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alexb
Charter Member
2120 posts 
Oct3007, 07:22 PM (EST) 

27. "RE: Geometry challenge"
In response to message #1

Peter Woo in the introduction to https://forumgeom.fau.edu/FG2001volume1/FG200119.pdf claims that Archimedes proved that IJHG is a a square. Although I could not locate the place Archimedes did that, the fact shows that P. Woo, at least was aware that a square is present. In the same paper he argues that DB passes through the midpoint of the missing big semicircle making it a bisector of angle at B. So that the existence of the other square is a consequence of that fact. 

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alexb
Charter Member
2120 posts 
Oct1607, 00:10 AM (EST) 

2. "RE: Geometry challenge"
In response to message #0

It's not a simple problem. Not at all. The following is just to document the progress. There is an awful amount of cyclic quadrilaterals. So using the intersecting chords theorem we can get (I refer to the applet illustration of Archimedes Lemma 6) CP·CB = CM·CK = CF·CH = CE·CG = CL·CI = CN·CA. Juxtaposing the first and the last product gives one of the required proportions: AC/BC = CP/CN. Several other proportions, but not all, are easy. 

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mpdlc
Member since Mar1207

Oct1707, 12:49 PM (EST) 

3. "RE: Geometry challenge"
In response to message #0

To proof “ghif” is a square it is not difficult. 1) Let us draw successive family of circumferences tangent to exterior circumference C1, the inner one C2 and the last one drawn as shown in the sketch. 2) Now we draw a circumference on inversion with center in “a” and passing thru the center of our last circumference drawn. 3) It is obvious that our last circumference remain unchanged in the inversion the outer circumference C1 and the inner one C2 will be transformed into line L1 and L2 and the other circumferences will be transformed into circumferences each one tangent to L1, L2 and the one above obviously all of them have same radius of the unchanged one. What means that the ordinate for circumference n will be 2nr , therefore for the first one the ordinate will be 2R equal to its diameter what force the rectangle “ghif” being a square I am working in the second square “ bedf ” but that one is not so immediate.
mpdlc 
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alexb
Charter Member
2120 posts 
Oct1807, 06:31 AM (EST) 

6. "RE: Geometry challenge"
In response to message #5

It'sure is. Is it your invention? Because of the applicability of inversion, at least, the squareness ijhg must have been known. Although I have no recollection of seeing it mentioned. I'd be curious to see how you proved it without inversion. Now, do you know that the perpendicular from the intersection of ah and cg onto ac falls on b? 

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Archi
guest

Oct2107, 12:11 PM (EST) 

8. "RE: Geometry challenge"
In response to message #7

Dear fellow geometers, Archimedes' Arbelos work, as you know, speaks for itself and does not need neither compliments nor complements. He seems to be sophisticated enough to work alone, without partners. There is a simple explanation why the fact of that square in Arbelos was not mentioned in serious geometry literature since his time. It looks rather silly to project a point from a right angle bisector onto the sides and anticipate something that differs significantly from a square, does it not? Big A. must have known it'since his middle school days, years before he started writing his fundamental Lemmas book. This was, in my opinion, the major reason why he did not ran naked again screaming on top of his ability: 'It is a square ! ' Cheers. Archibauld Epesshtein. 

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alexb
Charter Member
2120 posts 
Oct2107, 01:08 PM (EST) 

11. "RE: Geometry challenge"
In response to message #8

> Dear fellow geometers I am rather flattered. But please do not exaggerate. > He seems to be sophisticated enough > to work alone, without partners. He never published in a math magazine either, probably feeling a disdain for the trouble invloved. > There is a simple explanation why > the fact of that square in > Arbelos was not mentioned in > serious geometry literature since his time. That's a fine explanation really. However, it needs a proof and, candidly, some of the lemmas he did include among the survivng 15 are more trivial than the fact of a certain line being an angle bisector. But of course there could be a thousand other reasons for the omission of the squares in the literature. > This was, in my opinion, the major reason > why he did not ran naked again screaming > on top of his ability: 'It is a square ! ' Indeed, apparently such an event has not been documented; this does not say it never took place. 

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Greg Markowsky
guest

Oct2207, 01:57 PM (EST) 

16. "RE: Geometry challenge"
In response to message #8

How do you know that db is the angle bisector? It is, of course, but it'seems no easier to prove that then to prove that the shape is a square. 

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Greg Markowsky
guest

Oct2107, 12:11 PM (EST) 

10. "RE: Geometry challenge"
In response to message #7

I like the applet you made for this a lot. If you check the original link I posted https://www.umcs.maine.edu/~markov/GregMarkowskyShoemakerKnife.pdf you'll see that I added my solution, so you can see how I did it without inversion if you want. I hadn't noticed that the perpendicular from the intersection of ah and cg onto ac hits b, as you mention. That's another really pretty thing about this setup. It's amazing all the things that are true with this thing, including Archimedes' twins, etc. I have tried to play around with analogous setups, but nothing has ever worked out very well for me. One interesting fact that I learned from the Japanese Temple Geometry book is that if you replace one of the smaller semicircles with an isosceles triangle that touches the largest semicircle, then the center of the inscribed circle lies directly above b, which is pretty cool. The proof of that is quite involved, though. 

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Greg Markowsky
guest

Oct2207, 01:57 PM (EST) 

17. "RE: Geometry challenge"
In response to message #12

No prob if you want to put some of this stuff on your pages. That link you give is exactly the problem I had in mind, a beautiful problem. I don't remember everything I tried, this was > 1 year ago, but I know I tried with two isoceles triangles and couldn't even figure out the radius of the inscribed circle. I should try again. 

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mr_homm
Member since May2205

Oct2307, 08:51 AM (EST) 

21. "RE: Geometry challenge"
In response to message #7

Hi Alex, I have been following this discussion quietly, hoping to come up with a "neat" proof to present, so you can imagine my surprise at seeing myself credited with a piece of the solution. Are you sure it was me? I've been looking back through my older contributions to CTK, but none of them seems to have anything to do with this arbelos problem. Perhaps I am just missing the connection. Could you please give me a link back to the relevant page? I would love to take credit for assisting with this problem, but only if such credit is actually due! Thanks, Stuart Anderson 

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mpdlc
Member since Mar1207

Oct2107, 03:59 PM (EST) 

14. "RE: Geometry challenge"
In response to message #13

Referring to the sketch Let us look it from a different perspective. I think all we do agree that there is only a vertical diameter of C1 passing on z and lets call for now z1 the opposite end of such unique diameter. Since by construction we select z to draw the perpendicular to segment bd, and since angle bdz values 90 degree the extended segment of bd must cut C1 in z1 or z’ as I named in the drawing. The rest of problem concerning to the relations subject is also in the in the following link https://rapidshare.com/files/64190992/Shoemakerknife60.doc Sorry for the inconvenience but I am a fully inept to upload a readable attachment file in your site. By the way you did a heck of a work what evidences your deep knowledge of math and the rigorous of your proofs. For me just an amateur it take me a while to follow your reasoning, so I must resort in more intuitive methods but at the end I take fun putting my brain to work. I agree with you that Greg has presented us with a hard nut mpdlc
mpdlc 

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mpdlc
Member since Mar1207

Oct2207, 02:46 PM (EST) 

20. "RE: Geometry challenge"
In response to message #15

Below is another quick alternative to proof that rectangle bedf is a square using the well known property of preserving angles of the inversion. As it can be easily follow from the construction drawing of the inversion, point d it transforms in point g and point b become point j. And the rays origin in center of inversion remain unaltered. Therefore angle with vertex d (adb) diagonal of bedf transforms into angle with vertex in j (ajg) . which values 45 degrees since is the same angle of the diagonal in the square ighj. So angle adb values also 45 degrees therefore bedf is a square too.
mpdlc 
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