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Concyclicboy

Posts: 27
Location: República Independiente de Antiguo Cuscatlan
El Salvador

Posted: Oct 07, 2009, 3:19 pm • # 1

\item Two circles $\Gamma_1$ and $\Gamma_2$ intersect at points $A$ and $B$ . Consider a circle $\Gamma$ contained in $\Gamma_1$ and $\Gamma_2$ , which is tangent to both of them at $D$ and $E$ respectively. Let $C$ be one of the intersection points of line $AB$ with $\Gamma$ , $F$ be the intersection of line $EC$ with $\Gamma_2$ and $G$ be the intersection of line $DC$ with $\Gamma_1$ . Let $H$ and $I$ be the intersection points of line $ED$ with $\Gamma_1$ and $\Gamma_2$ respectively. Prove that $F$ , $G$ , $H$ and $I$ are on the same circle.

livetolove212

Posts: 598
Location: Hanoi
Viet Nam

Posted: Oct 08, 2009, 4:12 am • # 2

Construct a tangency $l$ of $(O)$ and $(O_1)$ then $\angle GHD=\angle (l,DG)=\angle CED$ thus $GH//CE$
Similarly, $FI//DC$
But $CG.CD=CF.CE$ then $DGFE$ is cyclic
Therefore $\angle GHD=\angle CED=\angle CGF=\angle CFI$ , which follows that $G,I,H,F$ are concyclic.

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luisgeometria

Posts: 1487
Location: Venezuela

Posted: Oct 09, 2009, 4:32 pm • # 3

The inversion with center $C$ and power $CB.CA$ takes $\Gamma_1$ and $\Gamma_2$ into themselves and $\Gamma$ into a common external tangent $\gamma$ to $\Gamma_1,\Gamma_2$ $\Longrightarrow$ $F,G$ are the contact points of $\gamma$ with $\Gamma_1,\Gamma_2$ . Since $\Gamma$ is internally/externally tangent to $\Gamma_1$ and $\Gamma_2$ because both cases hold , it follows that $D,E$ are inverse points through a positive inversion $(O,k^2)$ that maps $\Gamma_1 \to \Gamma_2$ , hence $H,I$ are inverse points and so are $F,G$ obviously, thus $OI.OH = OG.OF = k^2$ $\Longrightarrow F,G,H,I$ are concyclic.

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jayme

Posts: 1019

Posted: Oct 09, 2009, 10:06 pm • # 4

Dear Mathlinkers,
I can not resist to applied my favourite theorem..
1. Let (0) be the initial circle, (1) the one tangent at D, (2) the one tangent at E.
2. Let (3) the circle passing through D, G, E.
3. According to Monge or d'Alembert's theorem, it goes through F.
4. By Reim's theorem applied to (0) and (1), IF//DC
5. By a converse of the pivot theorem applied to (0), (1) and (3), GF is tangent to (1) at G
6. By a converse of Reim's theorem applied to (1) with IF//DG, F, G, H, I are cocyclics.
Sincerely
Jean-Louis

Moonmathpi496

Posts: 412
Location: Bangladesh

Blog: View Blog

Posted: Nov 23, 2009, 9:57 am • # 5

Concyclicboy wrote:

It is enough to prove that $\angle GHI=\angle GFI$ .

$DGFE$ is is a cyclic quadrilateral, as $CG\cdot CD=CA \cdot CB=CF\cdot CE.$ So, $\angle CFG=\angle D$
We know that a homothety with center $D$ maps $\Gamma_1 \to \Gamma$ . So, $GH \parallel EC$ , and for the same reason $FI \parallel FD$ .
We have, $\angle GHI=\angle E$ , and $\angle GFI=\pi-\angle CFG-\angle IFE=\pi-\angle D-\angle C=\angle E$ .
Q.E.D.

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diegu

Posts: 11

Posted: Dec 04, 2009, 12:53 pm • # 6

livetolove212 wrote:

Construct a tangency $l$ of $(O)$ and $(O_1)$ then $\angle GHD = \angle (l,DG) = \angle CED$ thus $GH//CE$
Similarly, $FI//DC$
But $CG.CD = CF.CE$ then $DGFE$ is cyclic
Therefore $\angle GHD = \angle CED = \angle CGF = \angle CFI$ , which follows that $G,I,H,F$ are concyclic.

why if $CG.CD = CF.CE$ then $DGFE$ is cyclic?
and on the picture where is the tangency $l$ ?
and why $\angle GHD = \angle (l,DG) = \angle CED$ ?

thanks

diegu

Posts: 11

Posted: Dec 05, 2009, 4:09 am • # 7

diegu wrote:

livetolove212 wrote:

Construct a tangency $l$ of $(O)$ and $(O_1)$ then $\angle GHD = \angle (l,DG) = \angle CED$ thus $GH//CE$
Similarly, $FI//DC$
But $CG.CD = CF.CE$ then $DGFE$ is cyclic
Therefore $\angle GHD = \angle CED = \angle CGF = \angle CFI$ , which follows that $G,I,H,F$ are concyclic.

why if $CG.CD = CF.CE$ then $DGFE$ is cyclic?
and on the picture where is the tangency $l$ ?
and why $\angle GHD = \angle (l,DG) = \angle CED$ ?

thanks

please...I don't understand

Joao Pedro Santos

Posts: 94
Location: Portugal-Alcanena/Lisbon

Posted: Mar 09, 2010, 1:08 pm • # 8

diegu wrote:

livetolove212 wrote:

Construct a tangency $l$ of $(O)$ and $(O_1)$ then $\angle GHD = \angle (l,DG) = \angle CED$ thus $GH//CE$
Similarly, $FI//DC$
But $CG.CD = CF.CE$ then $DGFE$ is cyclic
Therefore $\angle GHD = \angle CED = \angle CGF = \angle CFI$ , which follows that $G,I,H,F$ are concyclic.

why if $CG.CD = CF.CE$ then $DGFE$ is cyclic?
and on the picture where is the tangency $l$ ?
and why $\angle GHD = \angle (l,DG) = \angle CED$ ?

thanks

please...I don't understand

The cyclicity of $DGFE$ is a direct consequence of the Power of a Point Theorem's converse...
$\angle GHD = \angle (l,DG) = \angle CED$ is a direct consequence of the Half-Inscribed Angle Theorem...
But I think $\Gamma$ is inside $\Gamma_1$ and $\Gamma_2$ , and not the opposite... Anyway, the difficulty is the same.
Here's my solution:

Let $O$ , $O_1$ and $O_2$ be the centers of $\Gamma$ , $\Gamma_1$ and $\Gamma_2$ .
From the tangencies between the circles, we get the colinearities $D - O - O_1$ and $E - O - O_2$ .
Since $C$ lies on the radical axis of $\Gamma_1$ and $\Gamma_2$ , the points $D,E,F,G$ are concyclic.
With some angle chasing, it's easy to see it suffices to prove $\angle DGH = \angle EFI$ .
If we have the colinearity $O_1 - E - O - D - O_2$ , it's trivial, since $\angle DGH = \angle EFI = 90^{\circ}$ .
Otherwise it suffices to prove $\angle DO_1H = \angle EO_2I$ .
Let $D'$ be the reflection of $D$ w.r.t. $O$ .
Let $D''$ be the reflection of $D$ w.r.t. $O_1$ .
It's easy to see $[DD'E]$ and $[DD''H]$ are similar, so we have $\frac {DE}{DH} = \frac {DD'}{DD''} = \frac {DO}{DO_1}$ , therefore $[DOE]$ and $[DO_1H]$ are similar, so $\angle DOE = \angle DO_1H$ .
Analogously we prove $\angle DOE = \angle EO_2I$ , so $\angle DO_1H = \angle EO_2I$ , QED.

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Cyclic Quadrilateral

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