4 circles and non-trivial concurrency through the vertices

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Erken

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Erken

#1 h Sep 6, 2008, 10:43 AM • 2 Y

Y by Adventure10, Mango247

Let $\Gamma_1,\Gamma_2$ and $\Gamma_3$ be three non-intersecting circles,which are tangent to the circle $\Gamma$ at points $A_1,B_1,C_1$ ,respectively.Suppose that common tangent lines to $(\Gamma_2,\Gamma_3)$ , $(\Gamma_1,\Gamma_3)$ , $(\Gamma_2,\Gamma_1)$ intersect in points $A,B,C$ .
Prove that lines $AA_1,BB_1,CC_1$ are concurrent.

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The QuattoMaster 6000

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The QuattoMaster 6000

#2 h Sep 6, 2008, 3:12 PM • 2 Y

Y by Adventure10, Mango247

Erken wrote:

Let $\Gamma_1,\Gamma_2$ and $\Gamma_3$ be three non-intersecting circles,which are tangent to the circle $\Gamma$ at points $A_1,B_1,C_1$ ,respectively.Suppose that common tangent lines to $(\Gamma_2,\Gamma_3)$ , $(\Gamma_1,\Gamma_3)$ , $(\Gamma_2,\Gamma_1)$ intersect in points $A,B,C$ .
Prove that lines $AA_1,BB_1,CC_1$ are concurrent.

Solution

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plane geometry

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plane geometry

#3 h Sep 7, 2008, 12:04 AM • 1 Y

Y by Adventure10

Thank The QuattoMaster 6000 for a nice solution
but here I have a question: Monge's theorem states that for any three circles in a plane, none of which is inside one of the others, the three intersection points of the three pairs of external tangent lines are collinear.(http://en.wikipedia.org/wiki/Monge%27s_theorem)
in our problem the the three circles you mentioned don't satisfy "none of which is inside one of the others" ,could you tell me about the details?
Thank you in advance!

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The QuattoMaster 6000

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The QuattoMaster 6000

#4 h Sep 7, 2008, 2:39 AM • 2 Y

Y by Adventure10, Mango247

plane geometry wrote:

Thank The QuattoMaster 6000 for a nice solution
but here I have a question: Monge's theorem states that for any three circles in a plane, none of which is inside one of the others, the three intersection points of the three pairs of external tangent lines are collinear.(http://en.wikipedia.org/wiki/Monge%27s_theorem)
in our problem the the three circles you mentioned don't satisfy "none of which is inside one of the others" ,could you tell me about the details?
Thank you in advance!

Ah yes, sorry. Monge's Theorem might not be the correct name of the theorem I am stating. I am using some sort of extended verison of Monge's Theorem. I don't have a diagram, but I will try to explain. Consider three circles with centers $A$ , $B$ , and $C$ and with respective radii $x$ , $y$ , and $z$ (the circles can be internal or external.) Let the center of homothety between $(A)$ and $(B)$ be $Z$ , that between $(A)$ and $(C)$ be $Y$ , and that between $(B)$ and $(C)$ be $X$ . Then, $\frac {BX}{XC} = \frac {y}{z}$ . Similarly, $\frac {AY}{CY} = \frac {x}{z}$ and $\frac {AZ}{BZ} = \frac {x}{y}$ . Now, if you use the directed segments correctly, $X$ , $Y$ , and $Z$ should line up by Menelaus's Theorem (even when $(A)$ might be inside $(B)$ , for example.)

Sorry if some of my statements might be confusing... I am really bad at explaining directed segments without a diagram.

EDIT: I just remembered that the Geogebra Applet has been released

Here is a diagram, but it's quite inaccurate (this is my first time trying to use Geogebra.) I hope it at least helps you get the idea: [geogebra]6efb0d213b97e721d81eb95ffffc242da77cfc84[/geogebra]

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plane geometry

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plane geometry

#5 h Sep 7, 2008, 3:14 AM • 2 Y

Y by Adventure10, Mango247

The QuattoMaster 6000 wrote:

plane geometry wrote:

Thank The QuattoMaster 6000 for a nice solution
but here I have a question: Monge's theorem states that for any three circles in a plane, none of which is inside one of the others, the three intersection points of the three pairs of external tangent lines are collinear.(http://en.wikipedia.org/wiki/Monge%27s_theorem)
in our problem the the three circles you mentioned don't satisfy "none of which is inside one of the others" ,could you tell me about the details?
Thank you in advance!

Ah yes, sorry. Monge's Theorem might not be the correct name of the theorem I am stating. I am using some sort of extended verison of Monge's Theorem. I don't have a diagram, but I will try to explain. Consider three circles with centers $A$ , $B$ , and $C$ and with respective radii $x$ , $y$ , and $z$ (the circles can be internal or external.) Let the center of homothety between $(A)$ and $(B)$ be $Z$ , that between $(A)$ and $(C)$ be $Y$ , and that between $(B)$ and $(C)$ be $X$ . Then, $\frac {BX}{XC} = \frac {y}{z}$ . Similarly, $\frac {AY}{CY} = \frac {x}{z}$ and $\frac {AZ}{BZ} = \frac {x}{y}$ . Now, if you use the directed segments correctly, $X$ , $Y$ , and $Z$ should line up by Menelaus's Theorem (even when $(A)$ might be inside $(B)$ , for example.)

Sorry if some of my statements might be confusing... I am really bad at explaining directed segments without a diagram.

EDIT: I just remembered that the Geogebra Applet has been released

Here is a diagram, but it's quite inaccurate (this is my first time trying to use Geogebra.) I hope it at least helps you get the idea: [geogebra]6efb0d213b97e721d81eb95ffffc242da77cfc84[/geogebra]

thank you dear The QuattoMaster 6000!