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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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    See http://www2.stetson.edu/~efriedma/cirinsqu/ – 2017-01-27
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    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. – 2017-01-27
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    @watson there is absolutely no logic in providing this link... i posted the question since i want an explanantion – 2017-01-27
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    @kaynex hahhahah – 2017-01-27
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    Why did you think that the area of two circles should be equal to the area of the square? This may be helpful:https://en.wikipedia.org/wiki/Circle_packing_in_a_square – 2017-01-27
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    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. – 2017-01-27
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    @PaulSinclair thank u vvvvv much for ur thoughtful explanation – 2017-01-27
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    Obviously the longest distance between the center of two circles is on the diagonal of the square, therefore the largest circles are on the diagonal of the square. – 2017-01-27
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    @seyed hello seyed. the diagonal of the square is sqrt2, now how can that help meeee??????????? – 2017-01-27
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    It's interesting: the problem doesn't say that the circles have to be cut out in single pieces. For instance, you could cut four half-circles, each with its center on the midpoint of an edge, to assemble two equal circles, and perhaps do somewhat better than the "obvious" answer (I haven't actually checked!). We all know what the question's author *intended*, but what was said was actually a bit ambiguous. In the limit of clever cutting/and-gluing (for this interpretation), OP's answer is correct. – 2017-01-27
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    @JohnHughes yes i agree, however this question is from a uni so i doubt theyre asking us to cut them into pieces as it would be a v easy q – 2017-01-27
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    @exchangehelpforuni : my link provides shows some interesting generalizations, so it was perfectly fine. Moreover, saying "there is absolutely no logic in providing this link" is quite rude, in my opinion. – 2017-01-27

1 Answers 1

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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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    What tool did you use to draw the figure, please? – 2017-01-27
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    @JohnHughes: Geogebra (https://www.geogebra.org/apps/) – 2017-01-27
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    @JohnHughes: you are very welcome! – 2017-01-27
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    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? – 2017-01-27
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    @Vik78: you are right! I made a update as a hint. – 2017-01-27
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    @Vik78: I think it is not that difficult to finish. – 2017-01-27
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    It's a very nice picture, but this is still not a complete answer. I lack the geometric experience to see what you're hinting at. – 2017-01-27
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    @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. – 2017-01-27
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    @exchangehelpforuni: you are very welcome! – 2017-01-27
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    @arnaldo i think ur good with geomtery, may u plz help me with my other question? – 2017-01-27
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    @exchangehelpforuni: Sure! Post your question and I will try to help you! But don't forget to write what you tried so far! – 2017-01-27
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    Thank you!!! Here is the question, i already posted it http://math.stackexchange.com/questions/2116854/largest-equilateral-triangle-which-fits-inside-a-square?noredirect=1#comment4353433_2116854 – 2017-01-27
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    @exchangehelpforuni: Unfortunately I can't help you with that because it is marked as duplicated. You can check the solution here: http://math.stackexchange.com/questions/59616/find-the-maximum-area-possible-of-equilateral-triangle-that-inside-the-given-squ – 2017-01-27
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    @arnaldo i checked that answer but it does not contain anything valuabel – 2017-01-27
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    @Arnaldo pleaseeeee pleaseee is there any way you can help me with this question??? Can we comment on the other post??? Even though its 5 years old – 2017-01-27