The hollow circle is done in the same way, as the total length of concave arcs equals the total length of convex arcs,
all being the same radius.
It is then easy to get a rectangle in just a few cuts.
Then we have to dissect it into a square, by the usual techniques, and copy the cuts onto the circle.
The challenge is now to get less than 12 pieces !
Please note that, from another dissection, the hollow circle has the same area as the inscribed hexagon.
For dissecting into a square, the Dudeney's crescent is perfect : just overlay the square on a tape built by half crescents. We get then a 4 pieces only dissection. Easy...
The Sam Loyd's crescent is harder.
Considering the dissection of two squares 2×2 + 1×1 into one square gives the equivallent area of 5 units. Overlaying the tape of crescents and a tape of square pairs, with their own dissection (A-M-B) into one square, results then into an 8 pieces dissection.
For this dissection to work, the two tapes must be properly offset.
A minimum number of pieces is then obtained when the midline is evenly divided by squares and crescent edges. That is the crescent goes through point W.
This applet shows this :
The crescents can be shifted by dragging the yellow point, initialy at W.
We can choose the value of arc by dragging the center I, defining arc radius.
The limit is a semi-circle with diameter 5 times the crescent width, "Dudeney like".
Point M then comes exactly on a copy of the crescent (3² + 4² = 5²)
By increasing the radius, M is no more on that circle, an the piece MBP'M' has then a very small non null side MM'.
When P, or even P", comes in B, there is one less piece.
For large values of the radius, the "crescent" doesn't look a crescent anymore...
The crescent precisely defined by Sam Loyd has a radius = √26/2 times the width.
(This is for making the dissection into greek cross to work).
Gavin Theobald has found a dissection of
Sam Loyd's crescent into a square in just 7 pieces.
We might wonder why this dissection has not been found sooner. May be everybody was obscured by the 5 squares...
However this dissection seems afteral natural : "just" overlay a strip of half crescents and a strip of equivallent rectangles.
The dissection of this rectangle into a square is then "as usual" with a P-slide.
Here also, we may shift the strips to get infinitely many variants.
The following applet shows this dissection. Point I defines the radius.
The yellow point shifts the crescents relative to rectangle ABCD
The initial position in the applet corresponds to Gavin Theobald's dissection of the Sam Loyd's crescent. (type R to return to this position)
The dissection in 7 pieces works as long as the cuts of the P-slide don't interfere with the crescents edges.
The same method works when the height is not 5×witdh (point B is draggable).
It might then be worthy ... or not.
In the case of Dudeney's crescent (h = 2R = 4d) the rectangle is even ... directly the equivallent square !
For h ≤ 4d the P-slide just vanishes, it should be done in the other direction.
Don't forget the piece 4, little bit outside the cross, put on the other side in 4'.
That is 7 pieces.
The Dudeney's crescent is dissected into a greek cross by replacing the
overlayed tape of square by an equivallent tape of crosses.
The equivallent square is drawn as dotted lines.
To prove the dissection, we have just to prove that point X is on the circle.
That is AX = AB, which is quite obvious from the greek cross to square dissection.
Generalizing to a different number of petals is much harder.
Let's search for general n petals, here n = 5.
Although there might be different flower shapes (as in the 4 petals example above),
we'll assume that the big circle arc goes through the flower's center,
and the two small arcs are symmetric from there common point, located on the equivallent circle.
Of course the large green circle and the red equivallent circle have the same radius.
Arcs DB = BC' and BE = CB imply that arc DE = CC' = 2π/5 (2π/n in general case).
Also D' is deduced from D through a rotation by 2π/5, and C' is midpoint of D'E.
These last conditions define a unique location for D on circle ODBE, and all other points deduce.
The small arcs have any radius, just not overflowing...
Study this construction a little more with an applet
allowing to vary the arc CC', really BB'. The buttons define n.
The pattern is repeated 6 times (overlaying if n<6).
The n = 2 case (a digon) distorts the drawing of the arcs in the applet, but the construction is simplified in that case : D and D' come in O and the arcs are semi-circles !
Dragging B' to define arc BB' is totally free, but if not 2π/n, n integer, this doesn't result into a correct flower, petals overlap.
However the construction is 'locally' valid (for one petal).
Let A the center of circle ODBE, then <)DAE = 2π/n, and <)DOE = DAE/2 (inscribed angles)
As <)DOD' = 2π/n also, we deduce OD' is the reflection of OD through line OE.
Midpoint C' of D'E is then the reflection of F midpoint of DE, hence the construction :
Construct the apothem a of an inscribed n-gon, a circle centered in A with radius a intersects the equivallent red circle into F, then DE _|_ AF gives DE = side of n-gon (hence DAE = 2π/n) and F midpoint of DE.
Other points are easy deduced from D and E.
Remains to prove the dissection, by prooving that the such constructed arc BC' equals the arc BD.
Let A' the diametrally opposite point to A
Hence A'DFE in line (AFA' = π/2)
Let α = <)FA'O, and of course <)A'FO = α (A'OF isosceles) and <)FOA = 2α (inscribed angle).
Let β = <)DEO, here also <)DAO = 2β
In triangle FOE, <)FOE = π - β - (π - α) = α - β
<)C'OA = <)FOA - 2<)FOE = 2α - 2(α - β) = 2β
Angles <)C'OA and <)DAO are then equal, then arcs OD and AC' also, and finally arcs BD = BO - OD and BC' = BA - AC' are equal. QED.