Centre of circle, an antipode and tangency points are cyclic

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rah4927
Posts:110
Joined:Sat Feb 07, 2015 9:47 pm
Centre of circle, an antipode and tangency points are cyclic

Unread post by rah4927 » Tue Aug 02, 2016 10:15 pm

Let $\omega_1$ and $\omega_2$ be two orthogonal circles (definition given below). Let the centre of $\omega_1$ be $O$ . Diameter $AB$ of $\omega_1$ is selected so that $B$ lies strictly inside $\omega_2$. The two circles tangent to $\omega_2$ through both $O$ and $A$ touch $\omega_2$ at $F$ and $G$ . Prove that quadrilateral $FOGB$ is cyclic.

NOTE:

Let two circles with centres $P$ and $Q$ intersect at $X$ and $Y$. Then the two circle are orthogonal iff $\angle PXQ = \angle PYQ = 90 \circ $ .

tanmoy
Posts:312
Joined:Fri Oct 18, 2013 11:56 pm
Location:Rangpur,Bangladesh

Re: Centre of circle, an antipode and tangency points are cy

Unread post by tanmoy » Tue Aug 02, 2016 11:44 pm

Hint:
There are orthogonal circles.So,we can apply inversion 8-)
"Questions we can't answer are far better than answers we can't question"

rah4927
Posts:110
Joined:Sat Feb 07, 2015 9:47 pm

Re: Centre of circle, an antipode and tangency points are cy

Unread post by rah4927 » Wed Aug 03, 2016 10:07 pm

Invert with respect to $\omega_1$ . This keeps $\omega_2$ invariant, whilst the two tangent circles turn into lines tangent to $\omega_2$ that pass through $A$ . Let these tangency points be X and Y. Let the centre of $\omega_2$ be $O'$ . Then $AXO'Y$ is cyclic. Let $AO'$ intersect $\omega_1$ at $L$ .

We claim that $X,Y,L$ are collinear. To show this, invert once again wrt $\omega_2$ , and note that $X,Y$ remain in their respective places, but since the circles are orthogonal, $L$ is mapped to $A$. But since $P,X,O',Y$ are concyclic, hence $X,Y,L$ must be collinear. Let $B'$ be the other intersection between $\omega_1$ and $XY$. But $XY$ is perpendicular to $AO'$ , and therefore $\angle ALB$' is equal to $90$ degrees. Therefore $B'=B$ , and so $X,Y,B$ are collinear. We are done.

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