is transformed into the circle with diameter $A'B'.$ Let $P \equiv FA' \cap EB'.$ Since $EB' \perp FA'$ $\Longrightarrow$ $P$ is common point of $FA',EB'$ and the circles with diameters $A'B'$ and $EF.$ Hence, $\odot(OEA),\odot(OBF)$ and circles with diameters $AB,EF$ concur at the inverse $P'$ of $P.$

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#3 h Mar 17, 2010, 9:19 PM • 3 Y

Y by linpaws, Adventure10, Mango247

Ma solution is the following:
Denote $k_1$ , $k_2$ the circles with diametres $AB$ , $DE$ , respectively. Moreover, $k_1\cap k_2 = \{M,N\}$ , where $M$ is in the inner of angle $\angle{AOE}$ , and $N_1 = AE\cap k_1$ and $N_2 = AE \cap k_2$ .
As $\angle{AN_1B} = \angle{FN_2E} = 90^{\circ}$ due to Thales' theorem, and $\triangle{OBF}$ is isosceles with $\overline{OB} = \overline{OF}$ or $\angle{FBO} = \angle{OFB}$ , we must have $N = N_1 = N_2$ .
Thus, we have
$\angle{FEM} = \angle{FNM} = 90^{\circ} - \angle{MNA} = 90^{\circ} - \angle{MBA} = \angle{BAM} = \angle{OAM}\mbox{,}$
implying that $AOEM$ is cyclic.

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#4 h Feb 5, 2011, 10:45 AM • 2 Y

Y by Adventure10, Mango247

I have made it too much complicated

Let $m$ denote the circle with radius $AB$ and $n$ denote the circle with radius $EF$ .Draw tangents to $m$ at $A$ and $B$ .
The tangent at $E$ meets the tangent at $A$ at $M$ and The tangent at $F$ meets the tangent at $B$ at $N$ .The points $M$ and $N$ have equal powers w.r.t $m$ and $n$ .So $MN$ is the radical axis of $m$ and $n$ .Our required point must lie on $MN$ .Observe that $M$ lies on the circumcircle of $\bigtriangleup OAE$ .But this is not the required point,since its power w.r.t $n$ is $MF^2>0$ .Let $P$ be the projection of $O$ on $MN$ .This $P$ is the second intersection of the circle $OAE$ with $MN$ .We claim this one is the required point.
$\angle OPE=\angle EAO$ since $A,O,E,P,M$ are concyclic.and since $OA=OE$ we have $\angle OEA=\angle OAE=\angle OPE$ .
Also, $\angle FPN=\angle FON=\angle NOB$ since $PONF$ is cyclic and $FN=NB$ .Thus $\angle EOB=2\angle OPE=2\angle FPN$ .So $\angle FPE=\angle EPN+\angle FPN=\angle EPN+\angle OPE=\frac{\pi}{2}$ .So $P\in n$ and since $MN$ is the radical axis,P is also on $m$ .
QED

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