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Old 2003-06-24, 10:15   #1
hyh1048576
 
Jun 2003

26 Posts
Default Geometry puzzle(difficult)

Here is a triangle ABC.
Circle O1 touch AB,AC.and it's in the Angle BAC
Circle O2 touch AB,BC,Circle O1 and it's in the Angle ABC
Circle O3 touch AC,BC,Circle O2 and it's in the Angle ACB
Circle O4 touch AB,AC,Circle O3 and it's in the Angle BAC
Circle O5 touch AB,BC,Circle O4 and it's in the Angle ABC
Circle O6 touch AC,BC,Circle O5 and it's in the Angle ACB
Prove or disprove:Circle O6 touch Circle O1
I think it's right but I can't solve it :(
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Old 2003-06-24, 19:42   #2
NickGlover
 
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Aug 2002
Richland, WA

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Correct me if I'm wrong, but before you even consider whether O6 can touch O1, the existing arrangement you have given is not planar. If you represent the sides of the triangle and the circles as nodes in a graph and try to connect them all, you can see this.

Also consider it like this: How can both O1 and O4 touch AB and AC and both have a clear path for O1 to touch O2 and O4 to touch O3? No matter how you try to arrange it, one circle in each angle has to completely block the other from touching any other circles.
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Old 2003-06-24, 20:13   #3
NickGlover
 
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Aug 2002
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Ok, I realized I hadn't consider that the circles might overlap, so the problem is not as impossible as I initially thought.

However, the concept of circles "touching" seems vague. Does any intersection of circles count as touching or must the touching circles share a tangent line at the point of intersection?
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Old 2003-06-25, 00:54   #4
hyh1048576
 
Jun 2003

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Share a tangent line at the point of intersection。
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Old 2003-07-16, 13:30   #5
rpresser
 
Jul 2003

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Once P1 (center of O1) is chosen, all the rest of the circles are fixed, right?
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Old 2003-07-20, 02:13   #6
rpresser
 
Jul 2003

2·5 Posts
Default Solved it ... sort of

[I deleted the mess about GRACE because I found a better program.]

I believe I have demonstrated it, although I wouldn't call it a proof exactly.

I have a construction in Wingeom format. Unfortunately the construction wasn't able to identify that O1 and O6 intersect; but it does have a measure function, so I drew a segment between the centers (P1 and P6), intersected the segment with the circles, and measured the distance between the "intersection points". The measure remains zero - indicating that O1 and O6 are tangent - as you make changes to the triangle or the initial circle size as long as all six circles remain completely within the triangle. When one of the angles is very obtuse, it is sometimes not possible to construct one or more of the circles and stay within the triangle. (I'm not positive that this is right... would appreciate verification.)

http://member.newsguy.com/~rpresser/circles.jpg

Trivia about the construction: [list]Placing the first circle was my first return to geometry in many years. I solved it by using an external point to construct a point along the angle bisector of BAC, and using the constructed point as the center of O1.[/list:u][list]Once a circle is placed within the angle, constructing the next circle is the L-L-C case of the problem of Appolonius. I found an example of this construction here. The java applet JavaSketchpad, and the details of the construction were found in the page's source as a long text parameter to the applet. Some sweat and tears and text editing and I was able to reproduce the construction in Wingeom. (I used Wingeom because it's free ... couldn't convince wife I needed GSP).[/list:u]
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Old 2003-07-27, 06:46   #7
rpresser
 
Jul 2003

10102 Posts
Default Found another construction on the web

I found a page which has a Geometer's Sketchpad rendering of the construction. I think it proves it.
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