If all shaded blocks are squares derived from the largest one with edge length 1, does the area converge?

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    $\begingroup$ Can you write a series based on the combined areas of each size of square? $\endgroup$
    – Henry
    Jan 22, 2018 at 16:40
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    $\begingroup$ @ArnaudMortier if there’s no overlap , the area will be 1+2*1/2 + 4*1/4 ... not converging .but with overlap , I don’t know.. $\endgroup$
    – athos
    Jan 22, 2018 at 16:43
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    $\begingroup$ Do you know if the diameter of the whole picture tends to infinity? If it doesn't that would solve the problem. $\endgroup$ Jan 22, 2018 at 16:47
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    $\begingroup$ See en.wikipedia.org/wiki/Pythagoras_tree_(fractal) there's a section area. $\endgroup$
    – Atmos
    Jan 22, 2018 at 16:48
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    $\begingroup$ @JohnDoe I have to correct myself. Wiki says it's bouned by a $6 \times 4$ rectangle, so it converges. $\endgroup$ Jan 22, 2018 at 16:53

1 Answer 1


The horizontal distance from the midpoint of the largest square to the leftmost point can be computed as $$\frac12+\frac12+\left(\frac{1}{\sqrt2}\right)^2+\left(\frac{1}{\sqrt2}\right)^2+\left(\frac{1}{\sqrt2}\right)^4+\left(\frac{1}{\sqrt2}\right)^4+\cdots\\=1+2\left(\frac12+\frac14+\frac18+\cdots\right)=3$$ So the total width of the system is $6$. The total height is $$1+1+\left(\frac{1}{\sqrt2}\right)^2+\left(\frac{1}{\sqrt2}\right)^2+\left(\frac{1}{\sqrt2}\right)^4+\left(\frac{1}{\sqrt2}\right)^4+\cdots=4$$ So, as the wiki article says, it is bounded in a $6\times 4$ box, and so is finite in area.


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