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A man has a square piece of paper where each side has length $1$ m. Two equal circles are to be cut from this paper. What is the radius, in meters, of the largest possible circles?

This is what I did:

  • area of square: $1$

  • area area of circle: $2\pi(r^2)$

I multiplied by $2$ since they are $2$ circles. Now I made $2\pi r^2=1$ and solved for $"r"$, however the answer I got is completely off. May you please tell me what I did is wrong and how I can fix that?

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  • $\begingroup$ See www2.stetson.edu/~efriedma/cirinsqu $\endgroup$
    – Watson
    Jan 27 '17 at 17:57
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    $\begingroup$ How on earth are you going to cut those two circles such that they have the same area as the square? There's got to be wasted paper. $\endgroup$
    – Kaynex
    Jan 27 '17 at 18:04
  • $\begingroup$ @watson there is absolutely no logic in providing this link... i posted the question since i want an explanantion $\endgroup$ Jan 27 '17 at 18:04
  • $\begingroup$ Why did you think that the area of two circles should be equal to the area of the square? This may be helpful:en.wikipedia.org/wiki/Circle_packing_in_a_square $\endgroup$
    – Seyed
    Jan 27 '17 at 18:10
  • $\begingroup$ The logic behind Watson's posting of that link is blindingly obvious. Your circles have to fit inside the square without overlapping, as the 2nd picture on that page shows. What you did that was wrong was assume that the shape of the circles didn't matter and somehow it was possible to include all of the area of the square inside them. $\endgroup$ Jan 27 '17 at 18:16
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enter image description here

That is the picture that fits the problem.

See that

$$CE=\sqrt{2}=CA_1+A_1A+AE=\sqrt{2}r+2r+\sqrt{2}r \to r=\frac{\sqrt{2}}{2+2\sqrt{2}}=\frac{2-\sqrt{2}}{2}$$

EDIT

Hint

To prove that it is the maximum work with the picture below:

enter image description here

Work with variation of $\alpha$, the trapezium $EFGK$ and $DG+GK+KC=DC=1$.

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    $\begingroup$ What tool did you use to draw the figure, please? $\endgroup$ Jan 27 '17 at 18:40
  • $\begingroup$ @JohnHughes: Geogebra (geogebra.org/apps) $\endgroup$
    – Arnaldo
    Jan 27 '17 at 18:41
  • $\begingroup$ @JohnHughes: you are very welcome! $\endgroup$
    – Arnaldo
    Jan 27 '17 at 18:44
  • $\begingroup$ Intuitively this is what I thought the answer would be, but it's not exactly a proof. Can you prove that this is the optimal choice of circles? $\endgroup$
    – Vik78
    Jan 27 '17 at 19:18
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    $\begingroup$ @arnaldo thank YOU VERY much!!!! very very helpful!!! Btw u did not need that second picture titled "hint" i think some users r just challenging you. $\endgroup$ Jan 27 '17 at 20:31

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