It is definitely a curious fact that neither of the squares have been mentioned as such, although all the facts that led to the identification of the rectangles as squares have been known. See, for example,

https://www.retas.de/thomas/arbelos/arbelos.html

properties 7 and 8.

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.

https://www.cut-the-knot.org/Curriculum/Geometry/InversionInArbelos.shtml

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?

https://www.cut-the-knot.org/Curriculum/Geometry/SquaresInArbelos.shtml

In fact not very difficult. Really strange I never came across that before.

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.

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.

https://www.math.tamu.edu/~harold.boas/preprints/arbelos.pdf

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.

I will, and I shall probably borrow some of that for my pages, OK?

>I have tried to play around with analogous setups,

What are these?

>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.

This is probably what you have in mind

https://www.cut-the-knot.org/Curriculum/Geometry/CirclesAndRegularTriangle.shtml

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

mpdlc

Since there was problem last time uploading the sketch file to the website you can alternatively download from the address below

https://rs165.rapidshare.com/files/64139612/Shoemakerknife.rar

(Imagine starting with d' and following all the steps you applied to d. You'll obtain points, say, w (near your z') and w' (near your z). Why is it possible to claim that "near" is actually "is"?

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

I am probably missing something. This is exactly my problem. Why is that diameter vertical? It is no doubt bound to be vertical, but why?

If it is, the symmetry of course applies and works beautifully.

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.

I hope this time works

mpdlc

Since it did work, you may start calling yourself fully adept at uploading files at this forum. Just remember to never use the html extension. That's all. Thank you.