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I am reading the answer here by Hagen von Eitzen. Reading only the first paragraph.

His answer

Let $a\le 1$ be supremum of all finite disc packing areas in the unit square. Consider the square minus an inscribed disc. Because its boundary is a zero-measure set (or whatever argument also worked for the first part of the problem), it can be exhausted arbitrarily well by finitely many dyadic squares, that is for $\epsilon>0$ we find a finite set of squares such that the squares fill a proportion $1-\epsilon$ of this shape. For each small square find a finite disc filling that fills a proportion of $a-\epsilon$ of their respective area. Then the total area filled by all discs is $$\tag1\frac\pi4+\left(1-\frac\pi 4\right)(1-\epsilon)(a-\epsilon). $$ As $\epsilon\to 0$, the expression in $(1)$ goes $\to a+\frac\pi4(1-a)$ which must be $\le a$. Hence $a=1$.

My understanding:

Consider unit square minus inscribed disc. Call it $W$.

By similiar argument that worked for the first part of the problem:

$\color{blue}{\forall\ \epsilon >0, \exists\ \text{partition}\ P\ \text{such that:}}$

$($area of all square tiles in $W)>\left( 1-\dfrac{\pi}{4} \right)- \epsilon$

From here how shall I reach?

$\implies\ ($area of all discs in $W)=\left( 1-\dfrac{\pi}{4} \right) (1-\epsilon) (a-\epsilon)$

$\implies\ ($area of all discs in square$)\ =\dfrac{\pi}{4} + \left( 1-\dfrac{\pi}{4} \right) (1-\epsilon) (a-\epsilon)<a$

Applying limit $1 \leq a$

Also $a \leq 1$

Therefore $a=1$

EDIT

Comment by Hagen:

I use a scaled down (almost) optimal packing only for those tiny squares that are not coverd by the big disc enscribed ot the big square.

lorilori
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    It would help everyone if you stated, or summarised, the specific question in the title, rather than state that you are confused about it. – FShrike Dec 22 '22 at 15:02
  • By the way, it would make sense to comment under Hagen's post yourself to ask for clarification – FShrike Dec 22 '22 at 19:58
  • Whats up with the second link in your question? – Slugger Dec 22 '22 at 21:58
  • Nothing... I did that to color an important line... I tried to underline but couldn't – lorilori Dec 22 '22 at 22:16
  • While your problem is genuine, it's best to expand on the "I'm reading the answer here" part. If possible, isn't of simply quoting the whole para here, try if you can provide the crisp form instead (while retaining the link too). This'll save your question from getting closed. – VoidGawd Dec 24 '22 at 13:42
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    I picked the inscribed disk $\pi/4$, filled the remaining shape ($1-\frac\pi4$) almost (to a proportion of $1-\epsilon$) with squares, and each of these finitely many squares with finitely many disks to a proportion ($a-\epsilon$) close to the supremum $a$. - I cannot take the actual supremum $a$ because that cannot be reached with finitely many disks. Likewise, I cannot take the whole complement shape as that would require infinitely many squares. It is perhaps only a matter of taste whether one subtracts $\epsilon$ or multiplies with $1-\epsilon$ … – Hagen von Eitzen Dec 24 '22 at 14:51

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By definition of $a$, we know that for any square with area $S$ and for any $\epsilon > 0$, we could find a finite set of disjoint circles inside the square with area $> aS - \epsilon$.

Now use that result for each square tile in $W$. The total area of those tiles is $> \left(1-\frac{\pi}{4}\right) - \epsilon$, so by putting a finite collection of disjoint circles in each of those tiles, we can end up with total area of circles $> a \left[ \left(1-\frac{\pi}{4}\right) - \epsilon \right] - \epsilon$, which simplifies to $a \left( 1 - \frac \pi 4 \right) - a \epsilon - \epsilon$.

It looks like I applied the initial result in a slightly different way from Hagen, but the idea is the same and my result still works to let me finish the problem from there. In particular, we get $$\lim_{\epsilon \to 0} a \left( 1 - \frac \pi 4 \right) - a \epsilon - \epsilon = a \left( 1 - \frac \pi 4 \right).$$

David Clyde
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