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I have noticed Can I Haz My Eye Center'd? puzzle, which I know in slightly different formulation:

You have a disk on a table and a point A on it. You need to cut the disk into smallest possible amount of pieces and then glue them back into a disk, which has centre at A. But: 1. you are not allowed to take pieces out of the table plane (you can only slide them). 2. you need to use as little glue as possible (e.i. the cut length should be minimal).

Henning Makholm has already shown a way in his answer here, which allows to cut into the smallest amount of pieces, but the cut length still can be optimised and once you do it you will see this puzzle differently.

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  • $\begingroup$ Why not answering the first puzzle with your optimal solution instead of making a new one ? $\endgroup$ – Fabich Jul 27 '16 at 9:51
  • $\begingroup$ @Lordofdark, because people put puzzles on this site to let others enjoy process of thinking and discovery, not to answer them on their own... $\endgroup$ – klm123 Jul 27 '16 at 10:12
  • $\begingroup$ Just to be sure, the differences with the previous puzzle are : It is a point instead of a cat and you can only "slide" the pieces ? $\endgroup$ – Fabich Jul 27 '16 at 10:41
  • $\begingroup$ @Lordofdark, the differences are 1. you can only slide them. 2. you need to use as little glue as possible. Points or cats - its all the same $\endgroup$ – klm123 Jul 27 '16 at 10:47
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    $\begingroup$ @Ankoganit, rotation within table plane only is allowed. $\endgroup$ – klm123 Jul 27 '16 at 12:15
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Notation

Let $O$ be the centre of the disc, let $r$ be the radius of the disc and $d$ be the distance between point $O$ and point $A$.

Result

We can reduce the cut length of this problem whenever $\frac{d}{2r} < \alpha$ where $\alpha$ is the positive real solution to the equation $\cos (\pi x) = x$ (Cut 2) and whenever $\frac{d}{2r} > \beta$ where $\beta$ is the smallest positive real solution to the equation $\cos(2-2x) = x$ (Cut 3).

Proof
Cut 1 (Henning Makholm's solution)

We imagine a circle of radius $r$ and centre $A$ and cut along where this circle intersects our disc. Then slide the piece not containing $A$ to the shaded region

enter image description here

In this case, the length of the cut $C_1$ is equal to the length of the arc which subtends an angle of $2\theta$ at $O$

That is, $$ C_1 = 2r\theta . $$
Completing the isoceles triangle in the diagram by joining $A$ to the point of intersection of the circles, we find that

$$ \frac{r}{\sin \theta} = \frac{d}{\sin(\pi - 2\theta)} = \frac{d}{2\sin \theta \cos \theta}$$
In other words,
$$\theta = \arccos \left(\frac{d}{2r} \right),$$
and so
$$C_1 = 2r \arccos \left(\frac{d}{2r} \right).$$

Cut 2 (The circle cut)

This time we cut along the circle which has diameter $OA$, then rotate this circle through an angle of $\pi$ so that $A$ now coincides with the centre of the disc. enter image description here

Clearly, in this case, the length of the cut is

$$C_2 = \pi d$$.

NB We may want to be careful here and say that the points $O$ and $A$ should lie inside the circle. In that case, the length of our cut is $$C_2 = \pi d + \epsilon$$ where $\epsilon$ is an arbitrarily small positive quantity. This does not affect the result relating to the relationship between Cut 2 and Cut 1.

Then we have

$$C_2 < C_1$$ $$\Leftrightarrow \pi d < 2r \arccos \left(\frac{d}{2r} \right)$$ $$\Leftrightarrow \cos \left(\frac{\pi d}{2r} \right) > \frac{d}{2r}$$

where the last line follows because $\cos$ is decreasing on the range $\left(0, \frac{\pi}{2} \right)$.
This last inequality holds whenever $\frac{d}{2r}$ is less than the positive real solution to the equation $\cos \pi x = x$ and so the cut $C_2$ is shorter than $C_1$ whenever

$$\frac{d}{2r} < \alpha \approx 0.376967$$

Cut 3 (Florian F's cut)

In this case, we cut two arbitrarily close line segments parallel and equidistant to the line $OA$ such that the line segment extend from the circle, at points $P$ and $Q$ to a length of approximately $2r-d$, within the circle, as shown in the diagram. In fact, if the angle $POQ$ is $\theta$ then this length is $2r - d - 2r\sin \left(\frac{\theta}{2} \right) \sin\left(\frac{\theta}{4} \right)$. We then round off the cut with a reflection of the arc of the circle $PQ$ extending between the endpoints of the two line segments. We can slide out this piece containing $O$ and $A$ from this disc, rotate it through an angle of $\pi$ and slide it back in so that $A$ becomes the new centre of the disc.
enter image description here

The length of this cut is approximately given by $$C_3 = 4r-2d.$$ In fact, if $\theta$ is as before, then the formula for the length of the cut is $$C_3 = 4r - 2d + 2r\left(\frac{\theta}{2} - 2\sin \left(\frac{\theta}{2} \right) \sin\left(\frac{\theta}{4} \right) \right)$$ which is greater than $4r - 2d$ but we can make it arbitrarily close as $\theta \rightarrow 0$.

Then we have

$$C_3 < C_1$$ $$\Leftrightarrow 4r-2d < 2r \arccos \left(\frac{d}{2r} \right)$$ which holds whenever $\frac{d}{2r}$ is greater than the smallest positive solution to the equation $\cos(2-2x) = x$ and so cut $C_3$ is shorter than cut $C_1$ whenever $$\frac{d}{2r} > \beta \approx 0.445428$$

NB One thing that Florian F seemed to hint at in the comments was whether we could consider this cut to have length $2r-d$ in the limit as the lines get closer together. This is an interesting point but my suspicion would be that all cut pieces should be two-dimensional so this case is not allowed.

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  • $\begingroup$ This can be improved. $\endgroup$ – klm123 Aug 1 '16 at 17:05
  • $\begingroup$ There is the option to cut out a thin piece with 2 close parallel cuts. The piece starts from the border, enclose O and A and have length 2r-d. Both ends should have a curvature matching the disk's. I believe it is always shorter than Henning Makholm's cut. $\endgroup$ – Florian F Aug 1 '16 at 23:34
  • $\begingroup$ If the parallel cuts are close but not the same, shouldn't the cut be of length $4r - 2d$ plus some small $\epsilon$? $\endgroup$ – hexomino Aug 2 '16 at 9:13
  • $\begingroup$ I like the suggestion, I'll add in another bit covering this case. $\endgroup$ – hexomino Aug 2 '16 at 9:13
  • $\begingroup$ I have a solution using three pieces, but even less glue: Cut out a thin rectangle strip with O and A at the ends. This is possible with a $2d+\epsilon$ cut. However this piece can't be rotated inside the circle. To fix that, cut out another piece behind A to allow the rectangle to slide out. In total, this should take only $2r+\epsilon$ worth of glue. $\endgroup$ – Anon Aug 2 '16 at 13:47

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