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Topic: Lost Circles (Read 797 times) |
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FiBsTeR
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Lost Circles
« on: Apr 26th, 2008, 4:22pm » |
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Define a lost circle to be a circle that can be inscribed in a hexagon with side lengths 4, 8, 15, 16, 23, and 42, in some order; that is, the circle is tangent to each of the sides of the hexagon. Prove that no lost circle exists.
(Source: lingomaniac88, AoPS)
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| « Last Edit: May 3rd, 2008, 7:50am by FiBsTeR » |
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FiBsTeR
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Re: Lost Circles
« Reply #1 on: Apr 29th, 2008, 4:19pm » |
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Are these numbers too creepy for you guys? I can come up with some new ones instead, if you want. 
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FiBsTeR
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Re: Lost Circles
« Reply #2 on: May 1st, 2008, 2:33pm » |
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HINT:
You cannot inscribe a circle in a hexagon with side lengths 1,1,2,3,5,9, either.
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iono
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Re: Lost Circles
« Reply #3 on: May 1st, 2008, 5:43pm » |
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....cuz a hexagon can only have 6 sides?
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FiBsTeR
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Re: Lost Circles
« Reply #4 on: May 1st, 2008, 5:50pm » |
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Though I agree with your statement, I don't see how that proves these mysterious circles don't exist. 
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kindy_kid
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Re: Lost Circles
« Reply #5 on: May 1st, 2008, 8:26pm » |
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Does inscribed mean that the circle has to touch all 6 sides? Can it just touch 3, or 2, or 1 or none? 
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towr
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Re: Lost Circles
« Reply #6 on: May 2nd, 2008, 12:49am » |
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on May 1st, 2008, 8:26pm, kindy_kid wrote:| Does inscribed mean that the circle has to touch all 6 sides? Can it just touch 3, or 2, or 1 or none? |
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I think it has to touch all sides. 0,1,2 or 3 is easy, 4 and 5 are probably doable as well; I think the problem is touching the 6th side.
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Wikipedia, Google, Mathworld, Integer sequence DB
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FiBsTeR
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Re: Lost Circles
« Reply #7 on: May 3rd, 2008, 7:50am » |
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on May 1st, 2008, 8:26pm, kindy_kid wrote:| Does inscribed mean that the circle has to touch all 6 sides? |
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Yes, thank you; I edited the original post to remove the ambiguity.
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FiBsTeR
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Re: Lost Circles
« Reply #8 on: May 6th, 2008, 5:15pm » |
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 I thought this was interesting...
HINT 2 (Giveaway):
If you dig up your old high school geometry books, you'll probably find a problem that shows that if a circle can be inscribed inside of a quadrilateral ABCD, then AB+CD=BC+DA. The proof for this motivates a similar statement, which says that if a circle can be inscribed inside of a hexagon ABCDEF, then AB+CD+EF=BC+DE+FA.
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towr
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Re: Lost Circles
« Reply #9 on: May 7th, 2008, 1:48am » |
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I don't think I ever knew that; also I don't have my old highschool geometry book (or for that matter any but a few high school book; they belonged to the school).
It works for any 2N-gon.
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FiBsTeR
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Re: Lost Circles
« Reply #10 on: May 7th, 2008, 2:46pm » |
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on May 7th, 2008, 1:48am, towr wrote:
Yep; it comes right from TTT (the Two Tangents Theorem, as I learned it), which states that two tangents to a circle from an exterior point are equal in length.
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FiBsTeR
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Re: Lost Circles
« Reply #11 on: May 9th, 2008, 6:29pm » |
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Solution:
Suppose a lost circle existed in hexagon ABCDEF. Without loss of generality, let AB = 42. Also, let FA, AB, and BC meet the circle at points P, Q, and R, respectively. Since tangents drawn to a circle from an exterior point are equal, AP=AQ and BQ=BR. So 42 = AB = AQ+QB = AP+BR. But AP<FA and BR<BC, so 42=AP+BR<FA+BC. Since FA and BC are sides of ABCDEF, they are no greater than the two next-greatest side lengths, 16 and 23. But 42>16+23, a contradiction, and no lost circle exists.
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