MHB What is the total area of the infinite number of inscribed squares?

AI Thread Summary
The discussion revolves around calculating the total area of an infinite number of inscribed squares within a circle of radius R. Each square is inscribed within a circle, which in turn has a new square inscribed within it, creating a recursive pattern. The illustrations provided help visualize the geometric progression of the squares and circles. Participants confirm the correctness of the calculations and express appreciation for the visual aids. The conversation highlights the mathematical beauty of this infinite geometric series.
lfdahl
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Given a circle (radius $R$) with an inscribed square. Now inscribe a new circle in the square and then again a new square in the new circle etc. What is the total area of the infinite number of inscribed squares?
 
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lfdahl said:
Given a circle (radius $R$) with an inscribed square. Now inscribe a new circle in the square and then again a new square in the new circle etc. What is the total area of the infinite number of inscribed squares?
my solution:
the area of the 1st square=$2R^2$
the area of the 2nd square=$R^2$
the area of the 3rd square=$\dfrac {R^2}{2}$
the area of the 4th square=$\dfrac {R^2}{4}$
so the total area =$2R^2+R^2+\dfrac {R^2}{2}+\dfrac {R^2}{4}+------=4R^2$
 
Just for illustration purposes. (Smile)
\begin{tikzpicture}[very thick]
\newcommand\Square[1]{+(-#1,-#1) rectangle +(#1,#1)}
\draw[green] foreach \r in {0,...,16} { circle ({5*2^(-\r/2)}) };
\draw[blue!50] foreach \r in {1,...,16} { \Square{{5*2^(-\r/2)}} };
\fill circle (0.08);
\end{tikzpicture}
 
I like Serena said:
Just for illustration purposes. (Smile)
\begin{tikzpicture}[very thick]
\newcommand\Square[1]{+(-#1,-#1) rectangle +(#1,#1)}
\draw[green] foreach \r in {0,...,16} { circle ({5*2^(-\r/2)}) };
\draw[blue!50] foreach \r in {1,...,16} { \Square{{5*2^(-\r/2)}} };
\fill circle (0.08);
\end{tikzpicture}

Great illustration! Thankyou for your contribution, I like Serena!

- - - Updated - - -

Albert said:
my solution:
the area of the 1st square=$2R^2$
the area of the 2nd square=$R^2$
the area of the 3rd square=$\dfrac {R^2}{2}$
the area of the 4th square=$\dfrac {R^2}{4}$
so the total area =$2R^2+R^2+\dfrac {R^2}{2}+\dfrac {R^2}{4}+------=4R^2$

Thanks, Albert! Your result is - of course - right.
 
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