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        riddles >> easy >> Lost Circles
        
(Message started by: FiBsTeR on Apr 26^th, 2008, 4:22pm)

Title: Lost Circles
Post by FiBsTeR on Apr 26^th, 2008, 4:22pm

Define a lost (http://www.4815162342.com/) 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.

http://www.4815162342.com/images/4815162342.jpg
(Source: lingomaniac88, AoPS)

Title: Re: Lost Circles
Post by FiBsTeR on Apr 29^th, 2008, 4:19pm

Are these numbers too creepy for you guys? I can come up with some new ones instead, if you want. ;D

Title: Re: Lost Circles
Post by FiBsTeR on May 1^st, 2008, 2:33pm

HINT:

[hide]You cannot inscribe a circle in a hexagon with side lengths 1,1,2,3,5,9, either.[/hide]

Title: Re: Lost Circles
Post by iono on May 1^st, 2008, 5:43pm

....cuz a hexagon can only have 6 sides?

Title: Re: Lost Circles
Post by FiBsTeR on May 1^st, 2008, 5:50pm

Though I agree with your statement, I don't see how that proves these mysterious circles don't exist. ;)

Title: Re: Lost Circles
Post by kindy_kid on May 1^st, 2008, 8:26pm

Does inscribed mean that the circle has to touch all 6 sides? Can it just touch 3, or 2, or 1 or none? ::)

Title: Re: Lost Circles
Post by towr on May 2^nd, 2008, 12:49am

on 05/01/08 at 20:26:31, 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? ::)

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.

Title: Re: Lost Circles
Post by FiBsTeR on May 3^rd, 2008, 7:50am

on 05/01/08 at 20:26:31, kindy_kid wrote:

Does inscribed mean that the circle has to touch all 6 sides?

Yes, thank you; I edited the original post to remove the ambiguity.

Title: Re: Lost Circles
Post by FiBsTeR on May 6^th, 2008, 5:15pm

:'( I thought this was interesting...

HINT 2 (Giveaway):

[hide]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.[/hide]

Title: Re: Lost Circles
Post by towr on May 7^th, 2008, 1:48am

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.

Title: Re: Lost Circles
Post by FiBsTeR on May 7^th, 2008, 2:46pm

on 05/07/08 at 01:48:14, towr wrote:

It works for any 2N-gon.

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.

Title: Re: Lost Circles
Post by FiBsTeR on May 9^th, 2008, 6:29pm

Solution:

[hide]
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.
[/hide]