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## 'Blue and White' printed from http://nrich.maths.org/

We received correct
solutions (that the areas were all the same) from:

- Charles Blackham (Shrewsbury House School)
- Michael Brooker (home educated),
- Andrei Lazanu (School number 205, Bucharest),
- Chong Ching Tong, Chen Wei Jian and Teo Seow Tian (River Valley
High School, Singapore) and
- Chris Wells (Forres Academy).

Well done all of you.

I particularly
liked Michael's solution because he generalised the result and
I have used this as the basis of what follows. Although it was not
intended to be a trick question Michael!

With one circle in the square the diameter is the length of one
side of the square. The shaded area is ${\pi}{r^2}$ where $r$ =
radius of the largest circle.

With four circles in the square, the diameter of one circle is half
that of the large circle. The area of each small circle is
${\pi}({r/2}\: x \:{r/2})$. The total shaded area is
$4({\pi}{r^2}/4)$. This can be simplified to ${\pi}{r^2}$. \par
With nine circles in the square, the diameter of one circle is a
third that of the large circle. The area of each small circle is
$\pi(r/3\: x\: r/3)$. The total shaded area is $9({\pi}{r^2}/9)$.
This can be simplified to ${\pi}{r^2}$.

We can go one step further by saying that with $n$ circles the area
is $n({\pi}{r^2}/n)$ - which can again be simplified to
${\pi}{r^2}$. Therefore the answer is that the shaded area is the
same in each picture.